November 2020: Paper 3, Unified chemistry

Mark scheme

Examiner's comment

1a(i) Most candidates labelled some of the apparatus correctly and identified the reflux technique. A significant number showed water flowing in the wrong direction and ‘distillation’ was given as the name of the technique. The ‘condenser’ was sometimes labelled incorrectly, e.g. ‘condensation tube, ‘distillation tube’ and ‘water jacket’. Only just over half the candidates were given both marks. 1a(ii) Many diagrams were incomplete and it was comparatively rare for both of the two available marks to be given. Important labels were often missing. Some candidates drew diagrams of other techniques, such as distillation. Many responses were not credited with marks and this question was often omitted. Candidates need practice in recognising practical techniques and in drawing acceptable diagrams. 1b(i) Candidates were expected to explain the different boiling points of three amines. Most candidates linked increased branching with less surface contact between molecules, leading to weaker induced dipole interactions (London forces). Fewer candidates linked the decreasing boiling points to less energy being required to break the intermolecular bonds. The highest performing candidates recognised the large difference between the boiling points of the primary and tertiary amines. They then identified the cause: primary amines form hydrogen bonds but tertiary amines do not. This question was answered reasonably well with fewer candidates than in the past describing relative strengths of covalent bonds, a common misconception. 1(b)(ii) Most candidates rearranged the ideal gas equation correctly to make n the subject. They then substituted correct values for p, V, R and T into the equation, with correct unit conversions. Most candidates then calculated n correctly as 2.32 × 10–3 and combined this value with the mass of 0.202 g to derive the molecular formula as 87. Lower-attaining candidates often made an error in their conversion of cm3 into m3, by multiplying by 10–3 rather than 10–6. This error resulted in a value of n as 2.32 and the unrealistic molecular mass of 8.7 × 10–3, from which a molecular formula is impossible. Candidates in this position are advised to check back at the unit conversions in the working. Very few candidates did not convert 100ºC into 373 K. Candidates with the correct molecular formula of 87 usually suggested C5H13N as the molecular formula. A significant number quoted C5H11NH2. This response was not credited as it is not a ‘molecular’ formula. There are several possible structures of C5H13N with three 13C NMR peaks and any were credited. Error carried forward was applied from a different molecular mass for both the molecular formula and a structure. This emphasises the importance of working within calculation; credit will always be given for a correct method. Instead of using the ideal gas equation, some candidates used the molar gas constant at room temperature and pressure, 24.0 dm3, which gives a molecular mass of 67. This simplified approach could only gain the final 2 marks for this question: a molecular formula (with an error carried forward value of 67, e.g. C4H5N) and a structure of an amine with this molecular formula and 3 peaks in the 13C NMR spectrum. 1(c) Candidates were supplied with information about an unfamiliar reaction of an amino acid and asked to predict a possible equation. Many candidates suggested H2O as one product, being the difference in the formula of the amino acid and the C3H3NO cyclic organic product. Any cyclic structure of C3H3NO that met the bonding rules for C, H, N and O was credited. Examples included a 4-membered ring lactam and substituted cyclopropenes. A significant number of candidates showed an equation for the reaction of two molecules of the amino acid to form 2 H2O and a cyclic dipeptide. Although chemically feasible, the dipeptide could not be credited because the molecular formula was C3H3NO in the question. This error could have been avoided if the information in the question had been used.

Notes from the mark scheme and report

• water in at bottom and out at top AND condenser
• pear-shaped/round-bottom flask AND reflux
• DO NOT ALLOW conical flask, volumetric flask or beaker

Skills: PAG5 Synthesis of an organic liquid

Mark scheme

Examiner's comment

1a(i) Most candidates labelled some of the apparatus correctly and identified the reflux technique. A significant number showed water flowing in the wrong direction and ‘distillation’ was given as the name of the technique. The ‘condenser’ was sometimes labelled incorrectly, e.g. ‘condensation tube, ‘distillation tube’ and ‘water jacket’. Only just over half the candidates were given both marks. 1a(ii) Many diagrams were incomplete and it was comparatively rare for both of the two available marks to be given. Important labels were often missing. Some candidates drew diagrams of other techniques, such as distillation. Many responses were not credited with marks and this question was often omitted. Candidates need practice in recognising practical techniques and in drawing acceptable diagrams. 1b(i) Candidates were expected to explain the different boiling points of three amines. Most candidates linked increased branching with less surface contact between molecules, leading to weaker induced dipole interactions (London forces). Fewer candidates linked the decreasing boiling points to less energy being required to break the intermolecular bonds. The highest performing candidates recognised the large difference between the boiling points of the primary and tertiary amines. They then identified the cause: primary amines form hydrogen bonds but tertiary amines do not. This question was answered reasonably well with fewer candidates than in the past describing relative strengths of covalent bonds, a common misconception. 1(b)(ii) Most candidates rearranged the ideal gas equation correctly to make n the subject. They then substituted correct values for p, V, R and T into the equation, with correct unit conversions. Most candidates then calculated n correctly as 2.32 × 10–3 and combined this value with the mass of 0.202 g to derive the molecular formula as 87. Lower-attaining candidates often made an error in their conversion of cm3 into m3, by multiplying by 10–3 rather than 10–6. This error resulted in a value of n as 2.32 and the unrealistic molecular mass of 8.7 × 10–3, from which a molecular formula is impossible. Candidates in this position are advised to check back at the unit conversions in the working. Very few candidates did not convert 100ºC into 373 K. Candidates with the correct molecular formula of 87 usually suggested C5H13N as the molecular formula. A significant number quoted C5H11NH2. This response was not credited as it is not a ‘molecular’ formula. There are several possible structures of C5H13N with three 13C NMR peaks and any were credited. Error carried forward was applied from a different molecular mass for both the molecular formula and a structure. This emphasises the importance of working within calculation; credit will always be given for a correct method. Instead of using the ideal gas equation, some candidates used the molar gas constant at room temperature and pressure, 24.0 dm3, which gives a molecular mass of 67. This simplified approach could only gain the final 2 marks for this question: a molecular formula (with an error carried forward value of 67, e.g. C4H5N) and a structure of an amine with this molecular formula and 3 peaks in the 13C NMR spectrum. 1(c) Candidates were supplied with information about an unfamiliar reaction of an amino acid and asked to predict a possible equation. Many candidates suggested H2O as one product, being the difference in the formula of the amino acid and the C3H3NO cyclic organic product. Any cyclic structure of C3H3NO that met the bonding rules for C, H, N and O was credited. Examples included a 4-membered ring lactam and substituted cyclopropenes. A significant number of candidates showed an equation for the reaction of two molecules of the amino acid to form 2 H2O and a cyclic dipeptide. Although chemically feasible, the dipeptide could not be credited because the molecular formula was C3H3NO in the question. This error could have been avoided if the information in the question had been used.

Notes from the mark scheme and report

• Buchner flask with ONE side arm and Buchner/Hirsch funnel sealed on top
• label from side arm indicating reduced pressure (to pump/vacuum)
• DO NOT ALLOW just 'flask' or 'funnel' as labels

Skills: PAG6 Synthesis of an organic solid

Mark scheme

Examiner's comment

1a(i) Most candidates labelled some of the apparatus correctly and identified the reflux technique. A significant number showed water flowing in the wrong direction and ‘distillation’ was given as the name of the technique. The ‘condenser’ was sometimes labelled incorrectly, e.g. ‘condensation tube, ‘distillation tube’ and ‘water jacket’. Only just over half the candidates were given both marks. 1a(ii) Many diagrams were incomplete and it was comparatively rare for both of the two available marks to be given. Important labels were often missing. Some candidates drew diagrams of other techniques, such as distillation. Many responses were not credited with marks and this question was often omitted. Candidates need practice in recognising practical techniques and in drawing acceptable diagrams. 1b(i) Candidates were expected to explain the different boiling points of three amines. Most candidates linked increased branching with less surface contact between molecules, leading to weaker induced dipole interactions (London forces). Fewer candidates linked the decreasing boiling points to less energy being required to break the intermolecular bonds. The highest performing candidates recognised the large difference between the boiling points of the primary and tertiary amines. They then identified the cause: primary amines form hydrogen bonds but tertiary amines do not. This question was answered reasonably well with fewer candidates than in the past describing relative strengths of covalent bonds, a common misconception. 1(b)(ii) Most candidates rearranged the ideal gas equation correctly to make n the subject. They then substituted correct values for p, V, R and T into the equation, with correct unit conversions. Most candidates then calculated n correctly as 2.32 × 10–3 and combined this value with the mass of 0.202 g to derive the molecular formula as 87. Lower-attaining candidates often made an error in their conversion of cm3 into m3, by multiplying by 10–3 rather than 10–6. This error resulted in a value of n as 2.32 and the unrealistic molecular mass of 8.7 × 10–3, from which a molecular formula is impossible. Candidates in this position are advised to check back at the unit conversions in the working. Very few candidates did not convert 100ºC into 373 K. Candidates with the correct molecular formula of 87 usually suggested C5H13N as the molecular formula. A significant number quoted C5H11NH2. This response was not credited as it is not a ‘molecular’ formula. There are several possible structures of C5H13N with three 13C NMR peaks and any were credited. Error carried forward was applied from a different molecular mass for both the molecular formula and a structure. This emphasises the importance of working within calculation; credit will always be given for a correct method. Instead of using the ideal gas equation, some candidates used the molar gas constant at room temperature and pressure, 24.0 dm3, which gives a molecular mass of 67. This simplified approach could only gain the final 2 marks for this question: a molecular formula (with an error carried forward value of 67, e.g. C4H5N) and a structure of an amine with this molecular formula and 3 peaks in the 13C NMR spectrum. 1(c) Candidates were supplied with information about an unfamiliar reaction of an amino acid and asked to predict a possible equation. Many candidates suggested H2O as one product, being the difference in the formula of the amino acid and the C3H3NO cyclic organic product. Any cyclic structure of C3H3NO that met the bonding rules for C, H, N and O was credited. Examples included a 4-membered ring lactam and substituted cyclopropenes. A significant number of candidates showed an equation for the reaction of two molecules of the amino acid to form 2 H2O and a cyclic dipeptide. Although chemically feasible, the dipeptide could not be credited because the molecular formula was C3H3NO in the question. This error could have been avoided if the information in the question had been used.

Notes from the mark scheme and report

• less branching gives more points of contact and stronger London forces
• primary amines have hydrogen bonds but (CH3)3N has none
• H bonds stronger than London forces
• more energy to break
• DO NOT ALLOW 'more electrons' (same number)
• DO NOT ALLOW breaking covalent bonds
• IGNORE primary/secondary/tertiary comparison

Skills: M3.1 Translate information between graphical, numerical and algebraic forms

Mark scheme

Examiner's comment

1a(i) Most candidates labelled some of the apparatus correctly and identified the reflux technique. A significant number showed water flowing in the wrong direction and ‘distillation’ was given as the name of the technique. The ‘condenser’ was sometimes labelled incorrectly, e.g. ‘condensation tube, ‘distillation tube’ and ‘water jacket’. Only just over half the candidates were given both marks. 1a(ii) Many diagrams were incomplete and it was comparatively rare for both of the two available marks to be given. Important labels were often missing. Some candidates drew diagrams of other techniques, such as distillation. Many responses were not credited with marks and this question was often omitted. Candidates need practice in recognising practical techniques and in drawing acceptable diagrams. 1b(i) Candidates were expected to explain the different boiling points of three amines. Most candidates linked increased branching with less surface contact between molecules, leading to weaker induced dipole interactions (London forces). Fewer candidates linked the decreasing boiling points to less energy being required to break the intermolecular bonds. The highest performing candidates recognised the large difference between the boiling points of the primary and tertiary amines. They then identified the cause: primary amines form hydrogen bonds but tertiary amines do not. This question was answered reasonably well with fewer candidates than in the past describing relative strengths of covalent bonds, a common misconception. 1(b)(ii) Most candidates rearranged the ideal gas equation correctly to make n the subject. They then substituted correct values for p, V, R and T into the equation, with correct unit conversions. Most candidates then calculated n correctly as 2.32 × 10–3 and combined this value with the mass of 0.202 g to derive the molecular formula as 87. Lower-attaining candidates often made an error in their conversion of cm3 into m3, by multiplying by 10–3 rather than 10–6. This error resulted in a value of n as 2.32 and the unrealistic molecular mass of 8.7 × 10–3, from which a molecular formula is impossible. Candidates in this position are advised to check back at the unit conversions in the working. Very few candidates did not convert 100ºC into 373 K. Candidates with the correct molecular formula of 87 usually suggested C5H13N as the molecular formula. A significant number quoted C5H11NH2. This response was not credited as it is not a ‘molecular’ formula. There are several possible structures of C5H13N with three 13C NMR peaks and any were credited. Error carried forward was applied from a different molecular mass for both the molecular formula and a structure. This emphasises the importance of working within calculation; credit will always be given for a correct method. Instead of using the ideal gas equation, some candidates used the molar gas constant at room temperature and pressure, 24.0 dm3, which gives a molecular mass of 67. This simplified approach could only gain the final 2 marks for this question: a molecular formula (with an error carried forward value of 67, e.g. C4H5N) and a structure of an amine with this molecular formula and 3 peaks in the 13C NMR spectrum. 1(c) Candidates were supplied with information about an unfamiliar reaction of an amino acid and asked to predict a possible equation. Many candidates suggested H2O as one product, being the difference in the formula of the amino acid and the C3H3NO cyclic organic product. Any cyclic structure of C3H3NO that met the bonding rules for C, H, N and O was credited. Examples included a 4-membered ring lactam and substituted cyclopropenes. A significant number of candidates showed an equation for the reaction of two molecules of the amino acid to form 2 H2O and a cyclic dipeptide. Although chemically feasible, the dipeptide could not be credited because the molecular formula was C3H3NO in the question. This error could have been avoided if the information in the question had been used.

Notes from the mark scheme and report

• n = pV/RT with V = 72.0 × 10^-6 m3 and T = 373 K gives M = 87 so C5H13N
• structure must have 3 carbon environments
• use of 24 000 (not RTP) limits to 3 marks by ECF (M = 67.3, C4H5N)

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; M3.1 Translate information between graphical, numerical and algebraic forms

Mark scheme

Examiner's comment

1a(i) Most candidates labelled some of the apparatus correctly and identified the reflux technique. A significant number showed water flowing in the wrong direction and ‘distillation’ was given as the name of the technique. The ‘condenser’ was sometimes labelled incorrectly, e.g. ‘condensation tube, ‘distillation tube’ and ‘water jacket’. Only just over half the candidates were given both marks. 1a(ii) Many diagrams were incomplete and it was comparatively rare for both of the two available marks to be given. Important labels were often missing. Some candidates drew diagrams of other techniques, such as distillation. Many responses were not credited with marks and this question was often omitted. Candidates need practice in recognising practical techniques and in drawing acceptable diagrams. 1b(i) Candidates were expected to explain the different boiling points of three amines. Most candidates linked increased branching with less surface contact between molecules, leading to weaker induced dipole interactions (London forces). Fewer candidates linked the decreasing boiling points to less energy being required to break the intermolecular bonds. The highest performing candidates recognised the large difference between the boiling points of the primary and tertiary amines. They then identified the cause: primary amines form hydrogen bonds but tertiary amines do not. This question was answered reasonably well with fewer candidates than in the past describing relative strengths of covalent bonds, a common misconception. 1(b)(ii) Most candidates rearranged the ideal gas equation correctly to make n the subject. They then substituted correct values for p, V, R and T into the equation, with correct unit conversions. Most candidates then calculated n correctly as 2.32 × 10–3 and combined this value with the mass of 0.202 g to derive the molecular formula as 87. Lower-attaining candidates often made an error in their conversion of cm3 into m3, by multiplying by 10–3 rather than 10–6. This error resulted in a value of n as 2.32 and the unrealistic molecular mass of 8.7 × 10–3, from which a molecular formula is impossible. Candidates in this position are advised to check back at the unit conversions in the working. Very few candidates did not convert 100ºC into 373 K. Candidates with the correct molecular formula of 87 usually suggested C5H13N as the molecular formula. A significant number quoted C5H11NH2. This response was not credited as it is not a ‘molecular’ formula. There are several possible structures of C5H13N with three 13C NMR peaks and any were credited. Error carried forward was applied from a different molecular mass for both the molecular formula and a structure. This emphasises the importance of working within calculation; credit will always be given for a correct method. Instead of using the ideal gas equation, some candidates used the molar gas constant at room temperature and pressure, 24.0 dm3, which gives a molecular mass of 67. This simplified approach could only gain the final 2 marks for this question: a molecular formula (with an error carried forward value of 67, e.g. C4H5N) and a structure of an amine with this molecular formula and 3 peaks in the 13C NMR spectrum. 1(c) Candidates were supplied with information about an unfamiliar reaction of an amino acid and asked to predict a possible equation. Many candidates suggested H2O as one product, being the difference in the formula of the amino acid and the C3H3NO cyclic organic product. Any cyclic structure of C3H3NO that met the bonding rules for C, H, N and O was credited. Examples included a 4-membered ring lactam and substituted cyclopropenes. A significant number of candidates showed an equation for the reaction of two molecules of the amino acid to form 2 H2O and a cyclic dipeptide. Although chemically feasible, the dipeptide could not be credited because the molecular formula was C3H3NO in the question. This error could have been avoided if the information in the question had been used.

Notes from the mark scheme and report

• cyclic amide (four-membered lactam ring keeping C=C) + H2O
• products marked independently then balanced
• DO NOT ALLOW dimer
• structure must have correct number of bonds to each atom

Mark scheme, with the levels and indicative points

Examiner's comment

This question was assessed by level of response (LoR). The question required candidates to demonstrate their knowledge and understand of some important terms used in transition metal chemistry. Level 3 candidates showed complete definitions of ligand, coordination number and ligand substitution, supported by suitable equations with clear diagrams of complex ions. The responses were concise with 3D diagrams of 6- and 4-coordinate complex ions displayed with wedges, correct connectivity to ligand atoms and showing the role of lone pairs in the formation of the coordinate bonds. 4-coordination was shown as either or both of tetrahedral (usually CuCl42–) or square planar (e.g. platin). Ligand substitution was accompanied by a correct balanced equation, most commonly between [Cu(H2O6]2+ and NH3. Level 2 candidates usually gave definitions of ligand, coordination number and ligand substitution. There was usually a balanced equation for ligand substitution and one correct 3D diagram. A second diagram often had an unsuitable shape for the complex ion chosen (often CuCl42– shown as square planar). Level 1 candidates did produce definitions, but these were often incomplete. There was usually an attempt to show a 3D diagram or equation, but these often contained unsuitable examples. This question rewarded the candidates who had learnt their chemistry and the levels enabled the amount of knowledge and understanding to be assessed. The question discriminated well.

Notes from the mark scheme and report

• ligand donates a lone pair forming a coordinate bond to the metal ion
• coordination number = number of coordinate bonds
• octahedral e.g. [Cu(H2O)6]2+, tetrahedral e.g. CuCl4 2– or square planar Pt(NH3)2Cl2 with 3D diagrams
• balanced ligand substitution equation

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

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

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

• any two for 1 mark: heat loss to surroundings, incomplete combustion (IGNORE incomplete 'reaction'), evaporation of water
• IGNORE evaporation of C3H8

Skills: PAG3 Enthalpy determination

Mark scheme, with the levels and indicative points

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

• bonds broken 6488, bonds made 8542, ∆rH = –2054
• errors: 3 C–C used (–1707), C–C omitted (–2748)
• ∆cH needs H2O(l) so ∆cH = ∆rH – 4 × ∆vapH = –2217
• adding 4 × 40.65 (–1891) is incorrect

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

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

• M(s) + 2HCl(aq) → MCl2(aq) + H2(g) - state symbols required in overall equation only
• M → M2+ + 2e–
• 2H+ + 2e– → H2
• half-equations must only contain species that change
• 1 mark if both correct but oxidation/reduction swapped

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

• ALL fizzing has stopped
• ALL the metal has dissolved ('metal disappears' not quite enough)
• IGNORE constant mass and no temperature increase

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

• H+ + OH– → H2O
• IGNORE state symbols

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

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

• reweigh to constant mass
• IGNORE 'leave for longer' or 'wait until fizzing stops' without link to constant mass

Skills: 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/moles of CO2 lost would be smaller so n(X2CO3) appears smaller
• calculated molar mass would be greater

Skills: 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

• mark by row
• 1/T to 2 SF or more
• ln Kp whole numbers 105, 86, 72, 60
• DO NOT ALLOW whole-number errors e.g. 85 for 86

Skills: M0.1 Recognise and use expressions in decimal and ordinary form; 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

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

• equilibrium shifts to the left AND forward reaction is exothermic (Kp decreases as T increases)

Skills: M0.3 Estimate results

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

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

• extrapolate to y-intercept
• intercept = ∆S/R so ∆S = R × intercept
• ALLOW substituting ln Kp, 1/T and gradient into equation 5.1
• x-intercept loses first mark

Skills: M3.1 Translate information between graphical, numerical and algebraic forms; M3.3 Determine the slope and intercept of a linear graph

Mark scheme

Examiner's comment

6(a) Candidates were expected to use their knowledge and understanding of bond angles and shapes. Most candidates were given both available marks. A significant number responded with incorrect names of shapes, or chose the wrong shape, such as pyramidal for the O–C–O bond angle in the carboxylic acid. Candidates are recommended to learn this basic area of chemistry. 6(b) Most candidates completed a correct equilibrium equation and assigned the correct acid-base pairs. A significant number of candidates used ethanoic acid instead of water in the equation. The examiners allowed error carried forward in this case for the acid-base pairs. Candidates found it much more difficult to explain whether the sulfonic acid had a lower pH value. The higher-attaining candidates answered concisely. They usually identifying that the sulfonic acid would have a lower pH as the pKa value was lower (or the Ka value greater), leading to more dissociation than ethanoic acid. Lower-attaining candidates often produced unfocussed and lengthy responses, sometimes relating increased dissociation in an acid to a higher, rather than a lower, pH. Some candidates approached their explanation mathematically, calculating pH values for ethanoic acid and sulfonic acid from their concentrations and pKa values. If correct, this approach was fully credited. 6(c) Most candidates showed a good understanding and appreciation of drawing curly arrows. It must be stressed that curly arrows that do not start from a lone pair, negative charge or a bond cannot be credited. Lower-attaining candidates often drew imprecisely positioned curly arrows, curly arrows in the wrong direction or to the wrong atoms. For their response to be credited with marks, candidates should position curly arrows to ensure credit when outlining reaction mechanisms. If any of your students’ results are not as expected, you may wish to consider one of our review of results services. For full information about the options available visit the OCR website. If university places are at stake you may wish to consider priority service 2 reviews of marking which have an earlier deadline to ensure your reviews are processed in time for university applications. Our priority is supporting you and your students this autumn and to support you as you prepare for summer 2021 exams. We’ll update our website information regularly with resources, guidance and key information. • Teachers • Students • Exams officers • Assessment specialists We are sending 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. All our courses for the academic year 2020-2021 are being 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, Cambridge Technicals and Functional Skills qualifications. See the full list of available qualifications in the sign up form. 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 of results

Notes from the mark scheme and report

• 120° trigonal planar
• 104–105° non-linear (ALLOW bent, v-shaped, angular
• IGNORE planar)
• mark by row or column

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

Mark scheme

Examiner's comment

6(a) Candidates were expected to use their knowledge and understanding of bond angles and shapes. Most candidates were given both available marks. A significant number responded with incorrect names of shapes, or chose the wrong shape, such as pyramidal for the O–C–O bond angle in the carboxylic acid. Candidates are recommended to learn this basic area of chemistry. 6(b) Most candidates completed a correct equilibrium equation and assigned the correct acid-base pairs. A significant number of candidates used ethanoic acid instead of water in the equation. The examiners allowed error carried forward in this case for the acid-base pairs. Candidates found it much more difficult to explain whether the sulfonic acid had a lower pH value. The higher-attaining candidates answered concisely. They usually identifying that the sulfonic acid would have a lower pH as the pKa value was lower (or the Ka value greater), leading to more dissociation than ethanoic acid. Lower-attaining candidates often produced unfocussed and lengthy responses, sometimes relating increased dissociation in an acid to a higher, rather than a lower, pH. Some candidates approached their explanation mathematically, calculating pH values for ethanoic acid and sulfonic acid from their concentrations and pKa values. If correct, this approach was fully credited. 6(c) Most candidates showed a good understanding and appreciation of drawing curly arrows. It must be stressed that curly arrows that do not start from a lone pair, negative charge or a bond cannot be credited. Lower-attaining candidates often drew imprecisely positioned curly arrows, curly arrows in the wrong direction or to the wrong atoms. For their response to be credited with marks, candidates should position curly arrows to ensure credit when outlining reaction mechanisms. If any of your students’ results are not as expected, you may wish to consider one of our review of results services. For full information about the options available visit the OCR website. If university places are at stake you may wish to consider priority service 2 reviews of marking which have an earlier deadline to ensure your reviews are processed in time for university applications. Our priority is supporting you and your students this autumn and to support you as you prepare for summer 2021 exams. We’ll update our website information regularly with resources, guidance and key information. • Teachers • Students • Exams officers • Assessment specialists We are sending 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. All our courses for the academic year 2020-2021 are being 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, Cambridge Technicals and Functional Skills qualifications. See the full list of available qualifications in the sign up form. 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 of results

Notes from the mark scheme and report

• CH3SO2OH + H2O ⇌ CH3SO2O– + H3O+ with pairs labelled
• CH3SO2OH dissociates more/stronger acid (IGNORE 'more acidic')
• student correct AND lower pKa/higher Ka/greater [H+]
• ALLOW pH calculations for both acids

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

6(a) Candidates were expected to use their knowledge and understanding of bond angles and shapes. Most candidates were given both available marks. A significant number responded with incorrect names of shapes, or chose the wrong shape, such as pyramidal for the O–C–O bond angle in the carboxylic acid. Candidates are recommended to learn this basic area of chemistry. 6(b) Most candidates completed a correct equilibrium equation and assigned the correct acid-base pairs. A significant number of candidates used ethanoic acid instead of water in the equation. The examiners allowed error carried forward in this case for the acid-base pairs. Candidates found it much more difficult to explain whether the sulfonic acid had a lower pH value. The higher-attaining candidates answered concisely. They usually identifying that the sulfonic acid would have a lower pH as the pKa value was lower (or the Ka value greater), leading to more dissociation than ethanoic acid. Lower-attaining candidates often produced unfocussed and lengthy responses, sometimes relating increased dissociation in an acid to a higher, rather than a lower, pH. Some candidates approached their explanation mathematically, calculating pH values for ethanoic acid and sulfonic acid from their concentrations and pKa values. If correct, this approach was fully credited. 6(c) Most candidates showed a good understanding and appreciation of drawing curly arrows. It must be stressed that curly arrows that do not start from a lone pair, negative charge or a bond cannot be credited. Lower-attaining candidates often drew imprecisely positioned curly arrows, curly arrows in the wrong direction or to the wrong atoms. For their response to be credited with marks, candidates should position curly arrows to ensure credit when outlining reaction mechanisms. If any of your students’ results are not as expected, you may wish to consider one of our review of results services. For full information about the options available visit the OCR website. If university places are at stake you may wish to consider priority service 2 reviews of marking which have an earlier deadline to ensure your reviews are processed in time for university applications. Our priority is supporting you and your students this autumn and to support you as you prepare for summer 2021 exams. We’ll update our website information regularly with resources, guidance and key information. • Teachers • Students • Exams officers • Assessment specialists We are sending 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. All our courses for the academic year 2020-2021 are being 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, Cambridge Technicals and Functional Skills qualifications. See the full list of available qualifications in the sign up form. 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 of results

Notes from the mark scheme and report

• marks solely for curly arrows (6 correct = 4 marks, 3 correct = 1 mark)
• arrows from lone pair on –OH to S and S=O bond to O, O– lone pair reforming S=O and S–OCH3 bond to O, then –OCH3 lone pair to H and O–H bond to O
• IGNORE added charges/dipoles