Calculation questions

Mark scheme

Examiner's comment

Candidates answered this question very well. A few mistyped the numbers into their calculator after giving the correct calculation and even fewer made rounding errors. This type of question has featured on previous examinations, and it was encouraging to see how well it was answered.

Notes from the mark scheme and report

• answer to 2 decimal places (24.34)
• 1 mark ECF if percentages used with wrong isotopes once or one percentage transposed

Skills: M0.2 Use ratios, fractions and percentages; M1.2 Find arithmetic means; M3.1 Translate information between graphical, numerical and algebraic forms

Mark scheme

Examiner's comment

This question discriminated well. It requires the use of the graph to determine that 60 cm3 was the volume of gas evolved and hence calculate moles of H2. Many low scoring candidates did not read the volume of gas from the graph, using 24.5 cm3 instead. ‘Error carried forward’ marks could be given to candidates who subsequently multiplied an incorrectly calculated value by two to reflect stoichiometry and hence calculate moles of HCl reacted. The final mark needed use of moles = concentration x volume to calculate HCl concentration (as well converting to standard form and giving to 2 significant figures). Candidates who scored low on the whole paper tended to struggle to make an attempt on this question.

Notes from the mark scheme and report

• 2 sig figs AND standard form required
• final volume 60 cm3 read from graph
• common errors: using 24.5 cm3 as H2 volume, 60/24 (units), 0.204 not in standard form (2 marks each)
• 2:1 HCl:H2 ratio

Skills: M0.1 Recognise and use expressions in decimal and ordinary form; M0.2 Use ratios, fractions and percentages; M1.1 Use an appropriate number of significant figures; M3.1 Translate information between graphical, numerical and algebraic forms; PAG9 Rates of reaction – continuous monitoring

Mark scheme

Examiner's comment

This bond energy calculation proved difficult for some. Roughly a third of candidates obtained the correct enthalpy change of 463 kJ mol–1 for all 3 marks, but 0 marks was the next common score. Despite evidence that candidates had clearly practised this type of problem and they recognised that there were going to be three stages to the calculation, candidates frequently did not use the stoichiometry shown in the equation and therefore divided by 2 and 3 respectively for marking points 1 and 3 (rather than 4 and 6). Another common error was incorrect treatment of the enthalpy of reaction value (subtracting rather than adding 196 for the second marking point or omitting it altogether). Drawing out the total number of molecules helped students understand the numbers of bonds involved and should be encouraged. It is also worth reiterating the sign for bond enthalpies. As with all multi-stage calculations, error carried forward was applied, allowing credit to be given for a subsequent correct method.

Notes from the mark scheme and report

• 4 S=O broken in 2SO2 and 6 S=O made in 2SO3
• common errors: +578 (2 and 3 S=O bonds), +398 (wrong sign for 196), –463
• IGNORE sign of final answer

Skills: M2.2 Change the subject of an equation; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M2.4 Solve algebraic equations

Mark scheme

Examiner's comment

Following browsing of candidate responses, marks were given for correctly writing the Kc expression, substituting the terms with values, and finally, giving the calculated response to three significant figures and standard form. Most candidates scored either 2 or 3 marks. However, this question highlights the need for candidates to read the question carefully, with many responses not giving the Kc expression, or not giving the answer in standard form, or both.

Notes from the mark scheme and report

• square brackets required
• 3 sig figs AND standard form (DO NOT ALLOW 0.0261 × 10^0)
• ECF from inverted Kc only (3.83 × 10^1, 2 marks)
• IGNORE units

Skills: M0.1 Recognise and use expressions in decimal and ordinary form; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

Many candidates followed a well drilled method to apply the ideal gas equation to this calculation (including writing each term down and converting where necessary): • Moles of alcohol F used • Multiplication by 4 to give moles of CO2 • Rearranging the ideal gas equation • Substituting the correct values using the correct conversion to Kelvin for temperature • Converting from m3 to cm3 and giving the response to an appropriate number of significant figures (3 in this case) Most candidates scored at least 3 marks on this question. Common errors were using the number of moles of alcohol instead of the number of moles of carbon dioxide in the rearranged ideal gas equation, converting from m3 to cm3 incorrectly in the final step (usually only multiplying by 1,000 rather than 1,000,000) and not giving the response to an appropriate number of significant figures. Occasionally, candidates did not convert to Kelvin or used the pressure in kPa instead of Pa. This response shows an example of clear working. Values are assigned to each term and moles calculated correctly. Rearrangement and substitution of values is shown and correct. The appropriate number of significant figures is given, but the response only converted to dm3 not cm3. This response was given 4 marks.

Notes from the mark scheme and report

• must use pV = nRT
• common errors: no ×4 ratio (7240), no conversion to K (7760), not 3 sig figs (28943.8), using 25.9 as n
• DO NOT ALLOW 150 kPa unconverted in the final calculation
• Candidates should understand that the lowest level of significant figures in the raw data provided for a calculation will determine the number of significant figures that should be given in the final answer. More details can be found in Section M1.1 of the Mathematical Skills Handbook.

Skills: M0.0 Recognise and make use of appropriate units in calculations; M1.1 Use an appropriate number of significant figures; M2.2 Change the subject of an equation; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

Many well-structured calculations were seen, even if all the marks weren’t achieved. Almost all candi- dates calculated the number of moles involved. For those candidates’ hat didn’t get full marks, it was often caused by using an incorrect mass in Q = mcΔT. Some tried to calculate the mass of NaOH and HCl (from moles and Mr values) while others didn’t add the total volumes together i.e. used 40 g rather than 80 g. Some used moles × 2 i.e. 0.112 rather than per mole of water formed. One of the most common errors was the omission of the negative sign for the final enthalpy change. Some put a positive charge instead. Some lost a mark for the incorrect significant figures.

Notes from the mark scheme and report

• common errors (3 marks): –28.1 from using 40 g instead of 80 g or 2 × 0.056 mol, –1690 from adding 273 to the temperature change, +56.1 (no – sign)
• 3 SF AND – sign required
• ALLOW ECF throughout

Skills: M0.0 Recognise and make use of appropriate units in calculations; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; PAG3 Enthalpy determination

Mark scheme

Examiner's comment

Most were able to correctly calculate the number of moles with the molar mass given in the question. Many were then able to use the molar gas volume at RTP, given on the data sheet, to find the volume of gas. The most common errors seen here were an incorrect number of significant figures, not converting to cm3 but leaving as dm3, and some multiplied by 2. A few candidates attempted to use the ideal gas equation here and if this was done correctly marks were credited.

Notes from the mark scheme and report

• answer must be 3 SF and in cm3 (184)
• ALLOW ECF from incorrect n(BaCO3)
• ALLOW ideal gas equation with appropriate T and p

Skills: M0.0 Recognise and make use of appropriate units in calculations; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; PAG1 Moles determination

Mark scheme

Examiner's comment

Most scored here. A minority used all three titres for the mean and even fewer used 23.30 in the calcula- tion with either of the other two values.

Notes from the mark scheme and report

• mean of concordant titres only (23.05 and 23.15) = 23.10
• ECF titres must be within 0.1 cm3

Skills: M1.2 Find arithmetic means; PAG2 Acid–base titration

Mark scheme

Examiner's comment

Many responses were well constructed and showed most of the stages. This made it much easier to award error carried forward (ECF) marks if a correct response was not obtained. Around a quarter of candidates were able to score all 5 marks. Almost all candidates scored the first 2 marks. However, it was sometimes hard to follow the steps a candidate had taken to obtaining a final response, due to: • Steps not being labelled • Steps merging into one another (all equalling each other despite different processes) • Competing alternative methods described without crossing through incorrect working. Many candidates did not gain marks for unit conversions i.e. not converting volumes from cm3 to dm3 and incorrectly converting g to mg. Some did not gain marks for missing the × 10 step. Lower-attaining candidates often didn’t use the titre value obtained in (ii) but used 25 cm3 instead, muddling which vol- ume value should be attributed to which reactant. Maths Skills for AS Chemistry It is important for candidates to have the opportunity to develop their maths skills as part of the course. Particular attention should be paid as to how to: • Setting out the working for calculations • Keeping the full value of any step in their calculator to use in the next step • Common unit conversions, e.g. dm3 to cm3 • Giving the final answer to an appropriate number of significant figures or that asked for in the question • Checking their working for any errors Some support can be found in our Maths Skills Handbook: https://www.ocr.org.uk/Images/295468-chemistry-mathematical-skills-handbook.pdf This article from the RSC contains some useful resources: https://edu.rsc.org/resources/maths-skills-for-chemistry/4013899.article

Notes from the mark scheme and report

• common errors (4 marks): 0.413/4.13/41.3 incorrect conversion to mg, 1240 not divided by 3 tablets
• 1:2 acid:NaOH
• ×10 for 250 cm3
• 3 SF AND mg

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.2 Use ratios, fractions and percentages; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; PAG2 Acid–base titration

Mark scheme

Examiner's comment

Many found this unfamiliar calculation quite tricky. Some tried to use the equation q = mcT to try to find a mass. Others tried a Hess cycle. Some multiplied by the Mr for pentane rather than carbon dioxide. A common error was to use 1000 J in the calculation, despite the enthalpy change being given in kJ mol-1, giving a final answer of 62.7g.

Notes from the mark scheme and report

• 5 × 44.0 = 220 g CO2 per 3509 kJ so 0.0627 g
• ALLOW ECF from mole ratio in (b)(i)
• IGNORE significant figures

Skills: M0.2 Use ratios, fractions and percentages; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

Many candidates were given both marks with the correct answer of -25 kJ mol-1. The best responses showed clear working using a Hess’s law cycle. Common errors of +25 and not using the correct multiples were seen from a few candidates.

Notes from the mark scheme and report

• common errors (1 mark): +763 (no 3 × –394), +25 (sign), –247 (no 3 × –111), –2339 (added reactants)
• answer –25

Skills: M0.2 Use ratios, fractions and percentages; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

Many candidates were given both marks with the correct answer of 27 J K-1 mol-1. Common errors saw some candidates quoting +12 from reversing the subtraction and 54 through not dividing by 2.

Notes from the mark scheme and report

• 15 = 3(214) + 2x – 87 – 3(198) so S = 27
• common errors (1 mark): 241 (no 3 × 214), 12 (subtraction reversed), –171 (no 3 × 198)
• divide by 2 for 2Fe

Skills: M2.2 Change the subject of an equation; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M2.4 Solve algebraic equations

Mark scheme

Examiner's comment

Many candidates scored all three marks. They correctly calculated ΔG, made a feasibility statement and calculated the minimum temperature or 904K or 905K. The alternative methods found in the MS guidance were rarely seen. Nearly all candidates remembered to convert the units which is typically a common error. Less successful candidates missed the unit conversion (i.e. J K-1 mol-1 to kJ K-1 mol-1 for entropy or kJ mol-1 to J mol-1 for enthalpy) or incorrect temperature used (e.g. 298 K). Sometimes there was a lack of a clear feasibility statement.

Notes from the mark scheme and report

• ∆G = 491 – 373 × 0.543 = +288 kJ mol–1 (convert ∆S to kJ and °C to K)
• not feasible as ∆G > 0
• T = ∆H/∆S = 904 or 905 K, whole number required
• ALLOW ∆H > T∆S or ∆S(total) approach

Skills: M0.0 Recognise and make use of appropriate units in calculations; M2.2 Change the subject of an equation; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

This question required the calculation of two partial pressures. This question proved to be a challenge for most students. Common errors were to divide the 5476 by 2 instead of using the square root. Many were given a mark for recognising that the partial pressures needed to be the same. A common error was 602 from the incorrect rearrangement of the Kp expression and some candidates used the value for R (8.31) rather than 8.13 in the Kp expression. Less successful candidates attempted, unsuccessfully, to use an ICE table to work out the moles of everything at equilibrium.

Notes from the mark scheme and report

• p(H2O) × p(CO) = 211 × 211 ÷ 8.13 = 5476
• p(H2O) = p(CO) so take square root = 74 kPa for both

Skills: M2.2 Change the subject of an equation; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M2.4 Solve algebraic equations

Mark scheme

Examiner's comment

Lots of candidates were given 5 marks. They were able to use the experimental evidence to deduce the orders, write a rate equation and calculate k. The best candidates listed their reasoning for the orders rather than writing in large paragraphs of prose which sometimes got confused and led to contradictions. A good proportion of candidates recognised that 2.25 was 1.52, although a few thought that the 2nd order with respect to NO was a result of ‘2.25 being approximately 2 times 1.5’. Less successful candidates assumed that if they couldn’t identify what the relationship was between the concentration and the rate, that meant it was zero order. Less successful candidates were also unsuccessful in working out the units. A few candidates omitted k in their rate equations or didn’t write a rate equation. Many candidates were given marks as ECF from incorrect orders.

Notes from the mark scheme and report

• rate = k[NO]2[O2]
• k = 13.3
• units dm6 mol–2 s–1
• ALLOW ECF from incorrect orders and rate equation

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.2 Use ratios, fractions and percentages; M2.2 Change the subject of an equation; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; PAG10 Rates of reaction – initial rates

Mark scheme

Examiner's comment

Nearly all candidates were able to draw the line of best fit and linked the equation of a straight line to the equation. A few candidates drew a curve, and some included the anomalous result. Gradients were calculated with accuracy with a few candidates calculating the gradient using an inverted formula. A few didn’t notice the x-axis didn’t go to 0, so used Δx in their gradient at -7x10-4 and ended up with a gradient of about -12800 and Ea of 110 as ECF. Most candidates knew to multiply by R but a few incorrectly divided by R. Some candidates didn’t give to 2 SF and / or convert to kJ and gave a negative value. A few candidates tried to substitute values into lnk = - Ea/RT + lnA, rather than use the graph. Many realised that they didn’t have a y-intercept and then calculated the gradient. Those that were able to complete the maths calculated the correct answer within the range.

Notes from the mark scheme and report

• best-fit straight line excluding the anomalous point at ln k = 11.4
• gradient –28 300 to –31 300
• Ea = –gradient × 8.314
• 2 SF in kJ mol–1 (240–260)
• DO NOT ALLOW a negative Ea

Skills: M0.0 Recognise and make use of appropriate units in calculations; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M3.2 Plot two variables from experimental or other data; M3.3 Determine the slope and intercept of a linear graph

Mark scheme

Examiner's comment

Most candidates were successful in calculating pH. A few candidates gave the pH value to 1 or 3 decimal places, and few quoted 13.18 as if HNO3 were an alkali.

Notes from the mark scheme and report

• pH = –log(0.150) = 0.82
• 2 d.p. required

Skills: M0.4 Use calculators to find and use power, exponential and logarithmic functions; M2.5 Use logarithms in relation to quantities that range over several orders of magnitude

Mark scheme

Examiner's comment

Almost all candidates calculated the standard cell potential correctly. Common errors were -0.5V or -1.02V

Notes from the mark scheme and report

• –0.26 – (–0.76) = (+)0.50 V
• DO NOT ALLOW –0.5 V

Mark scheme

Examiner's comment

This calculation was answered well by many candidates. It is good to see that the level of communication throughout the multistep calculations is improving compared to previous series, although some candidates are still writing numbers with no indication as to what the number relates to. Annotation allows ECF to be given to a correct method, despite an incorrect value being given during the process. This also, more importantly, allows to candidate to remain focused on what is being calculated. The key steps were: • Calculating the number of moles of S2O32- • Using the molar ratio to link to the number of moles of Cu2+ • Calculating the number of moles in the original volume Calculating the mass by multiplying the moles of Cu2+ by the RFM of the hydrated copper sulfate. • Calculating the percentage in the mixture Common errors were that candidates: • Forgot to use the ratio within the equation. • Forgot to x 10 to calculate the moles in the original sample. • Used the incorrect RFM of copper sulfate or a combined RFM of the two salts CuSO4.5H2O and FeSO4.7H2O. • Divided the total mass of both salts (9.51) by the number of moles of CuSO4 to get a combined Mr. The candidate has produced a correct and fully annotated calculation. This clarity allows the candidate to remain focused on each calculation step. Any miscalculation could then easily gain ECF through a clear method.

Notes from the mark scheme and report

• common errors (4 marks): 65.2% (Mr 249), 41.8% (Mr of anhydrous CuSO4 159.6), 32.7% (second ratio not used), 6.54% (no ×10)
• 2Cu2+ ≡ I2 ≡ 2S2O3 2– so n(Cu2+) = n(S2O3 2–)
• 3 SF required

Skills: M0.2 Use ratios, fractions and percentages; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

Most candidates correctly calculated the number of H atoms as 7.04  1023. Some omitted to use 18 or 130 but the mark scheme allowed such omissions so that candidates could get partial credit, provided that clear working had been shown. A common error was to show the answer to more than three significant figures despite the requirement in the question. Overall, this was one of the more confidently answered questions on the paper.

Notes from the mark scheme and report

• × 18 H per molecule
• 7.04 × 10^23, 3 SF and standard form required
• common errors: 3.91 × 10^22 (no × 18), 2.17 × 10^21 (÷ 18)

Skills: M0.1 Recognise and use expressions in decimal and ordinary form; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

There were a variety of responses to this question, with many obtaining correct answers of 206 mg for ibuprofen and 494 mg for paracetamol. From the 2:1 stoichiometry in the provided equation, the amount of ibuprofen is 4.00  10–3 mol. Candidates then needed to multiply this value by the molar mass of ibuprofen (206 g mol–1). Most candidates obtained 2.00  10–3 mol for CO2. Common subsequent mistakes resulted from candidates ignoring the 2:1 stoichiometry, working out the molar mass incorrectly, not dividing by 4 for 1 tablet and not converting to mg. Some candidates used the ideal gas equation to work out the moles of CO2. The question clearly stated RTP and the molar volume at RTP is stated on the Data Sheet. When RTP is stated, candidates are wasting valuable time by using the ideal gas equation, which is harder and opens up calculation errors, particularly with powers of 10. Overall, the question was answered well. However, many calculations were unclearly displayed, with random numbers often sprayed across the page. This could make it very difficult to find answers, especially those containing errors.

Notes from the mark scheme and report

• 2 mol ibuprofen per mol CO2
• ÷ 4 for one tablet
• 206 mg and 494 mg
• common errors 103/597 (no × 2) and 824/1976 (no ÷ 4) score 3
• ECF throughout

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.2 Use ratios, fractions and percentages; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; PAG1 Moles determination

Mark scheme

Examiner's comment

This unstructured calculation required candidates to work out the concentration of Ba(OH)2, to scale the concentration for the 5 times dilution and to determine the concentration of OH– ions by multiplying by 2. Many different methods were seen for working out the moles of Ba (OH)2, but the ÷5 and 2 operations were often omitted. It was common for concentrations to be shown with the wrong number of powers of 10. Many candidates used Kw or pOH confidently to obtain a value for pH. Error carried forwards (ECF) could be applied provided that the numbers used had been generated using the numbers in the question. A good number of candidates determined the pH correctly to 2 decimal places as 13.04. The commonest incorrect answer was 12.74, the result of not using 2 to find [OH–]. As with Q2b(ii), responses were often unclearly laid out and it could be difficult to find answers.

Notes from the mark scheme and report

• g dm–3 to mol dm–3 (÷ 171.3)
• dilution ÷ 5
• no pH mark without Kw or pOH
• 13.04 to 2 DP
• common errors 12.74 (no × 2), 13.74 (no ÷ 5)

Skills: M0.2 Use ratios, fractions and percentages; M0.4 Use calculators to find and use power, exponential and logarithmic functions; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M2.5 Use logarithms in relation to quantities that range over several orders of magnitude

Mark scheme

Examiner's comment

This question was answered confidently with most candidates identified the conditions for a maximum equilibrium yield from the equation: a high pressure from fewer gaseous moles in the product and a low temperature from the exothermic reaction. This was all that the question asked but candidates often wrote far too much, considering operational conditions, rate, safely and expense. Unfortunately, candidates could end up contradicting themselves and wasting valuable time. Candidates are advised to read what the question is asking and to limit their response to this. The calculation was answered well, with most identifying that 2 significant figures is appropriate as the lowest number of significant figures present in the data provided in the question. Thus 0.48 mol dm–3 was commonly seen, rather than 0.476. The Kc expression was usually correct and most candidates rearranged this correctly to make [H2] the subject. The square root calculation showed that some candidates did not appear to have the required calculator skills. A significant number of candidates showed the Kc expression upside down and the mark scheme allowed ECF to be applied following this error.

Notes from the mark scheme and report

• high pressure because fewer gaseous moles on right (DO NOT ALLOW gaseous atoms)
• low temperature because forward reaction exothermic
• square brackets required in Kc
• square root needed
• 0.48 to 2 SF (0.476 loses a mark)
• ECF only from inverted Kc or no square root

Skills: M1.1 Use an appropriate number of significant figures; M2.2 Change the subject of an equation; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M2.4 Solve algebraic equations

Mark scheme

Examiner's comment

Most candidates gave either the correct mass of LiAlH4 or the correct percentage yield. Provided that working was shown, it was possible to award some marks form a correct method. This allows ECF to be applied where a candidate had produced an incorrect number early in their calculation. Common errors included: omission of the density, dividing instead of multiplying by the density. This question discriminated well and the presence of LiAlH4 in the problem introduced a harder feature to a convention percentage yield calculation. At with the unstructured calculations elsewhere on the paper, often numbers were seen randomly placed in the answer space, making it difficult to see where errors had arisen. Candidates may be penalising themselves if they do not show clear working. It was common to see the moles of LiAlH4 being confused with the moles of the alcohol in the percentage yield calculation with moles of LiAlH4. This error is shown in Exemplar 2. The candidate has calculated the moles of C6H5COOC2H5 and C6H5CH2OH correctly and has the determined the mass of LiAlH4. This has allowed 3 of the 4 marks to be given. For the percentage yield, the candidate has then used the moles of LiAlH4 instead of C6H5CH2OH. Notice that the candidate has also given their percentage yield to 4 significant figures (SF) rather than the 3 SF asked for in the question. If the correct moles of C6H5CH2OH had been used, this error would also have prevented the mark for percentage yield from being given.

Notes from the mark scheme and report

• mass of ester = volume × density (7.875 g, 0.0525 mol)
• LiAlH4 = 0.0525 × 2.5 × 37.9 = 4.97 g
• yield 77.6%
• common errors: no density (81.4%, 4.74 g), ÷ density (85.5%, 4.46 g)
• ECF each step
• In chemistry calculations, calculator values should be carried through all steps in a calculation. This is important to eliminate errors from rounding of intermediate values.
• Any rounding should take place at the very end of the calculation.

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.2 Use ratios, fractions and percentages; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

This question was answered well. Incorrect answers mainly resulted from the wrong sign for –1.24 or for 2.04 V, by adding rather than subtracting the 2 values given.

Notes from the mark scheme and report

• 0.40 – 1.64 = –1.24 V
• minus sign required

Skills: M2.2 Change the subject of an equation

Mark scheme

Examiner's comment

Candidates answered this question very well. A few mis-wrote the numbers and even fewer gave responses to the incorrect number of decimal places. This type of question has featured on previous examinations, and it was encouraging to see how well it was answered, even with more isotopes being included than previously.

Notes from the mark scheme and report

• answer to 2 DP (54.63)
• 1 mark ECF if %s used with wrong isotopes once or decimal places transposed for one %

Skills: M0.2 Use ratios, fractions and percentages; M1.1 Use an appropriate number of significant figures

Mark scheme

Examiner's comment

Most candidates were given at least one of the three available marks for calculating the amount of Fe(NO3)2 as 0.0200 mol. Candidates then needed to determine the moles of gas (NO2 + O2) as 0.0750 mol, which has a volume at RTP of 0.0750  24.0 = 1.80 dm3. The most common error was omission of ÷4 from the equation when working out the moles of gas and obtaining a total volume of 1.80  4 = 7.20 dm3. Such a response could still be given 2 marks by error carried forwards.

Notes from the mark scheme and report

• total moles of gas = 15/4 × n(Fe(NO3)3)
• DO NOT ALLOW 0.02 × 24 = 0.48 dm3 (1 mark only)
• ALLOW ECF
• ALLOW ideal gas equation with sensible p and T

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.2 Use ratios, fractions and percentages; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

Candidates are expected to use only concordant titres when working out the mean titre and the left-hand titre of 24.40 cm3 should be rejected. Most candidates did this to produce 24.20 cm3 as their mean titre. Use of 24.2 was allowed because rounding of a ‘0’ as the second decimal place had already been penalised in Question 21 (b) (i). Predictably, the most common error was to use all three titres to produce the incorrect mean of 20.27 cm3.

Notes from the mark scheme and report

• use concordant titres only (24.15 and 24.25)
• DO NOT ALLOW mean of all three titres
• ALLOW ECF from (b)(i)

Skills: M1.2 Find arithmetic means; PAG2 Acid–base titration

Mark scheme

Examiner's comment

Many candidates followed a well drilled method to analyse their titration results: • Moles of Na2CO3 in the mean titre • Moles of CH3COOH in 25 cm3 • Scaling 10 for moles of CH3COOH in 250 cm3 Candidates then needed to process their titration results further to determine the percentage composition: • Mass of CH3COOH in 250 cm3 • Percentage composition of CH3COOH to 3 significant figures. Most candidates were able to make some progress through the analysis. Common errors included: • Not 2 to obtain the moles of CH3COOH • Omission of the scaling stage. Some candidates ignored the titration results entirely, instead calculating the number of moles of CH3COOH in 6.5 g of the descaler as 0.1083 mol by assuming that all of the descaler was CH3COOH. This approach was flawed and could not be given any marks. A final comment must be made about the presentation of many of the responses. Numbers had often been sprayed across the page and it could be difficult to see how these related to a cohesive solution. It was often impossible to give marks for such responses. The question discriminated extremely well with some candidates given all 5 marks. Less successful responses demonstrated problems with approaching this type of question and some were given no marks at all. CH3COOH from 25 to 250 cm3. The result is a percentage composition of 8.94 % instead of 89.4 %. The clear presentation allowed the examiner to follow how the incorrect response had been obtained. Error carried forward allowed marks can be given for a correct method, giving a total of 4 out of 5 marks.

Notes from the mark scheme and report

• 2:1 ratio CH3COOH:Na2CO3
• ×10 scaling to 250 cm3 (8.94% = 4 marks)
• 3 SF required
• common errors: omitting ÷1000, using 25.0 cm3 instead of mean titre
• DO NOT ALLOW n(CH3COOH) = 6.50/60
• ALLOW ECF from mean titre

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.2 Use ratios, fractions and percentages; M1.1 Use an appropriate number of significant figures; PAG2 Acid–base titration

Mark scheme

Examiner's comment

This bond energy calculation contained more values than in recent papers and many candidates found analysing the data difficult. Despite this, the question discriminated extremely well with about half the candidates obtaining the correct enthalpy change of 391 kJ mol–1 for all 3 marks, despite an incorrect final response. Successful responses showed that some candidates had clearly practised this type of problem and they recognised that there were going to be three stages to the calculation: • use of 945 and 4 ´ –464 to obtain 2801 • correct incorporation of the other ∆H values to obtain 1564 for 4 N–H • division by 4 for 1 N–H bond to obtain 391 kJ mol–1 As with all multi-stage calculations, error carried forward was applied, allowing credit to be given for a correct method. The commonest errors were: • use of 2 O–H instead of 4 O–H in the 1st step. • use of 158 for N–N in the 1st step instead of 945 for NN (Did these candidates not recall that N2 has a triple bond?) • omission of some data for the second step.

Notes from the mark scheme and report

• common errors (2 marks): –391 (wrong sign), 159 (2 × O–H instead of 4), 681.5 (wrong sign for –581), 536.25 (∆H omitted)
• ALLOW ECF throughout

Skills: M2.2 Change the subject of an equation; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M2.4 Solve algebraic equations

Mark scheme

Examiner's comment

Just under half of the candidates scored all 3 marks. A range of responses were seen for this question but the most common error was to use the incorrect Mr for the hydrated which included either using the Mr for the anhydrous salt or the Mr for ZnSO4.H2O (179.5). It was evident that candidates struggle to understand what is meant by water of crystallisation. Some scored 2 marks for the correct Mr and finding moles but then were unsure how to proceed with the calculation. For example, some divided moles by 7 before multiplying by 161.5. Many candidates took the more complicated alternative route to find the mass of water before subtracting from mass of hydrated compound. Low-scoring candidates often struggled to correctly work out Mr values. Their working was set out so that individual steps were unclear, often with contradictory calculations.

Notes from the mark scheme and report

• M(ZnSO4•7H2O) = 287.5
• n = 0.04
• × 161.5 = 6.46 g
• no ECF if 161.5 used as molar mass of hydrate
• alternative via mass of water lost

Skills: M0.2 Use ratios, fractions and percentages; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

More candidates were able to correctly draw a tangent than seen in previous years with similar questions. A generous range was given for both tangent and gradient so many scored both marks. The most common reasons for losing marks was for having a gap between the curve and the tangent or calculating the gradient incorrectly, e.g. misreading scales, dx/dy, or by using interpolation rather than a tangent.

Notes from the mark scheme and report

• tangent at 50 s (±10 s)
• rate = gradient about 0.67 cm3 s–1
• DO NOT ALLOW interpolation
• ALLOW ECF from tangent or straight line of best fit

Skills: M3.5 Draw and use the slope of a tangent to a curve as a measure of rate of change; PAG9 Rates of reaction – continuous monitoring

Mark scheme

Examiner's comment

Over half of candidates scored all 3 marks here. However, around a quarter did not gain any credit at all. Some confused the volume of H2O2 for a volume for a gas and attempted to find moles using molar gas volume (i.e. used 50cm3 rather than 90cm3). This often lost all 3 marks as they then divided by 2 and to find concentration divided by 90/1000 instead. A few attempted to use the ideal gas equation but this rarely yielded a correct value. Candidates must be encouraged to set out working clearly, showing logical steps and preferably labelling each step with what is being calculated and giving units. Many wrote the calculation as series of steps which all equalled the previous. e.g. 90  24000 = 3.75 x 10-3 x 2 = 7.5 x 10-3  0.05 = 0.15

Notes from the mark scheme and report

• n(O2) = 90/24000
• × 2 for H2O2
• ÷ 0.0500 dm3 = 0.15
• common errors 0.075 (no × 2) and 150 (no cm3 to dm3)
• ALLOW ideal gas equation with sensible p and T
• Encourage students to assign information in the question to the correct chemical. One way to do this is to write out the equation and then underneath each species put correct volumes as given. It also helps to highlight ratios shown in the equation. It is important here to pay close attention to state symbols as it helps identify correct calculations to use.
• For example; 2H2O2 (aq) → 2H2O (l) + O2(g) 50.0cm3 90 cm3 Solution Gas at RTP Conc?

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.2 Use ratios, fractions and percentages; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

Almost all candidates had a good attempt at this calculation, with many gaining full marks. Most were able to calculate the entropy change. Almost all could reproduce the equation for free energy. Of those who did not get the correct final answer, the most common error was not converting the entropy value into kJ and / or the temperature to K. There were a few candidates who did not manipulate the equation correctly. A few candidates incorrectly calculated ΔS, obtaining the value of 165 J K-1 mol-1 or ΔH, obtaining -474 kJ mol-1. Candidates were given ECF in these cases.

Notes from the mark scheme and report

• ∆S in J K–1 mol–1 must be converted to kJ
• ½ mol O2 in ∆S
• 3 sig figs required

Skills: M0.0 Recognise and make use of appropriate units in calculations; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

The correct answer was seen frequently, along with lots of the common errors listed on the mark scheme. Candidates tended to misjudge the mole ratio and divided +249 by two. Some candidates applied the cycle incorrectly and therefore used the wrong sign for an enthalpy change, leading to them attaining one mark. Candidates should check for transcription errors as -3789 was often written for - 3798.

Notes from the mark scheme and report

• common errors: using –141 only, halving atomisation of O
• sign errors

Skills: M2.2 Change the subject of an equation; M2.4 Solve algebraic equations

Mark scheme

Examiner's comment

Candidates tend to find Kp calculations difficult and so a strategy to work their way through them could include: • Write the Kp expression using the molar ratio given in the question. Care should be taken not to change the molar ratio to help an easier calculation. Square brackets should not be used as these represent concentration. • Calculation of initial moles present, with careful consideration of the use of appropriate significant figures • Calculation of the change in moles present • Deduction of the number of moles present at equilibrium • Determination of total moles present at equilibrium These steps are often best completed as RICE tables (Ratio, Initial, Change, Equilibrium) and should look to use the appropriate amount of significant figures to avoid having a rounding error in the final answer.

Notes from the mark scheme and report

• equilibrium moles via ½ stoichiometry for O2
• mole fractions × total pressure
• units atm½
• Kp values are for the equation as stated. Candidates should recognise that changing the stoichiometry of the equation changes the Kp value.

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.2 Use ratios, fractions and percentages; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

Candidates made good progress with this calculation, many gaining 2 or 3 marks, including error carried forward. Common errors included, in various combinations: using -log[-1.95]; using an incorrect value for the concentration of ClCH2COOH; and using 10-Ka.

Notes from the mark scheme and report

• pKa = –log Ka
• 2 decimal places

Skills: M0.4 Use calculators to find and use power, exponential and logarithmic functions; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M2.5 Use logarithms in relation to quantities that range over several orders of magnitude

Mark scheme

Examiner's comment

Nearly all candidates were able to draw the line of best fit and linked the sharp vertical section of the graph with the volume of Ba(OH)2 needed to neutralise the ethanoic acid. Candidates should aim to produce a smooth line of best fit and avoid ‘tram’ lines when the pencil is taken off the paper and the curve started again. The line should go through most points. Some candidates misinterpreted the graph and used values of 8, 12, 12.2, 12.25 and 18. However, the remainder of the calculation was accessible, and most candidates scored well with ECF marks from this point. There was occasional division of 2 for the moles of ethanoic acid and dividing by the original volume of Ba(OH)2 rather than the 10cm3 of ethanoic acid. This candidate scored all available marks. This is a very good example of a candidate displaying their working. The response was well communicated indicating the end point, links were made to what was being calculated and how the next number was obtained.

Notes from the mark scheme and report

• smooth S-shaped best-fit curve through the majority of points
• equivalence at 12.5 cm3
• 1:2 ratio Ba(OH)2:CH3COOH

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.2 Use ratios, fractions and percentages; M3.2 Plot two variables from experimental or other data; PAG11 pH measurement; PAG2 Acid–base titration

Mark scheme

Examiner's comment

Although lots of candidates got the correct final answer, almost all achieved some credit from this calculation through error carried forward with marks spread across the available range. Almost all candidates were able to find the number of moles of iodide. A few candidates did not get the molar ratio and/or used the mass of just IO3- rather than KIO3. Some then did not realise the need to half this number to find the mass in 1 tablet, and multiplied by either 10 or 100 in order to convert g to mg.

Notes from the mark scheme and report

• 1:5 ratio IO3–:I–
• two tablets so halve
• answer in mg to nearest whole number (84)
• Candidates need to develop their ability to perform calculations that require them to convert between different units, e.g. mg to g. Each step of a calculation should be shown.

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.2 Use ratios, fractions and percentages; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

Most candidates managed to score at least one mark here, either for correctly determining the molar mass of the monomer, the repeat unit in the polymer or alternatively they multiplied a molar mass by 500. Many candidates gained 2 marks for either 73500 or 82500 but then struggled to account for the water lost. Some candidates lost marks due to errors in calculating the molar mass of the monomer or some tried to incorporate the use of Avogadro’s constant into the calculation. Many misunderstood what atoms would be lost during polymerisation. For example, a common incorrect response seen was found by subtracting 2 from the correct molar mass giving 163, followed by multiplication by 500 to give 81500 and finally adding of 2 to give 81502. Some struggled to understand what was meant by nearest whole number, e.g. rounding 73518 to 74000 or 82500 to 80000.

Notes from the mark scheme and report

• Mr of amino acid 165 (91 + 74)
• 500 × 165 minus 499 × 18 water = 73518
• 73500 scores 2
• common errors 36518 and 82500 (2 marks), 45500 (1 mark)

Skills: M0.2 Use ratios, fractions and percentages; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

Many candidates were well prepared for this question and secured all 3 marks. Some however were unable to calculate the correct Mr of either G or H or both. A small proportion gained credit from error carried forward. A significant number of lower attaining candidates used both masses directly or found the mass lost as a percentage. A few tried to use density information, provided earlier in the question, in their calculation. Very few lost marks due to the wrong number of significant figures, however some initially rounded the answer to 4 sf i.e. 53.85 then rounded this value to 3 sf giving 53.9.

Notes from the mark scheme and report

• n(G) = 8.97/138.0 = 0.0650
• n(H) = 5.32/152.0 = 0.0350
• 53.8% to 3 sig figs
• DO NOT ALLOW 59.3% (ratio of masses)

Skills: M0.2 Use ratios, fractions and percentages; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

Questions 1 (a), 1 (b) and 1 (c) required candidates to calculate the pH of a base, a diluted strong acid and a buffer solution. These different types of pH calculation form the basis of acid-base chemistry, all requiring the relationship pH = –log[H+] at some stage. Most candidates showed competency in their pH calculations. A number of candidates often did not appreciate the type of pH calculation required. Question 1(a) was a standard pH calculation of a strong base and this was an easy start to the paper. Most candidates used Kw and the concentration of NaOH to determine the H+ concentration of 2.5 × 10– 14 mol dm–3, from which the pH of 13.60 can be calculated using pH = –log[H+]. Some candidates first determined pOH as 0.40, and the pH using 14.00 – 0.40. This approach is based on sound chemistry and is acceptable. Several candidates often calculated the pH as –log 0.4 = 0.4 and gave this as their answer. NaOH is a common alkali and cannot have a pH <7 and this should have triggered that the candidate had made an error.

Notes from the mark scheme and report

• ALLOW pOH method
• ECF only for pH > 7

Skills: M0.4 Use calculators to find and use power, exponential and logarithmic functions; M2.2 Change the subject of an equation; M2.5 Use logarithms in relation to quantities that range over several orders of magnitude

Mark scheme

Examiner's comment

This question was answered well although nearly a half of candidates made errors. The initial solution had been diluted by 10 times and its concentration had been reduced from 0.750 to 0.0750 mol dm–3. From here pH = –log[H+] gives the correct answer of 1.12 to 2 decimal places, required by the question. Candidates found Question 1 (b) more difficult than 1 (a) or 1 (c). There seemed to be little pattern in candidate errors, the dilution being the difficult part of the question. Some did not dilute the initial concentration of 0.750 mol dm–3, giving 2.12. Others divided 0.750 by 2 instead of 10 or introduced 90 into the calculation as 90 cm3 of water would have been added. The most disappointing error was for a correct calculation to be displayed using the wrong number of decimal places. Two decimal places should be the norm for pH, reflecting the accuracy of most pH meters.

Notes from the mark scheme and report

• dilution ×10 gives 0.0750 mol dm–3
• 2 decimal places required (1.12)

Skills: M0.2 Use ratios, fractions and percentages; M0.4 Use calculators to find and use power, exponential and logarithmic functions; M2.5 Use logarithms in relation to quantities that range over several orders of magnitude

Mark scheme

Examiner's comment

This question was a standard calculation for the pH of a buffer solution and most candidates correctly identified the mixture as being a buffer. Most candidates used the expression for Ka and the concentrations of the weak acid and its salt to determine the H+ concentration of 5.25 × 10–5 mol dm–3. From here, the pH of 4.28 can easily be calculated. A common error was for the concentrations of the weak acid and its salt to be substituted into the Ka expression the wrong way round, producing an answer of 5.23. As there was only one error, this answer was given 1 mark with error carried forward (ECF). A few candidates first subtracted 0.1 from 0.3, using concentration values of 0.2 and 0.1 instead of 0.3 and 0.1. This gave an answer of 4.46 which was also given 1 mark. The substantial number of less successful responses fell into the trap of using the pH method for calculating the pH of a weak acid rather than a buffer, squaring the H+ concentration. This was the wrong method to apply for a pH buffer calculation and was not given any marks.

Notes from the mark scheme and report

• 2 decimal places required
• common errors: 5.23 from inverted [HA] and [A–] (1 mark)
• zero marks for the square-root (weak acid) approach
• ALLOW Henderson–Hasselbalch

Skills: M0.4 Use calculators to find and use power, exponential and logarithmic functions; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M2.5 Use logarithms in relation to quantities that range over several orders of magnitude

Mark scheme

Examiner's comment

This question differentiated well, with about half the candidates obtaining the correct bond enthalpy of +413 kJ mol–1. With ECF, few candidates scored 0 marks. Most candidates calculated the energy involved in making 3 H–H bonds and 1 C≡O bond as 2385 kJ mol–1. This value had to be incorporated with the energy associated with breaking bonds and the enthalpy change of –195 kJ mol–1. It was this step where problems arose. There were many errors with signs and the lowest attaining candidates sometimes omitted the enthalpy change completely in their calculation. The commonest error of 315.5 kJ mol–1 was the result of using the wrong sign for the enthalpy change. With ECF, this was still given 2 of the 3 available marks.

Notes from the mark scheme and report

• common errors: –413 (sign, 2 marks), 304 (2 mol H2), 316 (wrong sign for 195), 364 (∆H omitted, 1 mark), 529 (1 O–H instead of 2)
• DO NOT ALLOW negative final value

Skills: M2.2 Change the subject of an equation; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M2.4 Solve algebraic equations

Mark scheme

Examiner's comment

Unlike the standard bond enthalpy calculation in Question 2 (a) (ii), this question required candidates to apply their understanding of enthalpy changes to a new context relevant to energy. Candidates also had to interconvert units and provide an answer to 3 significant figures in standard form. Apart from the chemistry, there were various mathematical skills to use. Candidates found this question much more demanding than Question 2 (a) (ii), with only just over a half gaining marks. The first step involved working out that the number of moles of hydrogen gas in 60 m3 at RTP is 2500 mol. Candidates then had to apply their understanding of ∆fH to calculate the energy as 7.15 × 105 kJ. Common mistakes included: • multiplying 60 m3 by the enthalpy change of 285.8 kJ mol–1 (omitting to calculate the moles of H2 – a lack of understanding) • using more than 3 significant figures, e.g. 7.145 instead of 7.15 • making an error with powers of 10 in the mole calculation (usuallysually by not converting 60 m3 into 60,000 dm3, leading to 2.5 moles). This type of question looks simple but makes for an excellent way of developing application and mathematical skills.

Notes from the mark scheme and report

• convert m3 to dm3 (n = 2500 mol)
• 3 sig figs AND standard form required
• 1 mark for 7.15 × 10^x, 7.145 × 10^5 or 715000
• IGNORE sign

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.1 Recognise and use expressions in decimal and ordinary form; M1.1 Use an appropriate number of significant figures

Mark scheme

Examiner's comment

Most candidates determined the moles of AgNO3 and hence Ag+ as 1.50 × 10–3 mol. This was given 1 mark, but candidates then needed to use this amount to predict the identity of compound A. Most candidates could not see the way forward and many received only 1 mark. Many candidates had worked out ‘something’ from the supplied data, without knowing where this initial step would take them. Candidates needed to spot that the ratio of the element : Cl in compound A was 5 × 10–4 : 1.50 × 10–3 or 1 : 3. The correct formula of AlCl3 then follows. MgCl2 was a common error obtained by subtracting 5 × 10–4 from 1.50 × 10–3 to obtain a 1 : 2 ratio. This question would be a good exercise for improving the application skills of candidates.

Notes from the mark scheme and report

• n(AgNO3) = 1.50 × 10–3 is the essential mark
• ratio A:Cl = 1:3 so AlCl3 (ALLOW Al2Cl6 or PCl3)
• ALLOW ECF from formula of silver chloride in (b)(i)

Skills: M0.2 Use ratios, fractions and percentages

Mark scheme

Examiner's comment

This question was answered extremely well with about three-quarters of candidates securing both marks. Most candidates calculated the molecular masses of compounds C and D as 380 and 441.7 respectively, to obtain a difference of 61.7. Some candidates adopted a simpler different approach which gives the same correct answer, working out the difference between the masses of nitrogen and phosphorus (for C) and oxygen and sulfur (for D). ECF was applied to any incorrect molecular formulae from Question 2 (c) (i) from which both marks could be obtained.

Notes from the mark scheme and report

• 61.7 from Mr(D) 441.7 – Mr(C) 380, or 7 × (32.1 – 16) – 3 × (31 – 14)
• ALLOW ECF from incorrect formula in (c)(i)

Skills: M0.2 Use ratios, fractions and percentages

Mark scheme

Examiner's comment

Candidates are much better at calculating percentage uncertainties than in previous series. Almost all candidates realised that the uncertainty in the titre resulted from two readings and that the overall uncertainty in the titre would be 2 × 0.05 = 0.1 cm3, resulting in a percentage uncertainty of 0.46%. Comparatively few candidates ignored the ‘2’ and gave 0.23% as their answer. Some candidates worked out the percentage uncertainty from the mean titre but this was still given marks.

Notes from the mark scheme and report

• 2 × 0.05 because two readings
• 0.46%
• 2 decimal places minimum
• DO NOT ALLOW 0.50%

Skills: M1.3 Identify uncertainties in measurements and use simple techniques to determine uncertainty when data are combined

Mark scheme

Examiner's comment

In contrast to equation writing, candidates found this simple calculation far easier with the majority obtaining both marks for 26.72. Common incorrect percentages were 26.7 (wrong number of decimal places) and 8.91 (using 31 rather than 3 × 31 for the numerator).

Notes from the mark scheme and report

• 2 decimal places required (26.72
• 26.7 scores 1)
• 8.91 scores 1 (omitting the 3 P atoms)

Skills: M0.2 Use ratios, fractions and percentages; M1.1 Use an appropriate number of significant figures

Mark scheme

Examiner's comment

Candidates are well versed with calculating an Rf value, with nearly all candidates obtaining a value in the acceptable range of 0.12–0.18.

Notes from the mark scheme and report

• working required
• measure from origin line to solvent front (about 9.1 cm), not from the bottom of the plate
• ALLOW 0.12–0.18

Skills: M0.2 Use ratios, fractions and percentages; M3.1 Translate information between graphical, numerical and algebraic forms; PAG6 Synthesis of an organic solid

Mark scheme

Examiner's comment

Candidates are expected to use only concordant titres when working out the mean titre and the middle titre of 20.60 cm3 should be rejected. Most candidates did this to produce 20.30 cm3 as their mean titre. Predictably, the commonest error was to use all three titres to produce the incorrect mean of 20.40 cm3.

Notes from the mark scheme and report

• mean of the concordant titres only (20.30)
• DO NOT ALLOW mean of all three titres (20.40)

Skills: M1.2 Find arithmetic means; PAG2 Acid–base titration

Mark scheme

Examiner's comment

Many candidates followed a well drilled method to identify the unknown metal M as potassium: • Moles of H2SO4 in the mean titre • Moles of KOH in 25 cm3 • Scaling 10 for moles of KOH in 250 cm3 • Molar mass of the metal as 39.11 and identified as K. A number of candidates omitted the scaling stage to obtain a molar mass of 391.1. By ECF, the ‘correct’ identity would be caesium or francium. Some candidates then ‘fiddled’ their response, dividing by 10 to ‘identify’ the metal as K. This incorrect approach was not credited. A common error, illustrated in Exemplar 2, was for candidates to calculate 39.11 but to think that this was the mass of MOH and not M. They then subtracted 17 (for OH) from 39.11 to obtain a response of 22.11 and identified M as sodium instead of potassium. This error probably stems from candidates either not reading the question closely enough or confusion about the mole concept.

Notes from the mark scheme and report

• ×2 for MOH:H2SO4
• ×10 scaling to 250.0 cm3 (if absent Ar = 389, Cs/Fr)
• ALLOW ECF from incorrect mean titre in (c)(iii)
• use of 25.0 for acid volume loses first mark only

Skills: M0.2 Use ratios, fractions and percentages; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; PAG2 Acid–base titration

Mark scheme

Examiner's comment

This question discriminated extremely well with many obtaining the correct enthalpy change of –394 kJ mol–1 for 3 marks. There were 3 stages to the calculating: • Use of –824 and 3  –111 to obtain –1157 • Correct incorporation of –25 to obtain –1182 • Division by 3 for 1 mole of CO2 to obtain –394 kJ mol–1 Weak candidates usually gained the first marking point but did not get any further. Incorrect signs were common as was not dividing by 3 for the final marking point. The mark scheme allowed ECF to be applied, provided that there was some correct method in the response. Unfortunately, many responses were very difficult to follow, often shown as numbers almost placed at random in the answer space. This makes it very difficult to identify any logic and to apply ECF. Exemplar 3 serves as a model, clear response, showing what is possible.

Notes from the mark scheme and report

• common errors: –1182 (no ÷3) scores 2
• +394 scores 2
• –320 (no ×3 for CO) scores 2
• ALLOW ECF throughout

Skills: M2.2 Change the subject of an equation; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M2.4 Solve algebraic equations

Mark scheme

Examiner's comment

The Kc expression was written well by many inverted the expression or added rather than multiplying the values. Indices were usually included. Many candidates gave responses to more than 2 significant figures (2SF) and some did not give the response in standard form. Candidates should be made aware that the term in the question of ‘most appropriate’ means that the final response should be shown to the least number of significant figures in the supplied data. In this scenario, all values were provided to 2SF and so the final response should also be expressed to 2SF.

Notes from the mark scheme and report

• square brackets required
• 2 SF AND standard form (7.9 × 10^4)
• ECF only from inverted expression (1.3 × 10–5)
• DO NOT ALLOW + in denominator
• The Mathematical Skills Handbook provides guidance on the use of significant figures which can be shared with students.

Skills: M0.1 Recognise and use expressions in decimal and ordinary form; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

Most candidates scored both marks. Some lost marks for minor slips such as not giving their final answer to 2 decimal places or making calculator errors. A few didn't know how to attempt the calculation or calculated the average mass.

Notes from the mark scheme and report

• 2 DP required (47.92)
• 1 mark ECF from incorrect sum if answer between 46 and 50 to 2 DP

Skills: M0.2 Use ratios, fractions and percentages; M1.2 Find arithmetic means

Mark scheme

Examiner's comment

Candidates found this calculation quite challenging, with less than a quarter achieving full marks. The most common errors are highlighted on the mark scheme. Many that struggled were often given credit for the x4 ratio mark but only if it was possible to see this in the working. Many gave multiple, often contradictory attempts at the calculation. It was not always clear how the final answer had been obtained. Clear working enables us to follow the logic and give ECF where appropriate. Many divided 1000 g by the molar mass for TiCl4 and then found 72% of this. It was important here to read the question carefully to ensure complete understanding. This candidate achieved 3 out of the 4 possible marks. The steps in their calculation are logical and it is easy to follow their working and therefore spot the error in their calculation. They have divided by 4 rather than multiplying. It also shows the calculation can be performed in a different order to that on the mark scheme. All intermediate values are used in calculations as calculator values without rounding to ensure an accurate answer.

Notes from the mark scheme and report

• 3 SF AND kg required (2.67)
• common errors: 1.92 (yield ignored), 1.38 (yield inverted), 0.673 (Mr of TiCl4 used for Ti)
• ALLOW ECF throughout

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.2 Use ratios, fractions and percentages; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

This was a tricky calculation, made more challenging if candidates hadn’t been able to successfully complete (i). Many were able to calculate the moles of BaSO4 but often rounded their answer to only 2 significant figures at this stage i.e. 0.014. Many assumed a direct ratio between BaSO4 and NaCl so mass was found by multiplying moles by 58.5 (molar mass for NaCl) - if this was done then credit was given for ECF for the final marking point.

Notes from the mark scheme and report

• n(BaSO4) → mass Ba(NO3)2 (× 261.3) → mass NaCl by difference → 26.6% to 3 SF
• using n × 58.5 gives 16.4 (ECF only)
• ALLOW ECF from equation in (b)(i)
• The M1 section of the Mathematical Skills handbook contains useful information on handling data, including M1.1 use of significant figures.

Skills: M0.2 Use ratios, fractions and percentages; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; PAG1 Moles determination

Mark scheme

Examiner's comment

More than half of candidates gave the correct answer. Common mistakes included not multiplying the uncertainty by 2 for each measurement (0.81) or using the initial or final temperatures rather than the difference in their calculations.

Notes from the mark scheme and report

• two readings: 2 × 0.25/31 × 100 = 1.6%

Skills: M0.2 Use ratios, fractions and percentages; M1.3 Identify uncertainties in measurements and use simple techniques to determine uncertainty when data are combined; PAG3 Enthalpy determination

Mark scheme

Examiner's comment

Those that were more confident with the calculation in (ii) were more likely to get the answer here. ECF was occasionally seen from an incorrect mass or enthalpy change depending on method used. Many were able to work out q using 150g of water even if they had incorrectly calculated q in (ii) - such as previously using 1.31g.

Notes from the mark scheme and report

• 0.75 × 1.31 = 0.9825 g (ALLOW 0.98–0.99)
• ALLOW ECF from (b)(ii)

Skills: M0.2 Use ratios, fractions and percentages; M2.2 Change the subject of an equation

Mark scheme

Examiner's comment

Candidates’ working was often unclear and very difficult to follow for this question, making it challenging to find errors and award ECF where appropriate. Despite being given the displayed formula, many were unable to correctly count the bonds in the molecules. Missing some or all of C-C, C-O and 3 x O=O from the bonds broken was often seen. Some only counted 2 x C=O bonds in the products. Some subtracted the bonds broken from bonds made or omitted the enthalpy change from their calculation. Very few candidates gave a negative value for their final bond enthalpy.

Notes from the mark scheme and report

• +805
• +3220 (not divided by 4) scores 3
• common errors: –805 (sign), 718 (C–C missed), 431.5 (3 × O=O missed), 167 (rearrangement)
• 6 O–H made, 1 O–H broken

Skills: M2.2 Change the subject of an equation; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M2.4 Solve algebraic equations

Mark scheme

Examiner's comment

The correct was answer seen frequently, along with lots of the common errors. Candidates tended to forget the mole ratio and did not multiply either -107 or +296 by two. Some candidates applied the cycle incorrectly and therefore used the wrong sign for an enthalpy change, leading to them attaining 1 mark.

Notes from the mark scheme and report

• 1 mark for a single error: +1872 (sign), –2168 (296 × 1), –1765 (107 × 1), –3056
• DO NOT ALLOW any answer with two errors

Skills: M2.2 Change the subject of an equation; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M2.4 Solve algebraic equations

Mark scheme

Examiner's comment

Many candidates calculated the correct value of q and scored all three marks. Common errors saw some candidates using the incorrect mass, of either 0.133 or that added to or subtracted from the 150. The temperature change was given but a few candidates added the 10.5 to 273 in order to, incorrectly, convert to K. The final mark required the candidate to link the temperature increase to an exothermic value and include a minus sign.

Notes from the mark scheme and report

• q = 150 × 4.18 × 10.5
• n = 0.133/100
• – sign required
• common errors for 2 marks: +4950, –5077/–5064 (use of 2 SF)

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.2 Use ratios, fractions and percentages; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; PAG3 Enthalpy determination

Mark scheme

Examiner's comment

This calculation proved difficult. Many candidates did not deduce the correct cycle. Of those who did, a significant majority did not appreciate the stoichiometry and/or use the correct signs from the cycle. Many candidates scored 1 mark for this question, which was gained as an error carried forward, in the enthalpy cycle. -3163 was a common error. This candidate showed clarity in their working by drawing a correct Hess cycle and then applying correct multiples and directions for the enthalpy changes.

Notes from the mark scheme and report

• 186 = ∆cH(C9H20) – ∆cH(C5H12) – 2∆cH(C2H4)
• common errors for 1 mark: +3535, –4946 (1 × –1411), –3163 (+186)

Skills: M2.2 Change the subject of an equation; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M2.4 Solve algebraic equations

Mark scheme

Examiner's comment

Nearly all candidates were able to draw the line of best fit and linked the equation of a straight line to the equation for delta G. Some hadn’t used a ruler to draw a straight line and candidates should be reminded of this requirement. Candidates were able to recognise that the gradient corresponded to (-) delta S but some did not to use the scale correctly. Some candidates tried to use very small triangles to get the gradient and candidates would be advised to use the largest scale possible to minimise errors in the line of best fit. Most candidates recognised that the minimum temperature was the x-intercept (when delta G = 0) and that delta H was the y-intercept. Some did not use this approach and instead used the ∆G=∆H-T∆S equation, using previously obtained value(s), to calculate the minimum temperature, ∆S or ∆H. Some candidates used ∆H = -186 from a previous question or attempted to solve both ∆S and ∆H by simultaneous equations.

Notes from the mark scheme and report

• best-fit line extrapolating to y-axis 43–47 and x-axis 820–840
• ∆S = –gradient × 1000 (122–131)
• minimum T where ∆G = 0 (340–370 K)
• ∆H = y-intercept (43–47)
• ECF throughout

Skills: M0.0 Recognise and make use of appropriate units in calculations; M3.1 Translate information between graphical, numerical and algebraic forms; M3.2 Plot two variables from experimental or other data; M3.3 Determine the slope and intercept of a linear graph

Mark scheme

Examiner's comment

Candidates tend to find Kp calculations difficult and so a strategy to work their way through them could include: • write the Kp expression, with units, ensuring square brackets are not used. Common mistakes with units included atm-1 mol-1, mol-1 dm3, kPa-1 • calculation of initial moles present, with careful consideration of the use of appropriate significant figures • calculation of the change in moles present • deduction of the number of moles present at equilibrium • determination of total moles present at equilibrium. These steps are often best completed as RICE tables (Ratio, Initial, Change, Equilibrium) and should look to use the appropriate amount of significant figures: • calculation of mole fractions at equilibrium • calculation of partial pressures at equilibrium • inserting partial pressure vales into the Kp expression and avoiding any unnecessary unit conversions • writing an answer to the required number of significant figures.

Notes from the mark scheme and report

• DO NOT ALLOW square brackets in Kp
• n(N2O4) = 0.300 and total 5.70
• units atm–1
• 3 SF (1.17 × 10–2)
• common error 2.47 × 10–2 (0.6 mol N2O4) scores 2 calculation marks

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.2 Use ratios, fractions and percentages; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

This question was well answered by nearly all candidates and many scored all 5 marks. A number used the wrong units for the pressure and the volume so used both kPa and dm3. This resulted in the correct number of moles and scored 4 marks as error carried forward. Most candidates were able to find the formula from the molar mass and very few used the incorrect molar volume route.

Notes from the mark scheme and report

• V in m3, T = 348 K, p in Pa
• M = 108 so N2O5
• use of 24 dm3 gives only the last 2 marks by ECF
• DO NOT ALLOW N2O4

Skills: M0.0 Recognise and make use of appropriate units in calculations; M2.2 Change the subject of an equation; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

ocr.org.uk/science Oxford Cambridge and RSA CHEMISTRY A H432 For first teaching in 2015 A LEVEL Version 1 H432/01 Summer 2023 series Examiners’ report Contents Introduction .......................................................................................................................................... 4 Paper 1 series overview........................................................................................................................ 5 Section A overview ............................................................................................................................... 6 Question 1 .................................................................................................................................... 6 Question 2 ....................................................................................................................................... 6 Question 3 ....................................................................................................................................... 7 Question 4 ....................................................................................................................................... 7 Question 5 ....................................................................................................................................... 8 Question 6 ....................................................................................................................................... 8 Question 7 ....................................................................................................................................... 9 Question 8 ....................................................................................................................................... 9 Question 9 ..................................................................................................................................... 10 Question 10 ................................................................................................................................... 10 Question 11 ................................................................................................................................... 11 Question 12 ................................................................................................................................... 11 Question 13 ................................................................................................................................... 12 Question 14 ................................................................................................................................... 12 Question 15 ................................................................................................................................... 13 Section B overview ............................................................................................................................. 14 Question 16 (a) (i) ........................................................................................................................... 14 Question 16 (a) (ii) ......................................................................................................................... 15 Question 16 (b) .............................................................................................................................. 16 Question 17* .................................................................................................................................. 17 Question 18 (a) ............................................................................................................................... 18 Question 18 (b) .............................................................................................................................. 19 Question 18 (c) (i) .......................................................................................................................... 20 Question 18 (c) (ii) .......................................................................................................................... 21 Question 19 (a) (i) .......................................................................................................................... 23 Question 19 (a) (ii) ......................................................................................................................... 24 Question 19 (b) .............................................................................................................................. 24 Question 20 .................................................................................................................................... 25 Question 20 (a) (ii) ......................................................................................................................... 26 Question 20 (b) .............................................................................................................................. 26 Question 20 (c) (i) .......................................................................................................................... 27 Question 20 (c) (ii) .......................................................................................................................... 27 Question 20 (d) (i) .......................................................................................................................... 28 Question 20 (d) (ii) ......................................................................................................................... 28 Question 20 (e) (i) .......................................................................................................................... 29 Question 20 (e) (ii) ......................................................................................................................... 29 Question 20 (f) ............................................................................................................................... 30 Question 21 (a) (i) ........................................................................................................................... 31 Question 21 (a) (ii) ......................................................................................................................... 31 Question 21 (iii) .............................................................................................................................. 32 Question 21 (b) .............................................................................................................................. 33 Question 22 (a)* .............................................................................................................................. 34 Question 22 (b) (i) .......................................................................................................................... 36 Question 22 (b) (ii) ......................................................................................................................... 36 Question 22 (c) .............................................................................................................................. 37 Question 22 (d) .............................................................................................................................. 37 Introduction Our examiners’ reports are produced to offer constructive feedback on candidates’ performance in the examinations. Question 20 (a) (ii) This question proved difficult with only a few candidates able to state that in neutral water, [H+] = [OH-].

Notes from the mark scheme and report

• pH = 6.77 (2 DP)
• DO NOT ALLOW use of A– or X–

Skills: M0.4 Use calculators to find and use power, exponential and logarithmic functions; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M2.5 Use logarithms in relation to quantities that range over several orders of magnitude

Mark scheme

Examiner's comment

Most candidates wrote the correct equation. Common errors were using Sr2+ as reactant, not balancing the H2O and not having the H2 as second product. Most candidates calculated the moles of Sr(OH)2 correctly but fewer recognised that [OH-] = twice the [Sr(OH)2]. As a result, most candidates scored 3 calculation marks. A few candidates chose the incorrect Kw value.

Notes from the mark scheme and report

• Sr + 2H2O → Sr(OH)2 + H2
• Kw at 40 °C must be used
• common errors for 3 calculation marks: 11.98 (Kw = 1 × 10–14), 11.21 (no × 2), 10.91 (÷2)
• 2 DP

Skills: M0.2 Use ratios, fractions and percentages; M0.4 Use calculators to find and use power, exponential and logarithmic functions; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M2.5 Use logarithms in relation to quantities that range over several orders of magnitude

Mark scheme

Examiner's comment

This calculation proved difficult with many figures and sums appearing with little indication as to their relevance. Candidates should remember to provide written indications of what it is they’re working out – presenting the calculations without any annotations can make it harder for error carried forward marks to be given if there is an error in their calculation. Few candidates scored all 4 marks. Most found the concentration of H+ from the pH and the moles of KOH correctly but did not recognise they had to take away the moles of KOH from those of HA to find the remaining concentration of HA. Some candidates then used the [H+] as the [HA]. A few candidates tried a [H+] squared expression of a weak acid. The candidate has clearly set out the calculation so each numerical value can be linked and the steps understood. The only error is not calculating the excess acid (i.e. not subtracting moles OH- from initial moles acid) so 3 marks were given.

Notes from the mark scheme and report

• n(A–) = n(KOH) = 0.0604
• n(HA) remaining = 0.094 – 0.0604
• Ka = [H+][A–]/[HA] = 1.53 × 10–5
• common error 5.47 × 10–6 (KOH not subtracted from HA) scores 3

Skills: M0.4 Use calculators to find and use power, exponential and logarithmic functions; M2.2 Change the subject of an equation; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M2.5 Use logarithms in relation to quantities that range over several orders of magnitude

Mark scheme

Examiner's comment

Almost all candidates calculated the titres correctly. A significant number used all three titre values to derive their mean value.

Notes from the mark scheme and report

• titres 12.65, 12.95, 12.75
• mean of concordant titres only = 12.70
• Links to the legacy coursework tasks and PAG practice question sets can be found on Teach Cambridge and can help students prepare for practical-based questions like this one. Exam hints for students is useful to share with candidates.

Skills: M1.2 Find arithmetic means; PAG2 Acid–base titration

Mark scheme

Examiner's comment

ocr.org.uk/science Oxford Cambridge and RSA CHEMISTRY A H432 For first teaching in 2015 A LEVEL Version 1 H432/01 Summer 2023 series Examiners’ report Contents Introduction .......................................................................................................................................... 4 Paper 1 series overview........................................................................................................................ 5 Section A overview ............................................................................................................................... 6 Question 1 .................................................................................................................................... 6 Question 2 ....................................................................................................................................... 6 Question 3 ....................................................................................................................................... 7 Question 4 ....................................................................................................................................... 7 Question 5 ....................................................................................................................................... 8 Question 6 ....................................................................................................................................... 8 Question 7 ....................................................................................................................................... 9 Question 8 ....................................................................................................................................... 9 Question 9 ..................................................................................................................................... 10 Question 10 ................................................................................................................................... 10 Question 11 ................................................................................................................................... 11 Question 12 ................................................................................................................................... 11 Question 13 ................................................................................................................................... 12 Question 14 ................................................................................................................................... 12 Question 15 ................................................................................................................................... 13 Section B overview ............................................................................................................................. 14 Question 16 (a) (i) ........................................................................................................................... 14 Question 16 (a) (ii) ......................................................................................................................... 15 Question 16 (b) .............................................................................................................................. 16 Question 17* .................................................................................................................................. 17 Question 18 (a) ............................................................................................................................... 18 Question 18 (b) .............................................................................................................................. 19 Question 18 (c) (i) .......................................................................................................................... 20 Question 18 (c) (ii) .......................................................................................................................... 21 Question 19 (a) (i) .......................................................................................................................... 23 Question 19 (a) (ii) ......................................................................................................................... 24 Question 19 (b) .............................................................................................................................. 24 Question 20 .................................................................................................................................... 25 Question 20 (a) (ii) ......................................................................................................................... 26 Question 20 (b) .............................................................................................................................. 26 Question 20 (c) (i) .......................................................................................................................... 27 Question 20 (c) (ii) .......................................................................................................................... 27 Question 20 (d) (i) .......................................................................................................................... 28 Question 20 (d) (ii) ......................................................................................................................... 28 Question 20 (e) (i) .......................................................................................................................... 29 Question 20 (e) (ii) ......................................................................................................................... 29 Question 20 (f) ............................................................................................................................... 30 Question 21 (a) (i) ........................................................................................................................... 31 Question 21 (a) (ii) ......................................................................................................................... 31 Question 21 (iii) .............................................................................................................................. 32 Question 21 (b) .............................................................................................................................. 33 Question 22 (a)* .............................................................................................................................. 34 Question 22 (b) (i) .......................................................................................................................... 36 Question 22 (b) (ii) ......................................................................................................................... 36 Question 22 (c) .............................................................................................................................. 37 Question 22 (d) .............................................................................................................................. 37 Introduction Our examiners’ reports are produced to offer constructive feedback on candidates’ performance in the examinations. Question 21 (a) (i) Only a few candidates were given this mark. Question 21 (a) (ii) Almost all candidates calculated the titres correctly.

Notes from the mark scheme and report

• × 5 for Fe2+
• ÷ 0.0250
• ECF from incorrect titre
• common errors for 2 marks: 1.27 × 10–3 (no × 5), 2.46 × 10–2 (volumes transposed)

Skills: M0.2 Use ratios, fractions and percentages; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; PAG2 Acid–base titration

Mark scheme

Examiner's comment

Overall, this question was well answered with over half of candidates gaining all 3 marks. The use of a different molar volume confused some candidates. Some attempted to use PV=nRT or different combinations of the figures given with varying degrees of success. Lower attaining candidates typically struggled with unit conversions and were unable to make use of the units to help them work out the methodology to use.

Notes from the mark scheme and report

• use 40.0 dm3 mol–1 not 24: n = 1.80 × 10^-3, M = 480.5, C6H9Br5
• common error 24000 gives 288.3 (C6H12Br2) for 2 marks by ECF
• cm3 to dm3

Skills: M0.0 Recognise and make use of appropriate units in calculations; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M2.4 Solve algebraic equations

Mark scheme

Examiner's comment

Approximately half of the candidates scored all 3 marks and were confident calculating % yield. However, about a third did not gain any credit. A common response was 100% with both masses as 15 g. Some struggled with conversion of the volume of nitrobenzene to a mass using the density. Many candidates struggled with obtaining the correct molar mass values, especially if they were not sure on the correct structures. As with any calculation work, clear working is vital to gain error carried forward credit here. Three significant figures were required for final answer but many also lost marks as a consequence of incorrect rounding. For example, use of 0.089 moles gave a value of 73.0%. It was not uncommon to see contradictory calculations as candidates appeared to try to find a suitable value.

Notes from the mark scheme and report

• mass = volume × density first
• 73.2 to 3 SF
• common error 87.9 (density not used) scores 2
• no final mark if yield ≥ 100%

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.2 Use ratios, fractions and percentages; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

Nearly half the candidates were able to correctly calculate the number of PAS molecules in the maximum daily dosage. Some struggled with conversion from mg to g. Some made errors calculating the molar masses or did not calculate moles at all. Most were confident to multiply by NA at the end. As with other calculations some lost marks for significant figures and rounding errors.

Notes from the mark scheme and report

• 300 mg per kg × 20.0 kg = 6.0 g
• /153 = 0.0392 mol
• × 6.02 × 10^23 = 2.36 × 10^22
• ECF throughout

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.1 Recognise and use expressions in decimal and ordinary form; M0.2 Use ratios, fractions and percentages; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

This question was a good discriminator, producing marks across the whole 3 mark range. More successful candidates correctly calculated the energy change, moles of RbClO3 and enthalpy change of solution. However, there were pitfalls for many including the following: • calculating the energy change using the mass of water rather than the mass of the solution. This was despite the suppled information that the specific heat capacity of the solution is the same as for water. Candidates should understand that m in mc∆T is the mass of the substance that produces ∆T • calculating an incorrect value for the molar mass of RbClO3. Instead of 169, this was often seen as 120.5 (using the atomic number of 37 for Rb, rather than the mass number of 85.5) and 185 (for RbClO4) • using values of m at the wrong stages in the calculation. e.g. 2 g with the energy change and 102 g or 100 g with the moles calculation • calculating the correct numerical value for the enthalpy change of solution, but then placing a ‘–‘ sign in front of the value, despite ∆T being for a decrease in temperature. Finally, as with all multi-step calculations, candidates are advised to use calculator values throughout. Any intermediate rounding introduces rounding errors in the final value. The final value can be rounded either to the significant figures demanded by the question or to the lowest number of significant figures used in the provided data.

Notes from the mark scheme and report

• mass of solution 102 g (not 100 or 2.00 g)
• ∆T = 1.5
• n = 2.00/169
• +54.0 to 54.2
• IGNORE sign
• common errors: 100 instead of 102 (53.0) 2 marks, 102 and 2 swapped 1 mark, 8.314 used for c 2 marks
• ALLOW ECF

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.2 Use ratios, fractions and percentages; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; PAG3 Enthalpy determination

Mark scheme

Examiner's comment

The marks for this calculation were much more polarised than the calculation in Question 2 (a). Many candidates worked through the problem methodically to get an answer for the initial rate between 4.4  10–5 and 4.5  10–5 mol dm–3 s–1. A substantial number of candidates worked out [H+] using 10–pH for 1 mark but were then unable to progress any further. This was a novel calculation, requiring candidates to develop their own strategy for its solution. Candidates who found this question difficult often attempted a solution based on stock weak acid calculation, using [H+]2/[HA].

Notes from the mark scheme and report

• factor 0.0316/0.680 or k = rate/[HCl] = 1.40 × 10–3
• rate 4.4–4.5 × 10–5
• ECF only if derived from concentration and rate

Skills: M0.2 Use ratios, fractions and percentages; M0.4 Use calculators to find and use power, exponential and logarithmic functions; M2.5 Use logarithms in relation to quantities that range over several orders of magnitude

Mark scheme

Examiner's comment

This question required candidates to calculate the masses of two reactants that could be used to prepare a sample of the Tutton’s salt. Candidates were supplied with the formula of hydrated copper(II) sulfate but not the formula of ammonium sulfate, so candidates needed to work out its formula from ions that candidates are expected to be able to recall from the specification. Just over half the candidates obtained both correct masses but many obtained just one correct mass, usually that of CuSO4•6H2O. This exemplar shows a typical response with the incorrect formula of ammonium sulfate clear shown by the candidate, resulting in the incorrect mass of 2.8525 g. This incorrect formula and mass were seen in many responses and, from the initial crossings out, this candidate is clearly confused about how to tackle this simple mole calculation. The incorrect answer of 2.85 g was seen on almost as many scripts as the correct answer of 3.30 g. The moral is that candidates need to learn the formula and charges of the common ions encountered in chemistry. The comments here apply also to Question 6 (b) (i), where formulae need to be written using ions listed in the specification.

Notes from the mark scheme and report

• (NH4)2SO4 3.30 g
• CuSO4•5H2O 6.24 g (must include water of crystallisation)

Skills: M0.2 Use ratios, fractions and percentages; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

Compared with the application based Question 4, candidates answered this stock titration calculation well. Almost all candidates determined that the mean titre was 22.35 cm3 and went on to calculate the number of moles of HCl as 4.47  10–3 mol. Most scaled this value by 10 to determine the moles in 250 cm3. Most candidates then used the initial moles of HCl to determine the moles of aspirin in the 3 tablets as 5.30  10–3 moles. A significant number omitted this stage but they were able to be credited for the next stage of calculation using a correct method. Consequently over half the candidates were awarded 5 or 6 marks for this stock calculation.

Notes from the mark scheme and report

• mean of concordant titres 22.35 (not all three = 22.60, 5 marks max)
• n(NaOH) remaining × 10
• subtract from 0.0500 mol
• × 180 then ÷ 3 tablets = 318 mg
• common errors: no × 10 scaling, no subtraction from 0.05, omitting the titration calculation (3000 mg)

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.2 Use ratios, fractions and percentages; M1.2 Find arithmetic means; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; PAG2 Acid–base titration

Mark scheme

Examiner's comment

Almost all candidates identified X and Y as NO and H2O respectively, but the equation proved to be much more testing. Some candidates were careless, showing NO and H2O the wrong way round (credited with 1 out of these 2 marks) or with charges. For the equation, candidates needed to consider the oxidation number change of N from +5 to +2, This should have naturally led to 3e– being added on the left-hand side. Many candidates omitted the electrons entirely. Some did add 3e– but on the right. This suggests that candidates would benefit with practising the construction of half equations.

Notes from the mark scheme and report

• X = NO, Z = H2O (1 of 2 if in wrong order)
• HNO3 + 3H+ + 3e– → NO + 2H2O
• equation mark independent of identification

Mark scheme

Examiner's comment

As with the stock titration calculation in Question 5 (a), candidates were much more comfortable in tackling a stock Kc calculation. These have been assessed in several previous H432/01 exams and many excellent responses were seen, gaining all 5 marks. Error carried forwards allowed for an early slip in the calculation to be credited subsequently for a correct method. The hardest part of the calculation was the initial determination of the equilibrium moles and concentrations. The equilibrium moles of O2 was the main error, presumably from the 2:1:2 stoichiometry. Some candidates multiplied by 2 instead of dividing by 2 for the concentration. A common error was for candidates to use a partial pressure approach. These errors lost intermediate marks by credit could then be awarded for using the correct Kc expression to generate a Kc value with correct units. Very few candidates did not gain any marks.

Notes from the mark scheme and report

• equilibrium moles SO2 6.20 × 10–3 and O2 4.80 × 10–2
• ÷ 2.00 dm3 for concentrations
• Kc = 2930 dm3 mol–1
• common errors: no ÷ 2 (1465), original values used (21.6), SO3 not squared, inverted Kc
• units must match Kc expression used

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.2 Use ratios, fractions and percentages; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

Candidates answered this question very successfully. Few candidates mis-wrote the numbers and even fewer gave answers to the incorrect number of decimal places. This type of question has featured on previous examinations and it was pleasing to see how well it was answered, even with more isotopes being included than in previous instances.

Notes from the mark scheme and report

• answer to 2 decimal places (190.47)
• 1 mark ECF if percentages used with wrong isotopes once or one percentage transposed

Skills: M0.2 Use ratios, fractions and percentages; M1.2 Find arithmetic means; M3.1 Translate information between graphical, numerical and algebraic forms

Mark scheme

Examiner's comment

More successful candidates followed a well-rehearsed method for processing experimental enthalpy results to arrive at an enthalpy change. The result was usually the correct answer of –116 kJ mol–1. Most candidates combined the 2 volumes (25 cm3 and 50 cm3) to give 75 cm3 with m = 75 g and then calculated the correct energy change of 5799.75 J using mc∆T. Some candidates did not combine the volumes and used m = 25 g or 50 g instead. Most candidates worked out the amount in moles of Ba(OH)2 and HNO3. Unfortunately, these calculations were often scattered across the page with no indication of what the calculated values applied to. Some candidates incorrectly combined the moles (0.10 and 0.05) and divided this value (usually 0.15) into the energy change. The mark scheme accounted for these errors and allowed error carried forward marks to be allocated appropriately. Candidates are strongly advised to organise their calculation in a coherent way and to show what each calculated value applies to so that error carried forward can be applied for mistakes. misunderstanding the meaning of m in the equation q = mc∆T, used for finding the energy change q from experimental results. The information states that the density of the solutions is the same as for water, 1.00 g cm–3. Therefore, the mass m that changes temperature ∆T is 75.0 g and the energy change is 75.0  4.18  18.5 J. This candidate has correctly calculated the moles of one of the reactants, Ba(OH)2, as 0.0500 but has then calculated its mass as 75.0  4.18  18.5 J = 8.565 g and has used this in mc∆T. So the key message is that m in mc∆T is where the thermometer has been placed.

Notes from the mark scheme and report

• mass is 75.0 g (both solutions)
• divide by n(Ba(OH)2) = 0.0500 not n(HNO3)
• common errors –58.0, –38.7, –77.3 (3 marks)
• + sign loses 1 mark
• 3 sig figs needed
• In the equation q = mc∆T, mass m is the mass of the substance that changes temperature by ∆T. This substance has a specific heat capacity, c. Looking at the experimental results, the substance changing temperature is the mixture of the two solutions, which have volumes of 25 cm3 and 50 cm3. This is where the thermometer has been placed.

Skills: M0.0 Recognise and make use of appropriate units in calculations; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; PAG3 Enthalpy determination

Mark scheme

Examiner's comment

Many candidates correctly identified that neutralisation is the formation of 1 mole of water, whereas this equation forms 2 moles of water. Significantly fewer were then able to use their answer to the calculation in Question 25 (a) (i) to determine a value for the neutralisation enthalpy as half of that that value. This was a novel question, not used in previous examinations, and many candidates coped with the challenge admirably.

Notes from the mark scheme and report

• neutralisation refers to formation of 1 mol H2O (H2O essential) but equation forms 2 mol
• value = answer to (a)(i) ÷ 2
• IGNORE sign

Skills: M0.2 Use ratios, fractions and percentages

Mark scheme

Examiner's comment

In contrast with the mechanism in Question 26 (a), this ideal gas calculation was answered well by most candidates. Most had obviously learnt the method for successfully processing this type of calculation. More candidates than in previous series were able to convert ºC into K and cm3 into m3 to obtain n = 0.00241 mol. Many candidates were then able to achieve the correct molar mass of 136.9 g mol–1 and molecular formula of C4H9Br. Candidates with incorrect unit conversions could still collect method marks by error carried forwards. The most common error was to use  10–3 instead of  10–6 for the cm3 conversion to m3, unfortunately this error led to a molar mass of 0.1369 from which a molecular formula is impossible. Notwithstanding the success of obtaining the correct answers, most calculations were unclearly presented, as in Question 25 (a) (i). If any of your students’ results are not as expected, you may wish to consider one of our post-results services. For full information about the options available visit the OCR website. We send a weekly roundup to tell you about important updates. You can also sign up for your subject specific updates. If you haven’t already, sign up here. Attend one of our popular CPD courses to hear directly from a senior assessor or drop in to a Q&A session. Most of our courses are delivered live via an online platform, so you can attend from any location. Please find details for all our courses on the relevant subject page on our website or visit OCR professional development. ExamBuilder is the question builder platform for a range of our GCSE, A Level, Cambridge Nationals and Cambridge Technicals qualifications. Find out more. ExamBuilder is free for all OCR centres with an Interchange account and gives you unlimited users per centre. We need an Interchange username to validate the identity of your centre's first user account for ExamBuilder. If you do not have an Interchange account please contact your centre administrator (usually the Exams Officer) to request a username, or nominate an existing Interchange user in your department. Review students' exam performance with our free online results analysis tool. It is available for all GCSEs, AS and A Levels and Cambridge Nationals. It allows you to: • review and run analysis reports on exam performance • analyse results at question and/or topic level • compare your centre with OCR national averages • identify trends across the centre • facilitate effective planning and delivery of courses • identify areas of the curriculum where students excel or struggle • help pinpoint strengths and weaknesses of students and teaching departments. Find out more. Post-results services Keep up-to-date OCR Professional Development Signed up for ExamBuilder?

Notes from the mark scheme and report

• convert cm3 to m3 and °C to K
• M = 136.9–137 gives C4H9Br
• if 24000 cm3 mol–1 used instead of pV = nRT only last 2 marks available (M = 107, C2H5Br)

Skills: M0.0 Recognise and make use of appropriate units in calculations; M1.1 Use an appropriate number of significant figures; M2.2 Change the subject of an equation; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

Although most candidates gained some marks here, there were a significant number who did not gain all 4 marks. Some candidates need to practice reading burettes and recording their values – this can be checked during practical work. The most common errors included not recording values to 2 decimal places (especially if final number was 0), readings recorded the wrong way round in the table, and misreading 0.05 as 0.5. Most candidates were able to identify concordant results, but some still calculated the mean from all values. This led to an average that needed to be rounded to the nearest 0.05, as asked for in the question, resulting in them losing both the mean titre mark and mark for accuracy. Some had errors which lead to values which weren’t concordant – this should be a flag to students that they have made a mistake.

Notes from the mark scheme and report

• all six burette readings correct
• mean from concordant titres 27.30 and 27.40 only (27.35)
• all values to 2 d.p. ending in 0 or 5
• ALLOW ECF from incorrect readings but mean must use closest titres

Skills: M1.1 Use an appropriate number of significant figures; M1.2 Find arithmetic means; PAG2 Acid–base titration

Mark scheme

Examiner's comment

Most candidates managed to gain at least 1 mark for this question. The most common mark lost was for not multiplying by 10, having missed that only 25cm3 of 250 cm3 citric acid solution was used in the titration. Another mark that was often lost was for not dividing by 2 to find mass in 1 lime rather than the 2 used in experiment. Some candidates used 25cm3 to calculate their moles of NaOH rather than the titre value from (i). It is vital that candidates are given the opportunity to practice more complex multi-step calculations of this type, with modelling given for lower-attaining candidates. Identifying which information goes with each reactant is vital. All steps in the calculation should be separate and clearly labelled to help avoid confusion. Encourage candidates to keep full values in their calculators to avoid intermediate step rounding. When writing down intermediate values, ideally write down the full calculator value or where this is not possible the value must be given to at least 3 significant figures (correctly rounded). It is helpful to the examiner to know if calculator values are used and this could be indicated by using truncated answers followed by … , for example 0.00729… for n(A). The exemplar shows a response where each step of the calculation is shown clearly. The candidate has also used pictures to aid them, recognising that the NaOH is in the burette and lime juice in the conical flask. All values are identified and there is no intermediate rounding. All 5 marks are given here.

Notes from the mark scheme and report

• 1:3 ratio citric acid:NaOH
• ×10 scale-up from 25.0 to 250 cm3
• ÷2 because two limes used
• ALLOW ECF from titre in (b)(i) and through all steps

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.2 Use ratios, fractions and percentages; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; PAG2 Acid–base titration

Mark scheme

Examiner's comment

Most candidates gained this mark. The most common incorrect response was pH 7 with a few giving a pH value of less than 7.

Notes from the mark scheme and report

• any value in range 8–14

Mark scheme

Examiner's comment

Most candidates recognised that they needed to draw a tangent. Candidates need to be instructed to draw their line clearly and make sure that it covers several large squares on the graph – preferably extending to the axes. The range given for the calculated rate from the gradient of the tangent was generous, however some did not realise that the y-axis started at 91.0 so incorrectly calculated their gradient. Some candidates used interpolation so gained no marks.

Notes from the mark scheme and report

• tangent drawn at t = 50 s (±10 s)
• DO NOT ALLOW interpolation/direct reading - must be a gradient
• range 5.7 × 10^-3 to 6.9 × 10^-3
• IGNORE units and sign
• ALLOW ECF from incorrect tangent
• M3 section of the Maths Skills handbook contains useful information on use of graphs in chemistry, including M3.5 on drawing and using the slope of the tangent to a curve as a measure of a rate of change: https://www.ocr.org.uk/Images/295468-chemistry-mathematical- skills-handbook.pdf

Skills: M3.1 Translate information between graphical, numerical and algebraic forms; M3.5 Draw and use the slope of a tangent to a curve as a measure of rate of change; PAG9 Rates of reaction – continuous monitoring

Mark scheme

Examiner's comment

This question was well answered with most candidates scoring both marks. Common errors seen included: inverse Kc expression, final answer not to 3 significant figures, adding rather than multiplying concentrations, calculator error ([H2]2 instead of [H2]3) or transcription errors (e.g. using 0.11 rather than 0.111).

Notes from the mark scheme and report

• 3 SF (24.1)
• ALLOW ECF to 3 SF only from inverted expression (0.0414)
• DO NOT ALLOW + signs in expression
• IGNORE units

Skills: M0.2 Use ratios, fractions and percentages; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

Despite the structures being drawn out with displayed formula, a significant number of candidates did not correctly count all the bonds. Common errors included missing C-C from ethanol or O-H bonds in water. Most candidates set out their working clearly, so it was easy to identify any errors in calculation steps and award ECF appropriately. Some also had the calculation the wrong way round (made – broken) missing the essential idea that energy is needed to break bonds and released to make bonds, as well as link to endothermic changes being positive and exothermic changes being negative.

Notes from the mark scheme and report

• common errors for 2 marks: –198 (cycle reversed), –149 (missed C–C), –2586 (missed 6 O–H from 3H2O)
• DO NOT ALLOW – sign on final answer
• It is important to evaluate candidates understanding of enthalpy changes and build on prior knowledge from KS4. We have produced a transition guide for KS4 to KS5 on enthalpy: https://www.ocr.org.uk/Images/361305-enthalpy-changes-ks4-ks5.pdf. This includes a checkpoint task which is very useful for checking understanding of this topic, including the difference between endothermic and exothermic reaction, drawing profile diagrams, and calculating enthalpy changes from bond enthalpies.

Skills: M0.2 Use ratios, fractions and percentages; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

Despite the question asking for calculations to be included, many candidates didn’t include them and so lost both marks. Some gained one mark as recognised that 5.1 has 100% atom economy but either didn't or incorrectly calculated for 5.2 (30% was seen frequently). Care needs to be taken with rounding of final values.

Notes from the mark scheme and report

• Reaction 5.1 atom economy 136.9/217.8 × 100 = 62.9%
• Reaction 5.2 is 100% (addition, one product)

Skills: M0.2 Use ratios, fractions and percentages

Mark scheme

Examiner's comment

The most common errors were omitting the yield or inverting the yield, as given on mark scheme, resulting in 2 marks. Clear working was vital here to help marks to be given even if the final answer was incorrect. Many candidates did not gain the final mark due to incorrect significant figures. As with other multi-step calculations, rounding of intermediate values could also cause marks to be lost.

Notes from the mark scheme and report

• common errors for 2 marks: 5.41 g (no % yield), 3.62 g (yield inverted)
• 3 SF required (8.07
• 8.06 accepted from intermediate rounding)

Skills: M0.2 Use ratios, fractions and percentages; M1.1 Use an appropriate number of significant figures; M2.2 Change the subject of an equation; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

Many candidates were able to explain clearly the second order dependence on [NO] but many found it more challenging to identify and explain the first order relationship for [CO], where the candidate had to consider the second order dependence on [NO] to work out the effect on rate for the change in [CO]. Most gave a rate equation of the correct format based on the orders they had stated, and worked out a value for k. Although some did not determine the correct units, there was an encouraging number who managed this either from memory or by cancelling. The most common error was not including s-1

Notes from the mark scheme and report

• NO second order from expts 1 and 2
• CO first order needs both concentrations changing (expts 2 and 3: rate × 16)
• k = 3.37 × 10^6 dm6 mol–2 s–1
• common errors 4.65 × 10^9 mol–3 dm9 s–1 (CO second order) and 2446 mol–1 dm3 s–1 (CO zero order) score 4 with ECF

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.2 Use ratios, fractions and percentages; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; PAG10 Rates of reaction – initial rates

Mark scheme

Examiner's comment

Many candidates made some attempt at this unusual twist on a Q = mcΔT calculation, but only a minority of candidates produced a fully correct answer. A temperature of 84.9 oC was seen much more often than 52.4 oC. The calculation of n(AgNO3) was performed well. Common errors were not dividing the moles by 2, not recognising the need to calculate the energy released per mole of AgNO3, confusing the mass of water with the mass of the silver nitrate solid or of using an energy in kJ alongside a value of c based on J.

Notes from the mark scheme and report

• n(AgNO3) = 0.0400
• energy = 678 × 0.0400/2 = 13.56 kJ (2 mol AgNO3 per ∆H)
• ∆T = 32.4
• add 20.0 gives 52.4 °C to 3 sig figs
• common errors 32.4 (∆T only) and 84.9 (no ÷2) score 3

Skills: M0.0 Recognise and make use of appropriate units in calculations; M1.1 Use an appropriate number of significant figures; M2.2 Change the subject of an equation; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; PAG3 Enthalpy determination

Mark scheme

Examiner's comment

This calculation proved difficult. Many candidates did not deduce the correct cycle. Of those who did, a significant majority did not appreciate the stoichiometry and/or use the correct signs from the cycle. Many students scored 1 mark for this question which was gained as an error carried forward, providing all the values had been used in the enthalpy cycle. -90 was a common error.

Notes from the mark scheme and report

• 4∆fH(NO) = –1172 + 1716 – 184 = +360
• ÷4 = +90
• –90 scores 1

Skills: M2.2 Change the subject of an equation; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M2.4 Solve algebraic equations

Mark scheme

Examiner's comment

Almost all candidates had a good attempt at this calculation, with many gaining full marks. Most were able to calculate the entropy change. Almost all could reproduce the equation for free energy. Of those who did not get the correct final answer, the most common error was not converting the entropy value into kJ and/or the temperature to K There were also a few candidates who did not manipulate the equation correctly.

Notes from the mark scheme and report

• ∆S = +284 J K–1 mol–1 converted to 0.284 kJ
• T = 298 K
• ∆H = ∆G + T∆S = –2587
• common errors –2664.9 (25 °C), 81960 (∆S in J), –2756.6 (wrong sign of ∆S) score 2

Skills: M0.0 Recognise and make use of appropriate units in calculations; M2.2 Change the subject of an equation; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

This question proved demanding for candidates, with many simply quoting a molar value taken from the graph and converting into standard form. Of those who recognised the need to use the Avogadro constant, a few tried to multiply it by the molar ionisation enthalpy. For those who worked out the correct answer, several lost marks due to the requirement of 2 significant figures.

Notes from the mark scheme and report

• about 500/6.02 × 10^23 = 8.3 × 10^–22 kJ
• must be 2 sig figs AND standard form

Skills: M0.1 Recognise and use expressions in decimal and ordinary form; M0.3 Estimate results; M1.1 Use an appropriate number of significant figures

Mark scheme

Examiner's comment

Almost all candidates calculated the titres correctly, but a significant number were penalised for recording 22.70 as 22.7. A significant number also used this result and the trial titre to derive their mean value. There were also very small numbers of candidates who used all 4, or all of titres 1, 2 and 3, to calculate their mean.

Notes from the mark scheme and report

• titres to 2 d.p. (22.70 not 22.7)
• mean from concordant 22.45 and 22.55 = 22.50
• DO NOT ALLOW use of trial
• Links to the legacy coursework tasks and PAG practice question sets can be found on OCR Interchange. Exam hints for students can be found at: https://www.ocr.org.uk/Images/592305-exam-hints-for-students.pdf.

Skills: M1.2 Find arithmetic means; PAG2 Acid–base titration

Mark scheme

Examiner's comment

Candidates made good progress with this calculation, many gaining 4 or 5 marks, including error carried forward from incorrect titres. Common errors included, in various combinations: not converting the final answer into mg, not converting volume to dm3, missed ratio, multiplying the moles in 10cm3 acid by 10 instead of 25 and/or wrong Mr. Responses to Question 19 (a) (ii) often featured rows of figures and random sums without a single word about what the figures, or sums, were set to calculate. Candidates should remember to provide written indications of what it is they are working out – presenting the calculations without any annotations can make it harder for error carried forward marks to be given if there is an error in their calculation. The exemplar here shows a good use of annotation. There is a clear indication of the mathematical process so that the error carried forward is easily identified and the candidate gains the method marks.

Notes from the mark scheme and report

• ÷2 (dibasic)
• × 25 to scale 10.0 cm3 to 250.0 cm3
• × 118
• ÷4 tablets and convert to mg = 498
• common errors 996 (no ÷2), 19.9 (no × 25)
• 503 from titre 22.725 scores 5 by ECF

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.2 Use ratios, fractions and percentages; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; PAG2 Acid–base titration

Mark scheme

Examiner's comment

Most candidates were able to derive a value for Ka from pKa and calculate the number of moles of glycolic acid and potassium hydroxide reacting. Less were successful in determining the moles or concentrations present in the buffer solution causing many to get the common error of 3.57. Many candidates tried to calculate pH for the weak acid, without considering changes to concentrations or the buffering effect. Clarity of working is essential and in questions such as this, candidates are advised to include word descriptions of what they are calculating, even if it is abbreviations such as ‘n’ for number of moles.

Notes from the mark scheme and report

• n(HA) left = 0.0450 – 0.0250 = 0.0200
• n(A–) = 0.0250
• Ka = 10^–3.83
• pH = 3.93 (2 d.p.)
• common errors 3.57 (HA not reduced) scores 3, 3.75 (concentrations from question) scores 2

Skills: M0.4 Use calculators to find and use power, exponential and logarithmic functions; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M2.5 Use logarithms in relation to quantities that range over several orders of magnitude

Mark scheme

Examiner's comment

Although few candidates got the correct final answer, however almost all achieved some marks from this calculation through error carried forward, with marks spread across the available range. Almost all candidates were able to find the concentrations of hydrogen and hence hydroxide ions. A few candidates successfully used p[OH-] method. Most were able to calculate the moles of hydroxide ions in 250cm3. Many then did not realise the need to half this number to find the moles of barium, and/or used the Mr for barium hydroxide instead of barium oxide.

Notes from the mark scheme and report

• ÷4 for 250 cm3
• ÷2 for Ba(OH)2
• × 153.3 (BaO) = 2.53 g
• common errors 5.05 (no ÷2) and 2.82 (Mr of Ba(OH)2) score 4, 5.65 scores 3

Skills: M0.2 Use ratios, fractions and percentages; M0.4 Use calculators to find and use power, exponential and logarithmic functions; M1.1 Use an appropriate number of significant figures; M2.5 Use logarithms in relation to quantities that range over several orders of magnitude

Mark scheme

Examiner's comment

This proved a difficult question for most candidates. Most were able to correctly calculate the moles of fertiliser by converting kg to g. The next step was to deduce that 5 moles of calcium carbonate would be required for each mole of Z and multiply by 5, rather than the common error of dividing by 5. Few candidates were able to multiply by 100/95, to account for the impurities in limestone, with many multiplying by 95/100.

Notes from the mark scheme and report

• n(Z) = 1500/1080.5
• × 5 for CaCO3
• × 100.1 (DO NOT ALLOW 100)
• ÷ 0.950 = 731 g
• common errors 146 (no × 5), 660 (× 0.95), 29.3 (÷5) score 2

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.2 Use ratios, fractions and percentages; M1.1 Use an appropriate number of significant figures

Mark scheme

Examiner's comment

A variety of responses were seen in this demanding question. In general candidates adopted one of two approaches. The most common was to multiply the Mr of the repeat unit by 200 and then add the mass of H and OH at each end of the polymer. The other approach used the Mr of the monomer by 200 and then subtract the mass of the 199 water molecules removed in the polymerisation. Many candidates were successful with the first step of their approach, but the best responses included the second step taking into account the Mr of water. A significant number of candidates used an incorrect value for the Mr.

Notes from the mark scheme and report

• 28418: 400 × 89 minus 399 × 18 (water lost)
• 28400 scores 1 mark (end groups ignored)

Skills: M0.2 Use ratios, fractions and percentages

Mark scheme

Examiner's comment

Question 1 (b) presented candidates with a mass spectrum of two elements in an alloy. This novel question was answered well, and the introductory part (i) provided candidates with a hint of how to approach the harder part (ii). Almost all candidates analysed the percentages to show that the brass sample contains 66% Cu and 34% Zn, the 34 being required in Question 1 (b) (ii).

Notes from the mark scheme and report

• Cu 66% (45.64 + 20.36) and Zn 34%

Skills: M0.2 Use ratios, fractions and percentages; M3.1 Translate information between graphical, numerical and algebraic forms

Mark scheme

Examiner's comment

Most candidates produced a stock expression for a relative atomic mass calculation with the numerator comprising the four zinc isotopes with their respective percentage abundances, equating to 2224.28. The second mark required candidates to realise that the numerator needed to be divided by the total zinc percentage abundance of 34% to produce a value of 65.42. Less successful responses divided by 100% to obtain a value of 22.24. Almost all candidates gave their answers to the required two decimal places. The second mark proved to be a very good discriminator, with some candidates usually obtaining both marks with less successful responses still able to obtain the first mark for the numerator.

Notes from the mark scheme and report

• common errors: 22.24 (÷100 instead of ÷34, 1 mark), 64.23 (all 6 isotopes, no marks)
• divide by Zn % (34)
• 2 d.p. required (65.42)
• ECF from (b)(i)

Skills: M0.2 Use ratios, fractions and percentages; M1.2 Find arithmetic means

Mark scheme

Examiner's comment

This question required candidates to apply their knowledge and understanding of bond angles and electron pair repulsion of NH3 and NH4+ to amino acid salts. The best candidates rose to this challenge and secured all 3 marks for correct bond angles and explanations in terms of the numbers of bonded and lone pairs around the N atoms. Overall, candidates found this question quite difficult. Many different bond angles were predicted, with 120º being the commonest incorrect H–N–H bond angle in H2NCH2COONa. The explanation for 120º was in terms of three bonding pairs and no lone pairs. 104.5º was also seen, presumably relating H2N to H2O. The 109.5º bond angle was correct more often, as was its explanation in terms of 4 bonding pairs. Many successful responses showed working on diagrams in which bonded and lone pairs had been included. This strategy will have helped candidates in their conclusions.

Notes from the mark scheme and report

• 107° for NH2 (3 bonded pairs AND 1 lone pair) and 109.5° for NH3+ (4 bonded pairs)
• 'bonded' essential
• IGNORE electron regions/density
• IGNORE shapes

Skills: M4.1 Use angles and shapes in regular 2-D and 3-D structures

Mark scheme

Examiner's comment

Most candidates calculated the amounts of Cl2 and Ca(ClO)2 correctly as 17.5 mol and 8.75 mol respectively. Only the least successful did not use the equation’s stoichiometry to halve 17.5 to 8.75. For the final step in the calculation, candidates needed to convert 4.00 m3 into 4000 dm3 and to then determine the concentration to an appropriate number of significant figures and standard form. As all the data had been provided to 3 significant figures, the final answer was also required to 3 significant figures, as 2.09  10–3 mol dm–3. Answers such as 2.2  10–3, 2.1875  10–3 and 2.19  10–6 and 0.00219 illustrate errors in these areas.

Notes from the mark scheme and report

• common errors: 4.38 × 10^-3 (no ÷2) 2 marks
• wrong power of ten 2 marks
• 2.2 × 10^-3 2 marks (not appropriate SF)
• m3 to dm3 conversion
• 3 SF AND standard form

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.1 Recognise and use expressions in decimal and ordinary form; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

This question took the novel context introduced in 5b a stage further by considering the energy released during the combustion of this fuel. Most candidates were able to obtain some credit, and many obtained the correct energy of 341,850 kJ. The commonest error was for candidates to assume that the 8 kg mixture would contain 4 kg of octane and 4 kg of ethanol, rather than an equal moles of each. Such an approach could still be partly given marks by ECF, provided that the method was sound and clear.

Notes from the mark scheme and report

• 160 g per mole pair so 50 mol of each
• 341 850 kJ
• common error 310 800 (equal masses 4 kg each rather than equal moles) scores 2
• IGNORE sign

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.2 Use ratios, fractions and percentages; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

Most candidates added two asterisks to the diagram of vitamin C for the possible chiral centres. More successful responses usually realised that two chiral centres would give rise to 22 = 4 optical isomers, with 2 optical isomers being the commonest error.

Notes from the mark scheme and report

• 4 (two chiral centres)

Skills: M4.2 Visualise and represent 2-D and 3-D forms including 2-D representations of 3-D objects

Mark scheme

Examiner's comment

More successful responses determined that x = 15 and y = 31. The key to success here was to subtract the formula of vitamin C from the formula of the ester and to add the formula of water: C22H38O7 – C6H8O6 + H2O → C16H32O2 → C15H31COOH. A significant number of candidates did obtain one of these values, with 15/16 and 29/30/32 being common incorrect answers. Omitting part(s) of sequence above would result in these incorrect numbers.

Notes from the mark scheme and report

• x = 15, y = 31 (C22H38O7 + H2O – C6H8O6 = C16H32O2)

Skills: M2.4 Solve algebraic equations

Mark scheme

Examiner's comment

Most candidates had learnt a standard method for calculating the pH of a weak acid. with the correct answer of 2.16 being seen on very many scripts. Success required conversion of 0.150 moles of vitamin C in 250 cm3 to its concentration as 0.600 mol–3, calculation of [H+] using [H+] =  (Ka  [HA]) and determination of pH using –log[H+]. Common errors usually resulted from one mistake and could still be rewarded with 2 of the available 3 marks. Examples are shown below. • pH = 4.32 No square root of (7.94  10–5  0.600) • pH = 2.46 No conversion of 0.150 mol to 0.600 mol dm–3  (7.94  10–5  0.150) • pH = 2.76 ÷ 4 instead of  4 for concentration  (7.94  10–5  0.0375) pH calculations are common in A Level Chemistry There are four different types, and it is essential that the standard methods for determination of [H+] in the calculations are learnt: • pH of strong acids • pH of weak acids, using Ka and [HA] • pH of strong bases, using Kw and [OH–] • pH of buffers, using Ka and [HA]/[A–] It is extremely likely that at least one of these types of pH calculation will feature in at least one of the A Level units. This question was a standard titration calculation, set in a practical context. As with Question 6 (c) (i), this assessed one of the important principles encountered in A Level Chemistry. Success required three main stages: • Calculation of the amount, in mol, of I2 used in the titration of 25.0 cm3 of the diluted orange juice. • Determination of the amount in mol of vitamin C in the 250 cm3 solution (effectively scaling by 10) • Use of this value with the molar mass of vitamin C to determination of the mass of vitamin C, in mg, in the 150 cm3 serving of the orange juice. Most candidates calculated the amount of I2 (and vitamin C) in the titre as 2.16  10–5 mol. The next scaling stage by 10 to 2.16  10–4 mol introduced a problem. Many candidates were distracted by the ‘150 cm3’ serving of orange juice and they scaled further by a factor of 150/250 to give 1.296  10–4 mol. Candidates scaling corrected by 10 usually used the vitamin C molar mass of 176 g mol–1 to determine the correct mass of 38 mg in the orange juice serving. The common incorrect mass of 22.8 mg resulted from the extra scaling by 150/250 described above. This could still be given 3 of the available 4 marks as only one error had been made. Some candidates worked out an incorrect molar mass for vitamin C, with 174 often seen. Provided that this value was used with a correct method, ECF could still be applied for the final mass. If any of your students’ results are not as expected, you may wish to consider one of our post-results services. For full information about the options available visit the OCR website. We send a weekly roundup to tell you about important updates. You can also sign up for your subject specific updates. If you haven’t already, sign up here. Attend one of our popular CPD courses to hear directly from a senior assessor or drop in to a Q&A session. Most of our courses are delivered live via an online platform, so you can attend from any location. Please find details for all our courses on the relevant subject page on our website or visit OCR professional development. ExamBuilder is the question builder platform for a range of our GCSE, A Level, Cambridge Nationals and Cambridge Technicals qualifications. Find out more. ExamBuilder is free for all OCR centres with an Interchange account and gives you unlimited users per centre. We need an Interchange username to validate the identity of your centre's first user account for ExamBuilder. If you do not have an Interchange account please contact your centre administrator (usually the Exams Officer) to request a username, or nominate an existing Interchange user in your department. Review students' exam performance with our free online results analysis tool. It is available for all GCSEs, AS and A Levels and Cambridge Nationals. It allows you to: • review and run analysis reports on exam performance • analyse results at question and/or topic level • compare your centre with OCR national averages • identify trends across the centre • facilitate effective planning and delivery of courses • identify areas of the curriculum where students excel or struggle • help pinpoint strengths and weaknesses of students and teaching departments. Find out more. Post-results services Keep up-to-date OCR Professional Development Signed up for ExamBuilder?

Notes from the mark scheme and report

• common errors: 4.32 no square root
• 2.46 no ×4 (0.150 used)
• 2.76 ÷4
• concentration 0.600 mol dm–3
• 2 d.p. required (2.16)
• pH ECF only if [H+] derived from Ka and concentration

Skills: M0.4 Use calculators to find and use power, exponential and logarithmic functions; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M2.5 Use logarithms in relation to quantities that range over several orders of magnitude

Mark scheme

Notes from the mark scheme and report

• common errors: 22.81 mg scaling by 150/250 (3 marks)
• 42.24 mg using 25.0 instead of 22.50 (3 marks)
• 63.36 mg scaling by 250/150
• 1:1 ratio, ×10 for 250 cm3 flask which contains the whole 150 cm3 serving
• answer 38 mg

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.2 Use ratios, fractions and percentages; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Notes from the mark scheme and report

• answer to 3 DP (32.094)
• 1 mark ECF if %s used with wrong isotopes once or decimal places transposed for one %

Skills: M0.2 Use ratios, fractions and percentages; M1.1 Use an appropriate number of significant figures

Mark scheme

Notes from the mark scheme and report

• minimum 3 SF
• common errors: no ×3 for F atoms (2.408 × 10^22), no n(halothane) (1.806 × 10^24)
• ALLOW ECF from incorrect n

Skills: M0.1 Recognise and use expressions in decimal and ordinary form; M0.2 Use ratios, fractions and percentages; M1.1 Use an appropriate number of significant figures

Mark scheme

Notes from the mark scheme and report

• common errors (2 marks): wrong sign for 206 (570), wrong sign for 1077 (1150), missing 206 (501)
• DO NOT ALLOW – sign on answer
• ALLOW ECF

Skills: M2.2 Change the subject of an equation; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M2.4 Solve algebraic equations

Mark scheme

Notes from the mark scheme and report

• n(Sc) = 0.27/45 or n(O) = 0.144/16
• Sc2O3
• NO ECF

Skills: M0.2 Use ratios, fractions and percentages; PAG1 Moles determination

Mark scheme

Notes from the mark scheme and report

• n = pV/RT with V in m3 (9.39 × 10–3) and T = 293 K
• mass in g (1.69 × 10^3)
• M = 32 so O2
• ALLOW ECF throughout including a reasonable gas matching calculated M

Skills: M0.0 Recognise and make use of appropriate units in calculations; M2.2 Change the subject of an equation; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Notes from the mark scheme and report

• mean of concordant titres only (18.20 and 18.30) = 18.25

Skills: M1.2 Find arithmetic means; PAG2 Acid–base titration

Mark scheme

Notes from the mark scheme and report

• 2 × 0.05 = 0.1 uncertainty in a titre
• ALLOW ECF from (b)(i) or use of other titre values

Skills: M0.2 Use ratios, fractions and percentages; M1.3 Identify uncertainties in measurements and use simple techniques to determine uncertainty when data are combined; PAG2 Acid–base titration

Mark scheme

Notes from the mark scheme and report

• 1:2 ratio acid:NaOH
• scale × 10 to 250 cm3
• M = 132 then subtract 90 for two COOH and divide by 14
• whole number required
• ALLOW ECF throughout and from (b)(iii)

Skills: M0.2 Use ratios, fractions and percentages; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M2.4 Solve algebraic equations; PAG2 Acid–base titration

Mark scheme

Notes from the mark scheme and report

• rearrange Kc: 15.4 × 0.57 × 0.40^2 = 1.4
• ALLOW ECF from incorrect expression in (a)

Skills: M2.2 Change the subject of an equation; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Notes from the mark scheme and report

• silicon AND large increase between 4th and 5th IE
• 5th electron removed from a shell closer to nucleus/4 outer-shell electrons

Mark scheme

Notes from the mark scheme and report

• smooth best-fit curve
• tangent at t = 50 s
• gradient about 0.44 cm3 s–1
• DO NOT ALLOW interpolation (direct reading from graph)
• ALLOW ECF from incorrectly drawn tangent

Skills: M3.2 Plot two variables from experimental or other data; M3.5 Draw and use the slope of a tangent to a curve as a measure of rate of change; PAG9 Rates of reaction – continuous monitoring

Mark scheme

Notes from the mark scheme and report

• n = 0.26/87.6 then × 137.3 = 0.41 g
• 2 decimal places required

Skills: M0.2 Use ratios, fractions and percentages; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

Question 16(c)(iii): This was an enthalpy solution/hydration cycle.

Notes from the mark scheme and report

• –6 kJ mol–1
• 2 × hydration of F– needed
• 1 mark only

Skills: M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M2.4 Solve algebraic equations

Mark scheme

Examiner's comment

Question 18(d): This was a percentage by mass calculation.

Notes from the mark scheme and report

• n(Cu2+) = n(S2O3 2–) (2Cu2+ : I2 : 2S2O3 2–)
• 3 SF (1.35%)
• ECF from ratio in (b)
• ECF needs a calculated mass of Cu

Skills: M0.2 Use ratios, fractions and percentages; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; PAG2 Acid–base titration

Mark scheme

Notes from the mark scheme and report

• 0.40 – 1.23 = –0.83 V

Skills: M2.2 Change the subject of an equation

Mark scheme

Notes from the mark scheme and report

• T = ∆H/∆S = 776 K
• 3 SF required

Skills: M1.1 Use an appropriate number of significant figures; M2.2 Change the subject of an equation

Mark scheme

Examiner's comment

Question 20(c): This question asked the candidate to calculate Kp.

Notes from the mark scheme and report

• equilibrium moles 0.4, 0.9 and 1.2 (total 2.5)
• partial pressures = mole fraction × 1.21
• ALLOW 20.6 from 3 SF partial pressures
• 25.0 (moles used, no partial pressures) scores 3

Skills: M0.2 Use ratios, fractions and percentages; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

Question 21(b)(i): This question asked the candidate to determine the activation energy and to use the graph to calculate the temperature that an investigation was carried out..

Notes from the mark scheme and report

• gradient –5700 to –6100
• Ea = –gradient × 8.314
• convert to kJ and 3 SF
• missing × 10–3 on axis scores 2
• DO NOT ALLOW negative Ea

Skills: M0.0 Recognise and make use of appropriate units in calculations; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M3.3 Determine the slope and intercept of a linear graph; M3.4 Calculate rate of change from a graph showing a linear relationship

Mark scheme

Notes from the mark scheme and report

• ln k = –2.59
• read 1/T = 3.10 × 10–3 (ALLOW 3.09–3.11)
• T = 49.6 °C after converting from K

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.4 Use calculators to find and use power, exponential and logarithmic functions; M2.5 Use logarithms in relation to quantities that range over several orders of magnitude; M3.1 Translate information between graphical, numerical and algebraic forms

Mark scheme

Notes from the mark scheme and report

• 2 DP required (13.15)
• ALLOW pOH method
• ECF only if pH > 7

Skills: M0.4 Use calculators to find and use power, exponential and logarithmic functions; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M2.5 Use logarithms in relation to quantities that range over several orders of magnitude

Mark scheme

Examiner's comment

Question 22(b)(ii): This question asked the candidates to calculate the enthalpy change and then to deduce the value for the enthalpy change of neutralisation.

Notes from the mark scheme and report

• mass = 80.0 g (total volume)
• divide by n(H2SO4) (DO NOT ALLOW n(NaOH))
• ∆neutH = ∆H1/2
• – sign required

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.2 Use ratios, fractions and percentages; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; PAG3 Enthalpy determination

Mark scheme

Notes from the mark scheme and report

• mg to g
• 1.17 × 10^21, 3 SF essential
• ALLOW ECF from formula in (c)(i)

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.1 Recognise and use expressions in decimal and ordinary form; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

Q1(a) Partial pressure calculation Q1(d) Formula determination from percentage composition by mass Q2(b) Converting mass and volume into moles Q3(b) Unstructured titration calculation Q5(a)(ii) Ideal gas calculation Q5(b)(i) Calculation of the pH of a weak acid • Produced clear and concise responses for a Level of Response question based on a familiar context i.e., Q4(b) rates by continuous monitoring and graphical analysis. • Had difficulty when answering questions requiring application of knowledge and understanding on questions set in novel and unfamiliar contexts, e.g. Q1(a): Candidates were assessed on their understanding of partial pressure.

Notes from the mark scheme and report

• n(CO2) = 2.5 and n(O2) = 3.75
• mole fraction 0.4 × 50.0 = 20 atm
• ALLOW ECF from incorrect total moles only

Skills: M0.2 Use ratios, fractions and percentages; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

Q1(b): Candidates were required to apply their understanding of Kc to a novel context requiring some mathematical awareness.

Notes from the mark scheme and report

• square brackets required in Kc
• common errors: ÷2 instead of √ (2.00 × 10–4), inverse Kc expression (2.5)

Skills: M2.2 Change the subject of an equation; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M2.4 Solve algebraic equations

Mark scheme

Examiner's comment

Q1(a) Partial pressure calculation Q1(d) Formula determination from percentage composition by mass Q2(b) Converting mass and volume into moles Q3(b) Unstructured titration calculation Q5(a)(ii) Ideal gas calculation Q5(b)(i) Calculation of the pH of a weak acid • Produced clear and concise responses for a Level of Response question based on a familiar context i.e., Q4(b) rates by continuous monitoring and graphical analysis. • Had difficulty when answering questions requiring application of knowledge and understanding on questions set in novel and unfamiliar contexts, e.g. Q2(b): Candidates were required to analyse the experimental results to obtain a conclusion. Q2(d): Candidates found it much more difficult to evaluate the unfamiliar scenarios in Q2(c) and Q2(d) than the stock mole calculation in Q2(b).

Notes from the mark scheme and report

• n(H2) = 152/24000 = 6.33 × 10–3
• n(Eu) = 0.988/152 = 6.50 × 10–3
• ratio 1:1 so Equation 2
• conclusion only if the two amounts are approximately equal
• ALLOW ECF

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.2 Use ratios, fractions and percentages; PAG1 Moles determination

Mark scheme

Examiner's comment

Q1(a) Partial pressure calculation Q1(d) Formula determination from percentage composition by mass Q2(b) Converting mass and volume into moles Q3(b) Unstructured titration calculation Q5(a)(ii) Ideal gas calculation Q5(b)(i) Calculation of the pH of a weak acid • Produced clear and concise responses for a Level of Response question based on a familiar context i.e., Q4(b) rates by continuous monitoring and graphical analysis. • Had difficulty when answering questions requiring application of knowledge and understanding on questions set in novel and unfamiliar contexts, e.g. Q5(a)(ii): As with earlier quantitative questions, candidates are very comfortable in tackling stock mole- based calculations, here involving the ideal gas equation.

Notes from the mark scheme and report

• V = 16.0 × 10–3 m3 and T = 290 K
• n(N2) = 0.796
• × 2/3 for n(NaN3)
• 34.5 g to 3 SF
• common errors: 2/3 omitted (51.7/51.8) 4 marks, inverted gas equation (54.4) 4 marks

Skills: M0.0 Recognise and make use of appropriate units in calculations; M1.1 Use an appropriate number of significant figures; M2.2 Change the subject of an equation; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

Q1(a) Partial pressure calculation Q1(d) Formula determination from percentage composition by mass Q2(b) Converting mass and volume into moles Q3(b) Unstructured titration calculation Q5(a)(ii) Ideal gas calculation Q5(b)(i) Calculation of the pH of a weak acid • Produced clear and concise responses for a Level of Response question based on a familiar context i.e., Q4(b) rates by continuous monitoring and graphical analysis. • Had difficulty when answering questions requiring application of knowledge and understanding on questions set in novel and unfamiliar contexts, e.g. Q5(b)(i): Most candidates found this pH calculation easy and most obtained a pH of 2.75 for both marks.

Notes from the mark scheme and report

• pH = 2.75 to 2 DP
• common errors: no square root (pH 5.50) 1 mark, square root of only one term (3.20, 5.05) 1 mark

Skills: M0.4 Use calculators to find and use power, exponential and logarithmic functions; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M2.5 Use logarithms in relation to quantities that range over several orders of magnitude

Mark scheme

Examiner's comment

21(a) Most candidates could recall the number of electrons in the first four shells as being 2, 8, 18 and 32. However, many made errors for the third and fourth shell. Common incorrect responses included 2,8,18,18 and 2,8,8,8, perhaps linking back to GCSE coverage. 2,6,10,14 was also seen, confusing shells with subshells. 21(b) Almost all candidates were aware that different isotopes had different numbers of neutrons. For the similarities, ‘the same number of protons’ was usually identified, but the same number electrons was omitted by many candidates. This suggested that candidates were concentrating on the nucleus and did not read the question closely enough which asked for ‘atomic’ structures. 21(c)(ii) This was a high demand question and only the highest scoring answers identified that the peak at 72 was caused by a chlorine molecule containing the chlorine-35 and chlorine-37 isotopes shown in the supplied mass spectrum. Very few candidates went on to predict further peaks with m/z values of 70 (from 35Cl35Cl) and 74 (from 37Cl37Cl). There was little pattern in the incorrect responses, suggestion that many were guesses.

Notes from the mark scheme and report

• 2 DP required (35.48)
• ECF for 1 mark if % used with wrong isotope once or one % transposed and Ar between 35 and 37

Skills: M0.2 Use ratios, fractions and percentages; M1.2 Find arithmetic means; M3.1 Translate information between graphical, numerical and algebraic forms

Mark scheme

Examiner's comment

22(a)(i) Candidates answered this part well, showing good knowledge and understanding of electron configuration. When incorrect, it was usually errors with the number of electrons in the 3d and 4p subshells. 22a(ii) Most candidates were able to construct a correct balanced equation for this unfamiliar equation. The commonest error was use of Br instead of Br2 in the equation. 22(b) Structure and bonding continues to be a difficult concept for many candidates. This part did discriminate extremely well, with high-attaining candidates identifying that the compound would have a giant ionic lattice with fixed ions in the solid state which become mobile in the liquid state to conduct electricity. It was common for ‘ionic’ or ‘giant’ to be omitted in the name of the lattice. Most candidates used the ideal gas equation, which was usually rearranged correctly, with substituted values for p, V, R and T being added. A common mistake was for the volume in cm3 to be multiplied by 10–3 rather than 10–6 for the necessary conversion into m3. Low-attaining candidates often did not convert 100 ºC into 373 K. Unfortunately, candidates making an error in unit conversions then obtained an unrealistic molar mass for A and were then unable to derive a possible molecular formula. Well-prepared candidates carried out the calculation well with clear working. It was very encouraging to see the number of candidates able to predict the molecular formula of A as BrF5 from the calculated molar mass of 174.6 g mol–1. There were, however a significant number of incorrect responses, suggesting that some candidates found this type of question challenging. Some candidates used the molar gas volume at room temperature and pressure of 24.0 dm3 mol–1. Two of the five marks were still available for this response.

Notes from the mark scheme and report

• V converted to m3 and T to 373 K
• use of 24 dm3 gives only the final 2 marks by ECF (135, BrF3)
• ALLOW ECF throughout

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.1 Recognise and use expressions in decimal and ordinary form; M2.2 Change the subject of an equation; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

Unlike Q23(a), all intermediate calculations gave values to 3 significant figures. As with Q23(a), many candidates produced largely unreferenced numbers. 23(d) Candidates often find questions linked to practical work challenging.

Notes from the mark scheme and report

• common error 0.227 (no ×2 for OH–) scores 2
• 3 SF required
• ALLOW use of 171

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.2 Use ratios, fractions and percentages; M1.1 Use an appropriate number of significant figures

Mark scheme

Examiner's comment

23(a) This question required candidates to calculate the moles of Ba(OH)2 in 100 cm3, to scale this value up to 1 dm3 for the concentration of Ba(OH)2 and finally to multiply this value by 2 to obtain the concentration of OH– ions of 0.454 mol dm–3. This part discriminated extremely well. Lower-attaining candidates would often calculate an incorrect molar mass of Ba(OH)2 and scale up by either 10 or 2. The commonest incorrect answer seen was 0.227, half of the correct answer by omitting the ×2 stage. Most candidates did supply their final answer to the required 3 significant figures. As with all calculations, candidates should round once, at the end of the calculation. Rounding of intermediate values can lead to significant rounding errors. Sometimes, intermediate values were rounded to 2 significant figures when the final value required 3. This use of maths will almost certainly lead to a rounding error in the final answer. Candidates should show clear working so that credit can be given for such responses by applying error carried forward. Many candidates produced largely unreferenced numbers. 23(b) Many candidates seemed to be unfamiliar with the term ‘concordant titres’. Acceptable responses were ‘within 0.1 cm3 and within 0.05 cm3. Common incorrect answers for this important term included ‘the closest titres’, ‘similar titres’, ‘repeated titres’ and ‘within 0.5 cm3’. 23(c) Many candidates were able to calculate the amount of HNO3 in the titration as 4.28 × 10–3 mol. Most candidates were credited for the amount of Ba(OH)2 as 2.14 × 10–3 mol, half the calculated amount of HNO3. Candidates then need to scale up this value by 1000/25 to obtain the concentration as 0.0856 mol dm–3. Unlike Q23(a), all intermediate calculations gave values to 3 significant figures. Candidates found this stock titration calculation easier than the more unfamiliar 23(a), with many scoring all 3 marks. Discrimination was extremely good, but about a third of candidates did not receive any marks. Candidates should be encouraged to practise stock titration calculations as part of their preparation for the examinations. Candidates should show clear working so that credit can be given for such responses by applying error carried forward. As with Q23(a), many candidates produced largely unreferenced numbers. 23(d) Candidates often find questions linked to practical work challenging. The two reaction routes involving Group 2 reactions are drawn directly from the specification: the reaction of barium with water, and the reaction of barium with oxygen, followed by water. Most candidates found this difficult, and the one-step route using barium and water was seen far more often than the two-step oxygen then water approach. Only the highest-attaining candidates were able to suggest two routes, with scores of 3/4 or 4/4 marks being comparatively rare. As with 21(c)(ii) and 25c(iii), candidates had difficulty in interpreting provided information. OCR support Our new practical question packs, split up by PAG, provide additional opportunities for students to practise questions relating to practical work. They are available on Interchange: https://interchange.ocr.org.uk/Downloads/H432_H433_Chemistry_PAG_ Practice_Question_Sets.zip

Notes from the mark scheme and report

• DO NOT ALLOW 4.3 × 10–3 for n(HNO3) but ECF for remaining marks
• 1:2 ratio Ba(OH)2:HNO3

Skills: M0.2 Use ratios, fractions and percentages; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; PAG2 Acid–base titration

Mark scheme

Examiner's comment

24(a) Most candidates were able to make some progress with the enthalpy calculation. Most recognised that the provided ∆cH values had to be multiplied by the balancing numbers in the equation. Most candidates processed the data on either side of the equation correctly and subtracted the combined enthalpy change from one side of the equation from the other, although the subtraction was sometimes the wrong way round. This resulted in answers of +92, or –92 kJ mol–1, with the exothermic value being correct. An enthalpy change of –92 kJ mol–1 was the commonest final answer, with comparatively few candidates recognising that the enthalpy change of formation of NH3 is for the formation of 1 mol of NH3 and not 2 mol as in the equation. 24(b) This question was well-answered showing that most candidates were well-acquainted with the Boltzmann distribution. Some candidates used one diagram to show the effect of increasing the temperature and of using a catalyst. Other candidates showed two curves, one for temperature, the other for the presence of a catalyst. Either approach could produce a clear response. Most candidates knew that the activation energy was lower with a catalyst that without. A significant number of candidates limited their explanations to ‘successful collisions’ without referring to more molecules exceeding the lower activation energy in the presence of catalyst or at a higher temperature. The best responses secured all five marks from a well-drawn and annotated graph supported by a brief explanation. Candidate should make full use of annotated diagrams to communicate their answers.

Notes from the mark scheme and report

• common errors: –92 (no ÷2) scores 2
• +46 (wrong subtraction) scores 2
• +92 scores 1
• no ×3 for H2 or no ×2 for NH3 lose a mark each

Skills: M0.2 Use ratios, fractions and percentages; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M2.4 Solve algebraic equations

Mark scheme

Examiner's comment

1(a) A considerable number of candidates did not score the mark here despite knowing that it was an oxidation reaction. They either omitted to give the number of electrons lost or gave their justification using oxidation numbers. This question highlights the importance of reading the question carefully and answering the question asked, not one that may have been seen previously. 1(b)(i) Most candidates were able to make some progress with this calculation. Common errors included omitting to multiply by two to find the number of moles of HCl or not multiplying their volume in dm3 by 1000 to give an answer in cm3. 1(b)(ii) More candidates struggled with this part of calculation. A significant number of candidates used the number of moles rounded to 3 SF but then gave a final answer to 4 SF, giving an answer of 78.48 cm3. This was given credit as correct calculation had been carried out for one mark. A couple of candidates correctly used the ideal gas equation and were given credit for this, but made it more complicated than necessary for one mark. 1(b)(iii) Very few candidates managed to gain marks here. Most did not recognise that HCl was the limiting reagent in the reaction and suggested that a greater volume of H2 would be produced, as same mass of Mg would give a greater number of moles. 1(c)* Very few candidates managed to score full marks for this question. Even the highest-attaining candidates struggled with writing balanced chemical equations. The most successful candidates used ionic equations with state symbols in their responses. A large proportion of candidates gave unnecessary details such as testing for CO2 using limewater or the colours of other silver halide precipitates. The best responses broken down their response to cover each test in turn, giving clear and concise details for each.

Notes from the mark scheme and report

• 1:2 ratio Ni:HCl
• 3 SF required (43.6)

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.2 Use ratios, fractions and percentages; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

1(a) A considerable number of candidates did not score the mark here despite knowing that it was an oxidation reaction. They either omitted to give the number of electrons lost or gave their justification using oxidation numbers. This question highlights the importance of reading the question carefully and answering the question asked, not one that may have been seen previously. 1(b)(i) Most candidates were able to make some progress with this calculation. Common errors included omitting to multiply by two to find the number of moles of HCl or not multiplying their volume in dm3 by 1000 to give an answer in cm3. 1(b)(ii) More candidates struggled with this part of calculation. A significant number of candidates used the number of moles rounded to 3 SF but then gave a final answer to 4 SF, giving an answer of 78.48 cm3. This was given credit as correct calculation had been carried out for one mark. A couple of candidates correctly used the ideal gas equation and were given credit for this, but made it more complicated than necessary for one mark. 1(b)(iii) Very few candidates managed to gain marks here. Most did not recognise that HCl was the limiting reagent in the reaction and suggested that a greater volume of H2 would be produced, as same mass of Mg would give a greater number of moles. 1(c)* Very few candidates managed to score full marks for this question. Even the highest-attaining candidates struggled with writing balanced chemical equations. The most successful candidates used ionic equations with state symbols in their responses. A large proportion of candidates gave unnecessary details such as testing for CO2 using limewater or the colours of other silver halide precipitates. The best responses broken down their response to cover each test in turn, giving clear and concise details for each.

Notes from the mark scheme and report

• n(Ni) × 24000
• ALLOW ECF from incorrect n(Ni) in (b)(i)

Skills: M0.0 Recognise and make use of appropriate units in calculations; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

2 (a) Most candidates managed to gain at least one mark for this question having learnt the definition. Some lost a mark for stating one mole of substance gives one mole of product, clearly confusing their definitions. 2(b) Nearly all candidates managed to gain at least one mark for correctly calculating the moles of cyclohexane burnt and the majority also used q = mc∆T. However, very few candidates managed to gain all 4 marks, most frequently losing the final mark for SF and sign. Other common errors included using the mass of cyclohexane in calculation of q, using Gas Constant R rather than specific heat capacity of water, or not using ∆T but using 293 K instead. 2(c)(i) A significant number of candidates struggled to correctly calculate the % uncertainties for the apparatus given or did not back up their claims with any calculations. Even the highest-attaining candidates often did not account for two measurements for temperature, or tried to calculate the % uncertainty for each temperature measurement taken. 2(c)(ii) Most candidates gained at least one mark here from one of the three acceptable responses. There was some evidence of candidates being confused here by the use of average bond enthalpies to find values of enthalpy of combustion. Others discussed human errors, need for repeat measurements or precision of equipment, demonstrating a lack of understanding with regard to the language used in scientific measurement. 2(c)(iii) This was a challenging question. Some candidates gained one mark for reduced accuracy due to increased heat loss or water evaporation. The responses given demonstrated that candidates often do not understand the factors that influence the accuracy in scientific experiments. Some indicated that repeating results would improve accuracy. Another common response was that a longer time would allow for complete combustion, failing to understand that degree of combustion is due to the availability of oxygen, not the time taken.

Notes from the mark scheme and report

• –2680: 3 SF AND negative sign (+2680 scores 3)
• mass of water 200 g not mass of fuel in q = mc∆T
• common error –7.02 scores 3
• ALLOW ECF

Skills: M0.0 Recognise and make use of appropriate units in calculations; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; PAG3 Enthalpy determination

Mark scheme

Examiner's comment

4(a) This was a challenging example of predicting bond angles. Candidates often counted four pairs of electrons around the central atom, so based their bond angles on a tetrahedral arrangement (deducting 2.5 from 109.5o), losing this mark. Explanations were not always clear, but most candidates gained credit for describing electron pairs repelling. Many candidates state that they subtracted 2.5o rather than explaining why this is done in terms of electron pairs repelling. 4(b)(i) A wide range of responses, including addition signs, 2x, curved brackets, inverse. Candidates must take care to write the formulae correctly: Cl rather than Cl2 was seen a number of times. 4(b)(ii) Most recognised that the value for NO was 2 x Cl2 but some struggled to explain why. Answers needed to show clear evidence that they had taken into account the stoichiometry of the equation. Some gave responses such as using moles and volume to calculate or to put values into the Kc expression. 4(b)(iii) Those candidates that were able to give the correct expression in 4(b)(i) were often able to obtain the correct value here, demonstrating a good understanding of this topic. Some forgot to take the square root so only obtained one mark. 4(b)(iv) Candidates need to be well versed in how to tackle this type of question. Some, despite explaining what would happen, forgot to then state what would happen to the concentration of NO. Some said that the reaction would shift to the right but did not state that the reaction was endothermic. A small number of candidates thought that the reaction was exothermic, despite the + sign by the reaction, and some discussed other factors such as rate or changing pressure.

Notes from the mark scheme and report

• rearrange for [NOCl]2 then take the square root (1.3 scores 1)
• ALLOW ECF from inverted Kc expression

Skills: M2.2 Change the subject of an equation; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M2.4 Solve algebraic equations

Mark scheme

Examiner's comment

A number of arithmetic errors were seen for calculations, particularly Question 5(d)(i) where Mr values were added up incorrectly or final % did not equal the values given.

Notes from the mark scheme and report

• 88/190.9 × 100 = 46.1
• ALLOW ECF from incorrect Mr values

Skills: M0.2 Use ratios, fractions and percentages; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Notes from the mark scheme and report

• 3 ions per formula unit
• 3 sig figs required (1.71 × 10^22)
• common error 5.69 × 10^21 (no × 3) scores 2

Skills: M0.1 Recognise and use expressions in decimal and ordinary form; M0.2 Use ratios, fractions and percentages; M1.1 Use an appropriate number of significant figures

Mark scheme

Examiner's comment

OCR support Further support can be found in: The AS delivery guide ‘Theme: Identifying Unknowns’ https://www.ocr.org.uk/Images/208563-identifying-unknowns.pdf ‘Transition elements’: https://www.ocr.org.uk/qualifications/as-a-level-gce- chemistry-a-h032-h432-from-2015/delivery-guide/module-cam05-module-5- physical-chemistry-and-transition-elements/delivery-guide-cadg016- transition-elements The colours of the transition metal ions and complexes can be reviewed at: https://www.ocr.org.uk/Images/598991-colours-of-inorganic-ions-and- complexes-poster-.pdf Themes in candidate responses Calculation based questions Questions 16(d)(ii) and 16(d)(iii) This Born Haber cycle combined the traditional cycle with the enthalpy solution/hydration cycle.

Notes from the mark scheme and report

• –346.5 (–693 ÷ 2)
• 2 × atomisation and 2 × electron affinity of Br
• one sign or factor-of-2 error scores 1 (e.g. –693, +346.5, –509, –290.5)
• two errors score 0

Skills: M2.2 Change the subject of an equation; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M2.4 Solve algebraic equations

Mark scheme

Notes from the mark scheme and report

• Mg2+(g) + 2Br–(g) → MgBr2(s) with state symbols
• –2433
• ALLOW ECF from (d)(ii)

Skills: M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M2.4 Solve algebraic equations

Mark scheme

Notes from the mark scheme and report

• ∆S = –222 J K–1 mol–1 (2 × H2)
• convert to kJ or ∆H to J
• ∆G = +25.55 kJ mol–1 with units
• not feasible because ∆G > 0

Skills: M0.0 Recognise and make use of appropriate units in calculations; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

This was an unfamiliar expression linking ∆G and Kp. Many candidates correctly calculated this number. Some candidates calculated ln Kp as 10.0 but then put a – in front for the e calculation. The question required the answer to 3 significant figures. Higher-attaining candidates were able to work out the units as any pressure unit -2 .

Notes from the mark scheme and report

• ln Kp = –∆G/RT = 10.01
• Kp = 2.22 × 10^4 (3 sig figs)
• units atm–2 (or Pa–2 etc.)
• common errors 22400 (R = 8.31) and 4.50 × 10^–5 (sign) score 1

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.4 Use calculators to find and use power, exponential and logarithmic functions; M1.1 Use an appropriate number of significant figures; M2.2 Change the subject of an equation; M2.5 Use logarithms in relation to quantities that range over several orders of magnitude

Mark scheme

Examiner's comment

This question required the candidate to calculate the original concentration of ethanoate ions in the buffer. Higher-attaining students gained full credit. Most students calculated the concentration in the buffer solution but did not factor for the original solution. Lower-attaining candidates often scored the first two marking points but did not use the buffer equation.

Notes from the mark scheme and report

• n(CH3COOH) = 0.0816 in 1 dm3
• × 1000/600 = 2.4 × 10^–2
• common errors 8.7 × 10^–3 (600/1000 inverted), 3.6 × 10^–6 (ratio inverted), 3.6 × 10^–2 (no volumes)
• ALLOW Henderson–Hasselbalch

Skills: M0.2 Use ratios, fractions and percentages; M0.4 Use calculators to find and use power, exponential and logarithmic functions; M2.2 Change the subject of an equation; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Notes from the mark scheme and report

• 2H+ + ½O2 + 2e– → H2O
• E = 1.34 + (–0.11) = +1.23 V

Skills: M2.2 Change the subject of an equation

Mark scheme

Examiner's comment

Very few candidates were given full marks. Higher-attaining students calculated one half life in range but very few could come up with a second half life as the graph did not allow another successive half life to be obtained. Higher-attaining candidates chose alternative half lives from the data given.

Notes from the mark scheme and report

• two half-lives shown on graph (102–110 min) AND statement that half-life is constant
• values on the graph.
• Further guidance on rates of reaction can be found at: https://www.ocr.org.uk/Images/371956-experiments-on-rates-of-reaction.doc

Skills: M3.1 Translate information between graphical, numerical and algebraic forms; PAG9 Rates of reaction – continuous monitoring

Mark scheme

Examiner's comment

This question required the candidate to draw a line of best fit and then draw a tangent at t=0. Many candidates did not draw a line of best fit, and many did not get a tangent in the acceptable range. Very few candidates processed the gradient by using the correct subtraction on the y axis (scale was from 1 to 5) or by using the 10-3 on the axis label. Electrode potentials

Notes from the mark scheme and report

• tangent at t = 0 (gradient 2.9–4.0 × 10^–5) OR k = ln 2/t½ then rate = k × 5.0 × 10^–3
• note 10^–3 on y-axis
• ALLOW ECF from t½

Skills: M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M3.1 Translate information between graphical, numerical and algebraic forms; M3.5 Draw and use the slope of a tangent to a curve as a measure of rate of change; PAG9 Rates of reaction – continuous monitoring

Mark scheme

Examiner's comment

This question asked the candidate to calculate Kc. Higher-attaining students tended to gain full marks. Some candidates made full use of tables (e.g. RICE: Reaction, Initial concentration, Change in concentration, Equilibrium concentration) which allowed for credit to be given through error carried forward. Some candidates did not use 0.03 as the change, and lower-attaining candidates did not use water in the Kc expression. Candidates should remember to provide written indications of what it is they are working out – presenting the calculations without any annotations can make it harder for error carried forward marks to be given if there is an error in their calculation.

Notes from the mark scheme and report

• initial moles 0.3 and 0.2
• 0.17 mol reacted so 0.13, 0.03, 0.17, 0.17 at equilibrium
• Kc = 7.4
• V cancels but expression must be correct

Skills: M0.2 Use ratios, fractions and percentages; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Notes from the mark scheme and report

• n = 0.209/29.0 (given molar volume, not 24.0)
• Mr = 174.8
• BrF5
• formula dependent on Mr

Skills: M0.2 Use ratios, fractions and percentages; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Notes from the mark scheme and report

• Ni : S : N = 1 : 4 : 8 so x = 4
• N count 2 + x + y = 8 so y = 2
• ALLOW ECF for y from incorrect x

Skills: M0.2 Use ratios, fractions and percentages; M2.4 Solve algebraic equations

Mark scheme

Notes from the mark scheme and report

• convert 167 mg to g
• M = 294
• 22.4, 22 or 23 all score 3
• ALLOW ECF and alternative methods

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.1 Recognise and use expressions in decimal and ordinary form; M0.2 Use ratios, fractions and percentages

Mark scheme

Examiner's comment

3(a)(i) Most candidates analysed the raw practical results supplied to correctly calculate the energy change as 10,241 J using q = mc∆T, and the amount of propane burnt as 0.0550 mol. Candidates usually determined the energy change per mole of propane as 1862 kJ. The question asked for the enthalpy change of combustion to 3 significant figures, which gives –1860 kJ mol–1. The mark for this final value was often not given because either the – sign was omitted or the answer was not quoted to the required 3 significant figures. Candidates are advised to read the requirements of each question more closely to eliminate such errors. A double-check on significant figure requirements is recommended before moving on to the next question. Lower-attaining candidates often extracted the wrong data from the results table. Instead of the mass of water (100 cm3), the mass of propane burnt (0.242 g) was often used in the q = mc∆T calculation. Candidates are advised that the thermometer measures the temperature change of the material in which it is placed (the water). 3(a)(ii) One mark was available for identifying two reasons for the difference in enthalpy change. It was expected that candidates would respond with ‘heat loss’ and ‘incomplete combustion’. Most candidates suggested heat loss but the other reason often suggested was equipment error, e.g. the thermometer did not reading correctly’, ‘the balance gave incorrect readings’. Candidates are advised that responses of faulty apparatus are unlikely to be creditworthy. 3(b) This question was assessed by LoR. Level 3 candidates calculated the enthalpy change for Equation 3.1 correctly as –2054 kJ mol–1. They linked 4 × ∆vapH with the enthalpy change for Equation 3.1 in an energy cycle and calculated the enthalpy change of combustion as –2054 – (4 × 40.65) = –2216.6 kJ mol–1. Most Level 3 candidates drew out bonds in the molecules involved in Equation 3.1, which allowed them to visually count the bonds broken and bonds made. This strategy is recommended to all candidates for any calculation based on bond enthalpies and is much more likely to generate the correct enthalpy change than visualising the relevant bonds solely from the equation. Level 2 candidates often derived the correct enthalpy change for Equation 3.1 but then tried to link this value with ∆vapH instead of 4 × ∆vapH. Other Level 2 candidates calculated ∆H correctly for either the bonds broken or bonds made and combine this value with 4 × ∆vapH. A common error was the omission of the 2 C–C bonds in propane, use of 3 C–C, instead of 2 C–C, bonds, or use of 4, rather than 8 O–H bonds. These errors could have been avoided by drawing out the bonds as described above. Level 1 candidates typically attempted to use the bond enthalpies to calculate energies for bonds broken and bonds made but made several errors. They then combined this value with ∆vapH. This question discriminated extremely well across different abilities.

Notes from the mark scheme and report

• common errors (2 marks by ECF): m = 0.242 g (mass of propane) used in mc∆T, ∆T taken as 297.5 K
• –1862 scores 2 (not 3 SF)
• DO NOT ALLOW c = 4.2
• – sign AND 3 SF required

Skills: M0.0 Recognise and make use of appropriate units in calculations; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; PAG3 Enthalpy determination

Mark scheme

Examiner's comment

4(a)(i) This question required candidates to write an overall equation and half equations for oxidation and reduction. Many candidates made errors within one or more equations. The overall equation was often written without state symbols, despite the question instruction ‘with state symbols’. The oxidation half equation was more likely to be correct than the reduction half equation, which often used Cl instead of H+. When H+ was used, the half equation was often unbalanced or electrons had been omitted. It is recommended that candidates carefully use the chemical information in the question. 4(a)(ii) Most candidates identified that all the metal would have reacted when it had all disappeared and that gas bubbles from the reaction would have stopped. Some responses did not emphasise that these observations would have stopped and this prevented credit being given. 4(a)(iii) The ionic equation for neutralisation of an acid with an alkali was well known and this question was answered correctly by most candidates. 4(a)(iv) Candidates were presented with information about a back titration, a technique that they would be unlikely to have encountered during their course. The question stem to 4a(iv) suggested a three-step strategy. Many candidates followed this guidance and were credited with many of the available marks. Marks were given for a correct method (by error carried forward) even if there was an error or omission in the multi-step calculation. This emphasises the importance of clear working. Most candidates determined the correct mean titre of 27.25 cm3. A few candidates did take the mean of all three titres rather than the closest. Most calculated that 8.72 × 10–3 mol of NaOH reacted with the same number of moles of HCl in the titration and then scaled up the HCl by a factor of 10 to 8.72 × 10–2 mol in the 250 cm3 volumetric flask. These steps are standard for many titration calculations and gave a route to three of the six available marks. The more difficult back titration steps then followed and the higher-attaining candidates recognised the need to subtract this amount of HCl from the original amount of HCl used to react with metal M. These candidates then divide this value by two to find the moles of M that reacted (from the 1 : 2 stoichiometry of M : HCl). The correct calculation then gave a relative atomic mass of M as 112 and its identity as cadmium. It was common for candidates to omit the division by two and to arrive at a relative atomic mass of 56 for iron. The mark scheme shows the variety of metals that candidates identified from their calculations, the errors made, and the error carried forward marks that resulted. Many lower-attaining candidates did not follow the 3 steps in the stem, using only the original amount of HCl and ignoring the titration. This approach was not credited with marks. A large range of marks was seen, and the question discriminated extremely well. 4(b)(i) This straightforward mole calculation required candidates to determine the mass of CO2 as 2.75 g from the table of results and to divide this value by the molar mass of CO2 to arrive at 0.0625 mol. Most candidates were successful. As with other questions providing raw experimental results, lower-attaining candidates often chose the wrong data from the supplied results. 4(b)(ii) Many candidates used their value from 4b(i) with the mass of X2CO3 from the results table to obtain a value for the molar mass (233.13 from 14.57 ÷ 0.0625). Higher-attaining candidates obtained a value for the atomic mass of X (86.56 from 233.13) and identified X as the Group 1 metal rubidium. Some candidates did not recognise the significance of the formula of X2CO3 and identified X instead as strontium (Ar = 87.6). Lower-attaining candidates often used incorrect data from the supplied information and made little progress with this calculation. 4(c)(i) Most candidates suggested that the mass of the flask and contents could be reweighed until the mass no longer changed. An alternative creditworthy suggestion was to attach a gas syringe and to measure the gas volume until it no longer changed. Vague suggestions such as ‘leave for longer’ were not credited with marks. 4(c)(ii) Candidates found this question challenging. Success required candidates to appreciate that some CO2 dissolving would result in less mass of CO2 being recorded as lost, meaning that less moles of CO2 had apparently been formed and so less moles of X2CO3 had reacted. Using n = m/M, this would show that the calculated molar mass of X2CO3 appears to be greater. Only the highest-attaining candidates successfully navigated this question.

Notes from the mark scheme and report

• common errors: mean of all 3 titres 27.35 (5 marks)
• no ÷2 for n(M) gives Fe
• no subtraction from 0.210 gives Tb
• no ×10 scale-up gives Zn
• omitting titration calculation scores zero
• DO NOT ALLOW Ga or Sc (3+ ions only)
• mean from concordant 27.30 and 27.20

Skills: M0.2 Use ratios, fractions and percentages; M1.2 Find arithmetic means; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; PAG2 Acid–base titration

Mark scheme

Examiner's comment

4(a)(i) This question required candidates to write an overall equation and half equations for oxidation and reduction. Many candidates made errors within one or more equations. The overall equation was often written without state symbols, despite the question instruction ‘with state symbols’. The oxidation half equation was more likely to be correct than the reduction half equation, which often used Cl instead of H+. When H+ was used, the half equation was often unbalanced or electrons had been omitted. It is recommended that candidates carefully use the chemical information in the question. 4(a)(ii) Most candidates identified that all the metal would have reacted when it had all disappeared and that gas bubbles from the reaction would have stopped. Some responses did not emphasise that these observations would have stopped and this prevented credit being given. 4(a)(iii) The ionic equation for neutralisation of an acid with an alkali was well known and this question was answered correctly by most candidates. 4(a)(iv) Candidates were presented with information about a back titration, a technique that they would be unlikely to have encountered during their course. The question stem to 4a(iv) suggested a three-step strategy. Many candidates followed this guidance and were credited with many of the available marks. Marks were given for a correct method (by error carried forward) even if there was an error or omission in the multi-step calculation. This emphasises the importance of clear working. Most candidates determined the correct mean titre of 27.25 cm3. A few candidates did take the mean of all three titres rather than the closest. Most calculated that 8.72 × 10–3 mol of NaOH reacted with the same number of moles of HCl in the titration and then scaled up the HCl by a factor of 10 to 8.72 × 10–2 mol in the 250 cm3 volumetric flask. These steps are standard for many titration calculations and gave a route to three of the six available marks. The more difficult back titration steps then followed and the higher-attaining candidates recognised the need to subtract this amount of HCl from the original amount of HCl used to react with metal M. These candidates then divide this value by two to find the moles of M that reacted (from the 1 : 2 stoichiometry of M : HCl). The correct calculation then gave a relative atomic mass of M as 112 and its identity as cadmium. It was common for candidates to omit the division by two and to arrive at a relative atomic mass of 56 for iron. The mark scheme shows the variety of metals that candidates identified from their calculations, the errors made, and the error carried forward marks that resulted. Many lower-attaining candidates did not follow the 3 steps in the stem, using only the original amount of HCl and ignoring the titration. This approach was not credited with marks. A large range of marks was seen, and the question discriminated extremely well. 4(b)(i) This straightforward mole calculation required candidates to determine the mass of CO2 as 2.75 g from the table of results and to divide this value by the molar mass of CO2 to arrive at 0.0625 mol. Most candidates were successful. As with other questions providing raw experimental results, lower-attaining candidates often chose the wrong data from the supplied results. 4(b)(ii) Many candidates used their value from 4b(i) with the mass of X2CO3 from the results table to obtain a value for the molar mass (233.13 from 14.57 ÷ 0.0625). Higher-attaining candidates obtained a value for the atomic mass of X (86.56 from 233.13) and identified X as the Group 1 metal rubidium. Some candidates did not recognise the significance of the formula of X2CO3 and identified X instead as strontium (Ar = 87.6). Lower-attaining candidates often used incorrect data from the supplied information and made little progress with this calculation. 4(c)(i) Most candidates suggested that the mass of the flask and contents could be reweighed until the mass no longer changed. An alternative creditworthy suggestion was to attach a gas syringe and to measure the gas volume until it no longer changed. Vague suggestions such as ‘leave for longer’ were not credited with marks. 4(c)(ii) Candidates found this question challenging. Success required candidates to appreciate that some CO2 dissolving would result in less mass of CO2 being recorded as lost, meaning that less moles of CO2 had apparently been formed and so less moles of X2CO3 had reacted. Using n = m/M, this would show that the calculated molar mass of X2CO3 appears to be greater. Only the highest-attaining candidates successfully navigated this question.

Notes from the mark scheme and report

• mass loss 2.75 g ÷ 44 = 0.0625 mol

Skills: M0.0 Recognise and make use of appropriate units in calculations; PAG1 Moles determination

Mark scheme

Examiner's comment

4(a)(i) This question required candidates to write an overall equation and half equations for oxidation and reduction. Many candidates made errors within one or more equations. The overall equation was often written without state symbols, despite the question instruction ‘with state symbols’. The oxidation half equation was more likely to be correct than the reduction half equation, which often used Cl instead of H+. When H+ was used, the half equation was often unbalanced or electrons had been omitted. It is recommended that candidates carefully use the chemical information in the question. 4(a)(ii) Most candidates identified that all the metal would have reacted when it had all disappeared and that gas bubbles from the reaction would have stopped. Some responses did not emphasise that these observations would have stopped and this prevented credit being given. 4(a)(iii) The ionic equation for neutralisation of an acid with an alkali was well known and this question was answered correctly by most candidates. 4(a)(iv) Candidates were presented with information about a back titration, a technique that they would be unlikely to have encountered during their course. The question stem to 4a(iv) suggested a three-step strategy. Many candidates followed this guidance and were credited with many of the available marks. Marks were given for a correct method (by error carried forward) even if there was an error or omission in the multi-step calculation. This emphasises the importance of clear working. Most candidates determined the correct mean titre of 27.25 cm3. A few candidates did take the mean of all three titres rather than the closest. Most calculated that 8.72 × 10–3 mol of NaOH reacted with the same number of moles of HCl in the titration and then scaled up the HCl by a factor of 10 to 8.72 × 10–2 mol in the 250 cm3 volumetric flask. These steps are standard for many titration calculations and gave a route to three of the six available marks. The more difficult back titration steps then followed and the higher-attaining candidates recognised the need to subtract this amount of HCl from the original amount of HCl used to react with metal M. These candidates then divide this value by two to find the moles of M that reacted (from the 1 : 2 stoichiometry of M : HCl). The correct calculation then gave a relative atomic mass of M as 112 and its identity as cadmium. It was common for candidates to omit the division by two and to arrive at a relative atomic mass of 56 for iron. The mark scheme shows the variety of metals that candidates identified from their calculations, the errors made, and the error carried forward marks that resulted. Many lower-attaining candidates did not follow the 3 steps in the stem, using only the original amount of HCl and ignoring the titration. This approach was not credited with marks. A large range of marks was seen, and the question discriminated extremely well. 4(b)(i) This straightforward mole calculation required candidates to determine the mass of CO2 as 2.75 g from the table of results and to divide this value by the molar mass of CO2 to arrive at 0.0625 mol. Most candidates were successful. As with other questions providing raw experimental results, lower-attaining candidates often chose the wrong data from the supplied results. 4(b)(ii) Many candidates used their value from 4b(i) with the mass of X2CO3 from the results table to obtain a value for the molar mass (233.13 from 14.57 ÷ 0.0625). Higher-attaining candidates obtained a value for the atomic mass of X (86.56 from 233.13) and identified X as the Group 1 metal rubidium. Some candidates did not recognise the significance of the formula of X2CO3 and identified X instead as strontium (Ar = 87.6). Lower-attaining candidates often used incorrect data from the supplied information and made little progress with this calculation. 4(c)(i) Most candidates suggested that the mass of the flask and contents could be reweighed until the mass no longer changed. An alternative creditworthy suggestion was to attach a gas syringe and to measure the gas volume until it no longer changed. Vague suggestions such as ‘leave for longer’ were not credited with marks. 4(c)(ii) Candidates found this question challenging. Success required candidates to appreciate that some CO2 dissolving would result in less mass of CO2 being recorded as lost, meaning that less moles of CO2 had apparently been formed and so less moles of X2CO3 had reacted. Using n = m/M, this would show that the calculated molar mass of X2CO3 appears to be greater. Only the highest-attaining candidates successfully navigated this question.

Notes from the mark scheme and report

• mass of X2CO3 = 14.57 g
• M = 233.12
• subtract 60 and halve gives 86.6 so Rb
• DO NOT ALLOW Sr (wrong carbonate formula)
• ECF for X only if Group 1 metal or Ag

Skills: M0.2 Use ratios, fractions and percentages; M2.2 Change the subject of an equation; PAG1 Moles determination

Mark scheme

Examiner's comment

Question 5 related an unfamiliar mathematical relationship with the y = mx + c equation for a straight line. Use of y = mx + c is a mathematical requirement of A level chemistry and candidates were expected to apply their understanding of the Arrhenius equation (part of the specification) which makes use of this straight-line equation in a novel context. Many candidates were unable to use this mathematical concept. In contrast, higher-attaining candidates were able to be credited with marks throughout question 5. 5(a) Candidates were expected to complete values for 1/T and ln Kp from supplied T and Kp values. Candidates were supplied with an example and this enabled most candidates to obtain both available marks. 5(b) Most candidates were aware that a decrease in Kp with increasing temperature signals that the forward reaction is exothermic. The question also asked for the effect on the equilibrium position. A significant number of candidates omitted this part of the question. Candidates are advised to check back to all the requirements in a question. 5(c) Candidates were required to plot a graph using their calculated values from 5(a). The axes for the graph had been provided. It was expected that the plotting of 5 points, with a best-fit straight line, would be straightforward. Many candidates plotted one or more points incorrectly, particularly the point at 1/T = 1.25 × 10–3. Candidates then needed to recognise that the gradient is equal to –∆H/R from Equation 5.1, to measure the gradient, and then to determine ∆H. Most candidates recognised that the gradient needed to be measured but its value was then not taken any further. The higher-attaining candidates correctly multiplied the gradient by R but did not always convert the calculated J mol–1 value into kJ mol–1, or to express their value to 3 significant figures, as required in the question. Significantly, nearly a third of candidates did not collect any of the four available marks. The question was an excellent discriminator. 5(d) Mathematically able candidates used the y = mx + c equation for a straight line with the supplied mathematical relationship (Equation 5.1) to identify the y intercept as ∆S/R. They then stated that ∆S could be determined by multiplying the value of the y intercept by R. Many candidates found the mathematical requirements of parts 5(c) and 5(d) difficult. Responses for 5(d) were often in terms of the gradient instead of ‘intercept’.

Notes from the mark scheme and report

• all points plotted with best-fit straight line
• gradient ±57 000 to ±63 000
• ∆H = –gradient × 8.314
• answer in kJ mol–1 to 3 SF with – sign (range –480 to –530)

Skills: M0.0 Recognise and make use of appropriate units in calculations; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M3.1 Translate information between graphical, numerical and algebraic forms; M3.2 Plot two variables from experimental or other data; M3.3 Determine the slope and intercept of a linear graph

Mark scheme

Examiner's comment

Most candidates identified that the first and third titres were concordant and calculated the mean titre that should be used as 24.25 cm3. About a third of candidates calculated the mean of all 3 titres as 24.10 or 24.12 cm3. Normal practice in titrations would be to select the closest titres.

Notes from the mark scheme and report

• use concordant titres only
• DO NOT ALLOW mean of all three titres (24.10/24.12)
• The Practical Skills handbook contains guidance on correct practice for recording titration results and calculating average titre values in Appendix 4: Measurements, which can be shared with students: https://www.ocr.org.uk/Images/208932-chemistry-practical-skills- handbook.pdf.

Skills: M1.2 Find arithmetic means; PAG2 Acid–base titration

Mark scheme

Examiner's comment

Many candidates were able to calculate the amount of Na2CO3 in the titration as 0.00375 mol although a common error was to calculate the amount of Na2CO3 in the 250 cm3 volumetric flask as 0.0375 mol. Most candidates were credited for the amount of HCl: twice their calculated amount of Na2CO3. Candidates then need to scale up this value by 1000/mean titre to obtain the concentration as 0.309 mol dm–3, and to quote the answer to 3 significant figures. Many candidates scaled up using 50.0, the burette volume, rather than their mean titre, resulting in a concentration 0.15 or 1.5 mol dm–3. A significant number also rounded their value to 2 rather than 3 significant figures. Candidates are advised to show clear working so that credit can be given for such responses by applying error carried forward. The working shown in this response is clear. Many candidates working was more jumbled, with unreferenced numbers common.

Notes from the mark scheme and report

• 3 sig figs required
• 1:2 ratio Na2CO3:HCl
• ALLOW ECF from mean titre (0.311 from mean of all three titres)

Skills: M0.2 Use ratios, fractions and percentages; M1.1 Use an appropriate number of significant figures; M2.2 Change the subject of an equation; PAG2 Acid–base titration

Mark scheme

Examiner's comment

This part discriminated well. Most candidates were able to write the correct expression for Kc as the starting point of the calculation. Candidates often got into a muddle in calculating Kc, perhaps due to issues inputting the calculation into their calculators. The question asked for ‘an appropriate number of significant figures and in standard form’. As the provided data was all to 3 significant figures, this also indicates the required number of significant figures in the answer. A calculated value to 2 significant figures was often seen (see the response); also 0.0286 rather than the standard form: 2.86 × 10–2. Some responses showed Kc inverted or added, rather than multiplying the two reactants in the denominator. Other candidates wrote the correct equilibrium expression but were then used 2.752, rather than 2.753, to obtain the standard form answer of 7.786 × 10–2 or 0.0786 with no standard form. Candidates are advised to check back through calculations to see if they have made any such errors.

Notes from the mark scheme and report

• 3 sig figs AND standard form required
• ECF only from inverted Kc (3.50 × 10^1)
• DO NOT ALLOW + in denominator
• IGNORE units

Skills: M0.1 Recognise and use expressions in decimal and ordinary form; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

Most candidates were able to make some progress with the enthalpy calculation. Most recognised that the provided ∆fH values had to be multiplied by the balancing numbers in the equation. Correct processing of these values with ∆H for the reaction discriminated well between candidates. The response shows a response that was given 2 out of the 3 available marks. The candidate has correctly calculated +360 kJ mol–1 from their enthalpy cycle but has not realised that this value is for 4 mol NH3 and must be divided by 4 to obtain the enthalpy change of formation of 1 mol of NH3. It was impressive that the correct answer of +90 kJ mol–1 was achieved by about one-third of candidates. Question 25 (a) (i), (ii) and (iii) Part 25(a)(i) discriminated extremely well and rewarded well-prepared candidates. Most candidates recognised that B is a tertiary alcohol and will not react with acidified dichromate. The structure from A was often shown as an aldehyde rather than a carboxylic acid. It was also common for candidates to replace the OH group of A with the carboxyl COOH group, gaining a carbon atom in the chain in the process. The ketone oxidation product from C proved to be easier. Part 25(a)(ii) proved to be difficult. Candidates need to be careful in identifying the longest carbon chain to derive the stem of an organic name. Many candidates thought that alcohol C was a branched propanol, with 1-methylpropan-1-ol being seen very often instead of the correct name of butan-2-ol. In part 25(a)(iii), less than half the candidates wrote a correctly-balanced equation for this reaction. Although 4CO2 and 5H2O were usually seen for the products, oxygen was usually seen as 6½O2, rather than 6O2. Candidates need to look very closely at the formula of the organic compound so that the O in C4H10O is accounted for in the balancing.

Notes from the mark scheme and report

• common errors: –90 (sign), +360 (missing ÷4)
• if ∆H of –908 not used only 1st mark available

Skills: M2.2 Change the subject of an equation; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M2.4 Solve algebraic equations

Mark scheme

Examiner's comment

This question discriminated very well. It was encouraging to see the number of candidates who used 32.5 dm3 mol–1 as the molar gas volume under the experimental conditions to obtain 2.40 × 10–3 mol of gas. Many though used 24.0 dm3 mol–1 for RTP and obtained 3.25 × 10–3 mol. Error carried forward allowed such candidates to still secure the final 2 of the 3 marks available. Lower attaining candidates were unsure where to start and tried to do anything with the number provided. The result was often n unusable number.

Notes from the mark scheme and report

• common error: using 24000 instead of 32500 cm3 mol–1 gives M = 196 and C4H6Cl4 (ECF)
• molecular formula must be derived from a calculated M

Skills: M0.0 Recognise and make use of appropriate units in calculations; M2.2 Change the subject of an equation; M2.4 Solve algebraic equations

Mark scheme

Examiner's comment

Algebra was used very well here and allowed most candidates to obtain at least one mark, with the majority obtaining 2 marks

Notes from the mark scheme and report

• third isotope is 10.2%
• DO NOT ALLOW Sr-86 with no working

Skills: M0.2 Use ratios, fractions and percentages; M1.2 Find arithmetic means; M2.2 Change the subject of an equation; M2.4 Solve algebraic equations

Mark scheme

Examiner's comment

Most students managed to gain some marks on this question. The most common error was rounding to 6, something they have been taught to do for water of crystallisation. This caused them to lose a mark as the question asked for two significant figures. Many rounded too early so a variety of responses were seen.

Notes from the mark scheme and report

• use final constant mass 17.58 g
• answer must be to 2 SF (5.8)
• 5.83 scores 2
• ALLOW ECF from n(SrCl2) and n(H2O)

Skills: M0.2 Use ratios, fractions and percentages; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; PAG1 Moles determination

Mark scheme

Examiner's comment

This question was answered well, with the majority of the candidates obtaining 3 marks.

Notes from the mark scheme and report

• 2:1 ratio HNO3:Mg(NO3)2
• 3 SF required (38.5)
• ALLOW ECF

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.2 Use ratios, fractions and percentages; M1.1 Use an appropriate number of significant figures; M2.2 Change the subject of an equation

Mark scheme

Examiner's comment

This was a well answered question, with the majority of candidates obtaining all 3 marks

Notes from the mark scheme and report

• HCl is limiting
• 120 cm3 (no ÷2) or 240 cm3 (×2) score 2
• ALLOW ECF

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.2 Use ratios, fractions and percentages; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

This was mathematically challenging, but candidates who did not arrive at the correct answer were able to get partial credit for showing working.

Notes from the mark scheme and report

• third isotope abundance 11.01%
• answer must be to 5 SF (25.982)
• ECF for transcription errors in first sum

Skills: M0.2 Use ratios, fractions and percentages; M1.1 Use an appropriate number of significant figures; M2.2 Change the subject of an equation; M2.4 Solve algebraic equations

Mark scheme

Examiner's comment

Most candidates scored both marks. Candidates tended to forget the mole ratio of K meant its values should be multiplied by 2 in two places during this calculation.

Notes from the mark scheme and report

• 1 mark for one error with sign or ×2: +2277, –1858 (no 2 × 419), –2188 (no 2 × 89), –2152.5 (249 ÷ 2), –2280 (rounded to 3 SF)

Skills: M2.2 Change the subject of an equation; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M2.4 Solve algebraic equations

Mark scheme

Examiner's comment

Many candidates coped well with this multi-step calculation. The common errors were: • determining the mass of C12H22FeO14 in two tablets instead of just one tablet as required in the question • determining the mass of Fe in a tablet instead of the mass of C12H22FeO14 • failing to convert from grams to milligrams

Notes from the mark scheme and report

• ×5 for Fe2+
• ×10 scaling to 250 cm3
• ÷2 for one tablet and convert g to mg
• common errors 602 mg (not halved) and 37.7 mg (55.8 used instead of 445.8) score 4
• 3 SF

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.2 Use ratios, fractions and percentages; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; PAG2 Acid–base titration

Mark scheme

Examiner's comment

The calculation was done well. Some candidates found it difficult to convert ΔS values from J to kJ or temperature values from °C to K, and this was frequently seen in responses. The final mark was often lost through vague phrasing. ‘Yes’ was an insufficient answer, as it does not identify the reaction as feasible or non-feasible. ‘Gibbs free energy is negative’ was also insufficient, as it should be identified that ΔG is negative.

Notes from the mark scheme and report

• ∆S = –391 J K–1 mol–1 converted to kJ
• T = 293 K
• common errors –1117 (T = 20), +113438 (J/kJ mix), –1008 (298 K) score 2 calculation marks
• feasible as ∆G < 0 (ECF)

Skills: M0.0 Recognise and make use of appropriate units in calculations; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

This calculation proved very difficult. Many candidates did not deduce the correct cycle. Of those who did, a significant majority did not appreciate the stoichiometry of the equation and did not double the ΔcH values or half the initially calculated value of ΔfH(H2S).

Notes from the mark scheme and report

• both ∆cH values ×2
• subtract ∆rH
• divide by 2
• +20(.1) scores 2
• ALLOW –20.1

Skills: M2.2 Change the subject of an equation; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M2.4 Solve algebraic equations

Mark scheme

Examiner's comment

The equilibrium expression was correct in most responses. Occasionally incorrect square brackets were used instead of round brackets. Candidates should be reminded that square brackets indicate a concentration and are not appropriate for use in Kp expressions.

Notes from the mark scheme and report

• DO NOT ALLOW square brackets
• atm–1 (ALLOW any pressure unit)
• ECF atm if Kp inverted

Skills: M0.0 Recognise and make use of appropriate units in calculations

Mark scheme

Examiner's comment

Candidates tend to find Kp calculations difficult and so may need guidance in developing a strategy to work their way through them. A typical approach is: • Calculation of initial moles present • Calculation of the change in moles present • Deduction of the number of moles present at equilibrium • Determination of total moles present at equilibrium • Calculation of mole fractions at equilibrium • Calculation of partial pressures at equilibrium • Inserting partial pressure vales into the Kp expression • Writing an answer to the required number of significant figures.

Notes from the mark scheme and report

• initial SO2 from 10.2/24.0 and O2 from 12.0/32.0
• O2 used = 0.350/2
• total moles 0.625
• common errors 1.45 (n(SO2) = 2 × n(O2)) and 15.1 (12/16 for O2) score 4
• 3 SF

Skills: M0.2 Use ratios, fractions and percentages; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

Most candidates could calculate the pH of a weak acid although a significant number gave the answer as 3.0, presumably confusing the demand for two decimal points with two significant figures.

Notes from the mark scheme and report

• 2 DP required (2.98)
• 5.97 (no square root) scores 1
• 0.92 or 5.15 (wrong Ka) score 1

Skills: M0.4 Use calculators to find and use power, exponential and logarithmic functions; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M2.5 Use logarithms in relation to quantities that range over several orders of magnitude

Mark scheme

Examiner's comment

This calculation proved difficult with once again, many figures and sums appearing with little indication as to their relevance. Responses to 20bi/ii often featured rows of figures and random sums without a single word about what the figures were or sums were set to calculate. Candidates should remember to provide written indications of what it is they’re working out – presenting the calculations without any annotations can make it harder for error carried forward marks to be given if there is an error in their calculation.

Notes from the mark scheme and report

• excess OH– = 0.0025 mol in total volume 70.0 cm3
• common errors 12.81 (45 cm3 volume) and 11.40 (moles not concentration) score 3
• ALLOW pOH method

Skills: M0.4 Use calculators to find and use power, exponential and logarithmic functions; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M2.5 Use logarithms in relation to quantities that range over several orders of magnitude

Mark scheme

Examiner's comment

Candidates are familiar with the calculation of Rf values and this question was answered very well. Although almost all candidates showed appropriate working as part of their response some candidates gave the final answer to only one significant figure and did not receive credit. Candidates were expected to give their answer to a number of significant figures appropriate to the measurements they had made, so a minimum of two significant figures was expected. The mark scheme allowed a response within a range, rather than a specific value, and most candidates scored in this part.

Notes from the mark scheme and report

• 0.33–0.35
• ALLOW 0.3

Skills: M3.1 Translate information between graphical, numerical and algebraic forms; PAG6 Synthesis of an organic solid

Mark scheme

Examiner's comment

A LEVEL CHEMISTRY A Examiners’ report H432/02 Summer 2019 series Version 1 H432 For first teaching in 2015 www.ocr.org.uk/science Contents Introduction .............................................................................................................................................. 4 Paper 2 series overview ........................................................................................................................... 5 Section A overview ................................................................................................................................... 6 Question 3 ............................................................................................................................................ 6 Question 4 ............................................................................................................................................ 6 Question 5 ............................................................................................................................................ 7 Question 10 .......................................................................................................................................... 8 Question 12 .......................................................................................................................................... 9 Question 13 ........................................................................................................................................ 10 Question 14 ........................................................................................................................................ 10 Question 15 ........................................................................................................................................ 11 Section B overview ................................................................................................................................. 12 Question 16 (a) (i) ............................................................................................................................... 12 Question 16 (a) (ii) .............................................................................................................................. 12 Question 16 (b) (i) ............................................................................................................................... 13 Question 16 (b) (ii) .............................................................................................................................. 14 Question 16 (b) (iv) ............................................................................................................................. 14 Question 16 (b) (v) .............................................................................................................................. 15 Question 17 (a) (i) ............................................................................................................................... 17 Question 17 (a) (ii) .............................................................................................................................. 19 Question 17 (b) (i) ............................................................................................................................... 20 Question 18 (a) (i) ............................................................................................................................... 21 Question 18 (a) (ii) .............................................................................................................................. 22 Question 18 (b) ................................................................................................................................... 23 Question 18 (c) ................................................................................................................................... 23 Question 18 (d) ................................................................................................................................... 24 Question 18 (e) (i) ............................................................................................................................... 24 Question 18 (e) (ii) .............................................................................................................................. 26 Question 19 (a) (i) ............................................................................................................................... 27 Question 19 (a) (ii) .............................................................................................................................. 29 Question 19 (b) (i) ............................................................................................................................... 30 Question 19 (b) (iii) ............................................................................................................................. 32 Question 19 (b) (iv) ............................................................................................................................. 34 Question 20 (a) (i) ............................................................................................................................... 36 Question 20 (a) (ii) .............................................................................................................................. 37 Question 20 (a) (iii) ............................................................................................................................. 37 Question 20 (b) (i) ............................................................................................................................... 38 Question 21 ........................................................................................................................................ 39 Introduction Our examiners’ reports are produced to offer constructive feedback on candidates’ performance in the examinations. Question 18 (d) Question 18 (e) (i) This question was marked using a level of response mark scheme.

Notes from the mark scheme and report

• 200 × 138 – 199 × 18 = 24018
• 27600 (no water removed) scores 1
• ALLOW ECF from incorrect Mr

Skills: M0.2 Use ratios, fractions and percentages

Mark scheme

Examiner's comment

This question tested an understanding of ‘partial pressure’ as a concept. Most candidates obtained the correct response of 21,000 Pa or 2.1 × 104 Pa. Scaling proved to be a common error with 2.1 × 10–4 Pa being seen, presumably from dividing, instead of multiplying, 0.21 by 1 × 104. In calculations, candidates are advised to think about whether their answer is sensible, rather than relying just on the answer displayed on the calculator.

Notes from the mark scheme and report

• 0.21 × 1.0 × 10^5 = 2.1 × 10^4 Pa

Skills: M0.1 Recognise and use expressions in decimal and ordinary form; M0.2 Use ratios, fractions and percentages

Mark scheme

Examiner's comment

This part required candidates to work out the amount in moles of propane that combusted, followed by a calculation of the volume of CO2. The initial mole calculation required candidates to first convert the volume of 42.0 m3 into dm3 or cm3. A correct calculation shows that 1750 mol of propane was combusted. Many struggled with the unit conversion, with 0.175 and 1.75 × 10–3 often being seen. This value then needed to be multiplied by 3 (from the stoichiometry of the equation) and 44 to produce 231,000 g. Here, ×3 was sometimes omitted. Some candidates ignored RTP in the question and used the ideal gas equation to calculate the moles of propane. This approach was acceptable provided that sensible values had been chosen for the conditions at room temperature and pressure (e.g. 298K, 293K, 100 kPa,101 kPa). Candidates are advised to practice volume unit conversions – important for any calculation involving gases.

Notes from the mark scheme and report

• common errors (1 mark): no conversion of m3 to dm3 (231), cm3/dm3 confusion (0.231), no 3 × for CO2 (77 000)
• ALLOW ideal gas equation with sensible T and p

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.2 Use ratios, fractions and percentages; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

This part tested an understanding of pH as a logarithmic scale and the relationship between rates and order. This part discriminated extremely well. A pH of 3 meant that the H+ concentration would be 100 times less than at a pH of 1. Being a first order reaction with respect to H+, the initial rate is simply 100 times less than at a pH of 1: 2.4 × 10–5 mol dm–3 s–1. The higher-ability candidates identified the relationships and often wrote the correct answer on the answer line with no visible working (presumably doing the calculation in their head). Others used their well-learnt equations to calculate the same correct answer. Many candidates found the calculation difficult and 7.2 × 10–3 (3 × (2.4 × 10–3)) was a very common incorrect answer. If candidates had fully scrutinised this answer, they may have realised that a more dilute solution cannot produce a faster rate.

Mark scheme

Examiner's comment

This part required candidates to initially work out the amount in moles of P2O5 in 4.26 g of P2O5, followed by multiplication by 5 (for O) and the Avogadro constant to produce 9.03 × 1022 atoms. The initial moles calculation proved to be a relatively easy mark which most candidates were awarded. Many then struggled, sometimes omitting the ×5 factor to account for the O atoms, producing the common incorrect answer of 1.806 × 1022.

Notes from the mark scheme and report

• minimum 3 SF
• common error: no × 5 (1.806 × 10^22) 1 mark
• ALLOW ECF from incorrect n(P2O5)

Skills: M0.1 Recognise and use expressions in decimal and ordinary form; M0.2 Use ratios, fractions and percentages; M1.1 Use an appropriate number of significant figures

Mark scheme

Examiner's comment

Candidates are well practised with percentage yield calculations with about half obtaining the correct percentage yield of 33.8 or 33.9% to secure all 3 marks. Many were able to secure partial credit for incorrect answers, provided that the working was laid out clearly. Some responses showed a simple percentage of the two masses with no consideration of moles or molar masses. Such a response received no credit.

Notes from the mark scheme and report

• mass from volume × density
• 3 SF
• common errors (2 marks): no use of density (35.2%), ÷ density instead of × density (36.5%)
• ALLOW ECF for each step

Skills: M0.2 Use ratios, fractions and percentages; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

Most candidates were able to work out the masses of Pb and O atoms in the lead oxide from the supplied experimental results, and to then attempt to find the molar ratio of Pb : O. The 1 : 1.33 ratio was not always identified and a significant number of candidates approximated this to PbO or Pb2O3 instead of scaling up the ratio to 3 : 4 to show that the formula is Pb3O4 .

Notes from the mark scheme and report

• masses Pb 3.132 g and O 0.322 g
• Pb3O4 must come from masses
• NO ECF from incorrect masses

Skills: M0.2 Use ratios, fractions and percentages; PAG1 Moles determination

Mark scheme

Examiner's comment

Candidates produced some very good calculations. It was common for candidates to determine the correct Ka value of 1.71 × 10–10 mol dm–3, after first calculating the hydrogen ion and weak acid concentrations. The initial calculation of the hydrogen ion concentration from 10–pH proved to be the easiest mark. The weak acid concentration proved to be more difficult with a significant number unable to determine the correct molar mass of chloroxylenol as 156.5. Most candidates recognised that the molar concentration in mol dm–3 was required from the mass of chloroxylenol in 100 cm3. Almost all candidates divided the mass of 4.80 by their molar mass but the scaling by ×10 was not always successful. Some candidates used the mole equations that they had learnt to obtain the ×10 scaling in a very roundabout way. The final mark was often given by ECF (error carried forward) from an incorrect concentration. The question did also ask for an equation, using molecular formulae, for the acid dissociation. This was sometimes omitted or written using C8H8ClOH as the molecular formula of chloroxylenol, instead of C8H9ClO. Even if the equation had been omitted, the mark could still given from a correct Ka expression using molecular formulae. The usual advice of ‘read and answer the question’ still holds. equation has been written using molecular formulae and all stages of the calculation are clear to see. Most candidates worked out the hydrogen ion concentration as 7.244 × 10–6 mol dm–3 before carrying out the final Ka calculation. This candidate had combined these stages.

Notes from the mark scheme and report

• molecular formulae required in equation (C8H9ClO ⇌ H+ + C8H8ClO–)
• common error: no squaring of [H+] (2.36 × 10–5) 3 of 4 calculation marks
• NO ECF from incorrect formula in equation

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.4 Use calculators to find and use power, exponential and logarithmic functions; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M2.5 Use logarithms in relation to quantities that range over several orders of magnitude

Mark scheme

Examiner's comment

Just over half the candidates obtained the correct temperature change of 1°C. A common error was a temperature change of 0.5°C, the result of not considering that two temperature readings are made when calculating a temperature change.

Notes from the mark scheme and report

• two readings so 0.2/20 × 100 = 1.0 °C
• IGNORE direction of temperature change

Skills: M1.3 Identify uncertainties in measurements and use simple techniques to determine uncertainty when data are combined; M2.2 Change the subject of an equation; PAG3 Enthalpy determination

Mark scheme

Examiner's comment

This was a testing question. This question discriminated very well but lower ability candidates struggled to score marks. This unfamiliar scenario required candidates to first decide on a strategy for solving the problem. Candidates needed to use the Gibbs’ equation and supplied data to determine the entropy change. They then needed to combine this value with other supplied data to determine the unknown standard entropy. The data provided were all standard, signified by the standard sign. Candidates were expected to use 298 K as the standard temperature to first determine the entropy change. Although many candidates chose 298 K, many different temperatures were seen, with 273 K and 286.5 K (the temperature change in 5a(i)) being common. Some candidates were obviously thrown by the absence of a stated temperature and many omitted T from their Gibbs’ equation, using ∆G = ∆H – ∆S instead. This calculated value for ∆S (correct: –0.133 kJ) then needed to be combined with the supplied entropy data to determine the unknown standard entropy (correct: 156 J) to three significant figures. ECF could be applied during marking but this second calculation first required a unit conversion from kJ to J. Many candidates did not make this conversion. Some candidates made errors combining this information, e.g. incorrect signs, adding rather than subtracting and no unit conversions. Finally, some candidates, successful in their calculation, did not follow the instruction to quote their answer to three significant figures. first in kJ and then, in preparation for the second part of the determination, in J. The ∆S value is combined with the supplied entropy data to obtain the unknown entropy change, shown by ‘x’ in the response. This calculated value is then rounded to the required three significant figures.

Notes from the mark scheme and report

• use of 298 K
• ∆S = (–55.8 – –16.1)/298 = –0.133 kJ K–1 mol–1 (sign required), convert to J
• S = 372.4 – 5 × 69.9 + 133 = 156
• 3 SF
• ALLOW ECF from incorrect T or ∆S

Skills: M0.0 Recognise and make use of appropriate units in calculations; M1.1 Use an appropriate number of significant figures; M2.2 Change the subject of an equation; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

Almost all candidates followed a well-learnt procedure to complete the calculation. Despite being in the rubric to the question, some candidates did not give an answer to two decimal places. Others made a rounding error in reducing 28.1086 to two decimal places, with 28.10 and 28.12 being common errors.

Notes from the mark scheme and report

• answer to 2 DP (28.11)
• 1 mark ECF if %s used with wrong isotopes once or decimal places transposed for one %

Skills: M0.2 Use ratios, fractions and percentages; M1.1 Use an appropriate number of significant figures

Mark scheme

Examiner's comment

Most candidates are competent at answering questions based on the mole. Almost all candidates were able to calculate the amount of H3PO4 as 0.062 mol. Candidates then needed to use the 2:3 mole stoichiometric ratio to show that 0.093 mol of Mg reacts, which has a mass of 2.26 g to the required 3 significant figures. The commonest errors were use of the inverse 3:2 ratio to obtain 1.00 g Mg, or to omit the ratio to obtain 1.51 g Mg, as shown in the exemplar. Candidates are advised to show clear working so that credit can be awarded for such responses by applying error carried forward.

Notes from the mark scheme and report

• 3 SF required
• ALLOW ECF
• common errors: 3:2 ratio omitted (1.51 g) or inverted (1.00 g)

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.2 Use ratios, fractions and percentages; M1.1 Use an appropriate number of significant figures

Mark scheme

Examiner's comment

Almost all candidates realised that the calculation required the ideal gas equation. Most candidates correctly rearranged the equation and used the data from the question to obtain a value for the volume of phosphine. The most common errors were with conversion of units into Pa and m3. It is recommended that candidates learn how to carry out these conversions. In their calculations, many candidates used the amount of phosphoric acid, 3.20 × 10–3 mol, rather than 8.00 × 10–3 mol of phosphine, obtaining a volume of 1258 cm3. Error carried forward ensured that 3 of the available 4 marks could be credited, provided that the working was clear. The exemplar shows such a response. Answer = 315 cm3

Notes from the mark scheme and report

• common errors: using n(H3PO3) without dividing by 4 (1260 cm3)
• no °C to K conversion (133)
• no kPa to Pa conversion
• no m3 to cm3 conversion
• IGNORE use of 24000 molar volume

Skills: M0.0 Recognise and make use of appropriate units in calculations; M1.1 Use an appropriate number of significant figures; M2.2 Change the subject of an equation; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

Candidates are well-versed with the relationship q = mc∆T and most were able to calculate that 2.926 kJ of energy was released in this reaction. It was also common to see the amount of AgNO3 correctly calculated as 1.28 × 10–3 mol. Candidates were expected to determine the amount of energy released from 1 mol AgNO3 as 229 kJ and finally to multiply this value by 2 for the molar quantities in the equation to match the ‘enthalpy change of reaction’. It was common to see –229 given as the final answer but this was rarely multiplied by 2. The question also required the final answer to be given to an appropriate number of significant figures. Many candidates seemed to be unaware that this reflects the least significant figure provided in the data, in this case 3 significant figures. The exemplar shows a typical response for 3 of the available 4 marks. Many omitted the negative sign in their ∆H value to consider the exothermicity of the reaction. Candidates are also advised to only round at the end of a multi-step calculation. Rounding of intermediate values introduces rounding errors in the final answer. Answer = –457 kJ mol–1

Notes from the mark scheme and report

• 3 SF and – sign needed
• ±229 scores 3 (not doubled for 2 mol AgNO3)
• mass used in q is 25.0 g of solution
• ALLOW ECF throughout

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.2 Use ratios, fractions and percentages; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; PAG3 Enthalpy determination

Mark scheme

Examiner's comment

Most candidates were able to accurately record the burette readings and made the correct subtractions. Despite the examination question requesting the mean titre to be recorded to the accuracy of the burette, many candidates did not do this. A common error was taking a mean of all three readings instead of only the concordant results; this led the candidates to give an answer of 26.58 which lost them 2 marks.

Notes from the mark scheme and report

• readings to 2 d.p. ending in 0 or 5
• mean from concordant titres only (26.45)
• DO NOT ALLOW 26.5
• ALLOW ECF from incorrect burette readings

Skills: M1.2 Find arithmetic means; PAG2 Acid–base titration

Mark scheme

Examiner's comment

A good attempt by many candidates but some did not know how to calculate this or did not multiply by 2.

Notes from the mark scheme and report

• two readings so 2 × 0.05
• ALLOW ECF from (c)(i) or use of any titre

Skills: M0.2 Use ratios, fractions and percentages; M1.3 Identify uncertainties in measurements and use simple techniques to determine uncertainty when data are combined; PAG2 Acid–base titration

Mark scheme

Examiner's comment

This was a good discriminator. Many candidates were credited the first two marks, and higher ability candidates went on to obtain 3 or 4 marks. Common errors were not rounding to a whole number at the end and not scaling up by 250/27.30. Many candidates did not scale up at all, obtaining an answer of 1081, or merely multiplied by 250/25 and obtained a molar mass of 108.

Notes from the mark scheme and report

• acid A is in the burette so scale by 250.0/27.30
• 1:2 ratio A:NaOH
• common ECF answers 108, 1081, 59, 114, 99 (3 marks)

Skills: M0.2 Use ratios, fractions and percentages; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; PAG2 Acid–base titration

Mark scheme

Examiner's comment

Most candidates made a good attempt at this question. The most common mistake was to use the wrong sign when incorporating the enthalpy change into the calculation, or not to incorporate it at all.

Notes from the mark scheme and report

• 431.5
• –431.5 scores 1 (wrong sign)
• 247.5 scores 1 (wrong expression)
• must divide by 2

Skills: M2.2 Change the subject of an equation; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M2.4 Solve algebraic equations

Mark scheme

Examiner's comment

Most candidates were able to convert from tonnes to moles and then went on to complete the majority of the calculation steps. Many omitted to divide by 4 and were credited 3 marks. Some candidates lost marks by not stating the answer to standard form or quoted their answer to more than 3 significant figures. A number of candidates attempted to use Q = mc∆T and did not get very far in the calculation.

Notes from the mark scheme and report

• tonnes to g
• ÷4 for per mole NH3
• 3 sig figs AND standard form (6.79 × 10^7)
• common errors 2.72 × 10^8 (no ÷4), 1.09 × 10^9 (×4), 6.79 × 10^1 (no tonnes to g)

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.1 Recognise and use expressions in decimal and ordinary form; M0.2 Use ratios, fractions and percentages; M1.1 Use an appropriate number of significant figures

Mark scheme

Examiner's comment

This question was well answered, but a significant number of candidates incorrectly used the Mr of butan-1-ol when calculating the mass of 1-bromobutane.

Notes from the mark scheme and report

• 3 sig figs required (12.6)
• common error 33.4 (dividing by yield)
• DO NOT ALLOW 6.82 (Mr of butan-1-ol)

Skills: M0.2 Use ratios, fractions and percentages; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

Most candidates knew they had to draw a tangent to the curve but calculating the gradient led to careless errors. Many drew tiny triangles and mis-read readings of coordinates. It is far easier, and more accurate, to use large triangles using the axes for coordinates. Lower ability candidates just interpolated the graph at 30s and were not credited marks for this.

Notes from the mark scheme and report

• tangent drawn at t = 30 min (±10)
• DO NOT ALLOW interpolation (direct reading)
• ALLOW ECF from incorrect tangent
• IGNORE units and sign

Skills: M3.1 Translate information between graphical, numerical and algebraic forms; M3.5 Draw and use the slope of a tangent to a curve as a measure of rate of change

Mark scheme

Examiner's comment

Most candidates were able to do this relatively straightforward calculation by rearranging the values of the enthalpy changes associated with the energy cycle diagram. However, a significant number forgot to divide –1008 by 2 to score full marks.

Notes from the mark scheme and report

• –504
• –1008 scores 1 (not divided by 2)
• common sign errors +2694, –2113, –2126, –517, +504 score 1
• minus sign needed

Skills: M2.2 Change the subject of an equation; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M2.4 Solve algebraic equations

Mark scheme

Examiner's comment

This relatively straightforward calculation depended upon a candidates’ knowledge that the sum of average bond enthalpies of bonds formed subtracted from the sum of average bond enthalpies of bonds broken was equal to ∆H (– 598 kJ mol–1), and that a rearrangement of this sum gave the value of the bond enthalpy associated with 2 mol of F–F bonds (+316 kJ mol–1). Thus, the bond enthalpy of the F–F bond required division by two.

Notes from the mark scheme and report

• +158
• common errors for 2 marks: –158, 316 (no ÷2), 622 (2 × 464), 457 (omitting –598), 756 (+598)
• 970 scores 1

Skills: M2.2 Change the subject of an equation; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M2.4 Solve algebraic equations

Mark scheme

Examiner's comment

Higher ability candidates realised that the gradient was equivalent to –Ea/R and determined a gradient within the range ± 800 to ± 1040, depending upon the line drawn. Credit was given to gradients of ± 0.800 to ± 1.040 resulting from calculations which omitted 10–3 on the x-axis. The negative gradient was multiplied by R to determine Ea with a value rounded to 3 significant figures. Common errors were omission of 10–3 in the calculation which led to Ea values between 6.65 and 8.65 J mol–1, or not rounding to 3 significant figures.

Notes from the mark scheme and report

• gradient in range –800 to –1040 (K)
• Ea = –gradient × 8.314
• 3 sig figs AND use of 10^–3 on x-axis
• omission of 10^–3 can score first 2 marks
• ALLOW lines that do not reach the y-axis

Skills: M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M3.2 Plot two variables from experimental or other data; M3.3 Determine the slope and intercept of a linear graph

Mark scheme

Examiner's comment

Higher ability candidates realised that the y-intercept of ln k was equivalent to ln A and A was equivalent to e(ln A) and were able to process this readily in their calculators. Common errors were to mis-read the intercept. For example, 31.5 was frequently seen as 31.05. Other candidates assumed the y-intercept was log A and tried to determine A by 10(log A).

Notes from the mark scheme and report

• ln A = y-intercept (extrapolate line to 1/T = 0) about 31.4
• A = e^intercept about 4.3 × 10^13
• ALLOW substitution of a point into ln k = –Ea/RT + ln A
• ALLOW ECF from incorrect ln A

Skills: M0.4 Use calculators to find and use power, exponential and logarithmic functions; M2.5 Use logarithms in relation to quantities that range over several orders of magnitude; M3.3 Determine the slope and intercept of a linear graph

Mark scheme

Examiner's comment

The expression and the units were almost universally known by the candidates.

Notes from the mark scheme and report

• square brackets essential
• units dm3 mol–1
• ALLOW mol dm–3 as ECF from inverted expression

Skills: M0.0 Recognise and make use of appropriate units in calculations

Mark scheme

Examiner's comment

There were three steps to this calculation: Conversion of molar quantities of NO and O2 to molar concentrations. Insertion into the Kc expression and determination (via a square root calculation) of the molar concentration of NO2. Conversion of the molar concentration of NO2 to a molar quantity. Steps 1 and/or 3 of the calculation were occasionally omitted but if the calculation was presented in a coherent manner, even here, partial credit was awarded.

Notes from the mark scheme and report

• convert mol to concentrations (÷ 4.0) first
• square root of 0.090
• × 4 to get mol
• common errors 9.6 (moles used, 3 marks), 0.36 (no square root, 3 marks), 2.4 (2 marks)

Skills: M2.2 Change the subject of an equation; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M2.4 Solve algebraic equations

Mark scheme

Examiner's comment

This three-step calculation was successfully completed by almost all candidates. The common errors were to omit giving the final answer to 2 decimal places or to use [H+] rather than [H+]2 in the calculation, leading to a pKa of 2.35.

Notes from the mark scheme and report

• Ka = [H+]2/0.870 = 1.74 × 10^–5
• pKa = 4.76
• 2 decimal places required

Skills: M0.4 Use calculators to find and use power, exponential and logarithmic functions; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M2.5 Use logarithms in relation to quantities that range over several orders of magnitude

Mark scheme

Examiner's comment

This proved a more difficult calculation than expected, but higher ability candidates realised that [H+] (determined from the pH) divided by the given concentration of CH3COOH was required. Answers had to be expressed to three significant figures in order to receive credit.

Notes from the mark scheme and report

• 3 sig figs required

Skills: M0.2 Use ratios, fractions and percentages; M1.1 Use an appropriate number of significant figures

Mark scheme

Examiner's comment

To help candidates, on this occasion early rounding was ignored and consequently most candidates scored full marks in this multi-step calculation. However, candidates should be advised not to round in the early stages of calculations such as this, as this introduces rounding errors into the final answer. Candidates should be encouraged to indicate what they are attempting to calculate in unstructured calculations such as this. The first step was frequently seen as 10–13.48 = 3.31… x 10–14 which most examiners could take to be [H+]. However, it is clearer to write [H+] = 10–13.48 = 3.31… x 10–14 mol dm–3. Even inclusion of units would help some candidates achieve partial credit as this might allow examiners to determine what a candidate is attempting to do.

Notes from the mark scheme and report

• mass NaOH = 0.302 × 0.100 × 40.0 = 1.21 g
• 95.9%
• ALLOW pOH method
• ALLOW ECF throughout

Skills: M0.2 Use ratios, fractions and percentages; M0.4 Use calculators to find and use power, exponential and logarithmic functions; M1.1 Use an appropriate number of significant figures; M2.5 Use logarithms in relation to quantities that range over several orders of magnitude