Edexcel A-Level Chemistry AS Paper 1, June 2024: Question 6

Question

Acceptable Answer Additional Guidance Mark

Number

6(a)(i) Example of calculation (3)

• percentage of oxygen (1) 100 – 26.7 – 2.2 = 71.1%

Allow 71%

• conversion of % to moles (1) C 26.7 ÷ 12 = 2.225

H 2.2 ÷ 1 = 2.2

O 71.1 ÷ 16 = 4.444

• divide smallest into the others to get a ratio 2.225 ÷ 2.2 = 1

and 2.2 ÷ 2.2 = 1

empirical formula (1) 4.444 ÷ 2.2 = 2

CO2H

Allow elements in any order

No TE

Question

Acceptable Answer Additional Guidance Mark

Number

6(a)(ii) Example of calculation (1)

• relative atomic mass (90) ÷ empirical mass (45) CO2H = 45

and 90 ÷ 45 = 2

molecular formula (C2O4H2) and

2 × CO2H = C2O4H2

Correct answer with no working scores (1)

Question

Acceptable Answer Additional Guidance Mark

Number

6(b) Example of calculation (2)

• correct calculation (1) (50 × 4.31) + (52 × 83.76) + (53 × 9.55) + (54 × 2.38)

= 52.0569

• relative atomic mass = 52.06

and

final answer corrected to 2 DP (1) Correct final answer with no working scores (2)

Allow TE

If units given, allow g mol-1 / AMU units only

Question

Answer Mark

Number

6(c) The only correct answer is C (p = 12 , n = 12, e = 10) (1)

A is not correct because the number of electrons and the number of protons is the same, so this is a neutral atom

B is not correct because the number of electrons and the number of protons is the same, so this is a neutral atom

D is not correct because the number of electrons exceeds the number of protons, so this is an anion

Question

Acceptable Answer Additional Guidance Mark

Number

6(d) An answer that makes reference to the following point: (1)

• relative molecular mass = 114 Ignore units, even if incorrect

(Total for Question 6 = 8 marks)

Question

Acceptable Answer Additional Guidance Mark

Number

7(a)* An answer that makes reference to the following points: Guidance on how the mark scheme should be applied: (6)

This question assesses a student’s ability to show a coherent

and logically structured answer with linkages and fully- The mark for indicative content should be added to the mark

sustained reasoning. for lines of reasoning. For example, an answer with five

Marks are awarded for indicative content and for how the indicative marking points that is partially structured with some

answer is structured and shows lines of reasoning. linkages and lines of reasoning, scores 4 marks (3 marks for

The following table shows how the marks should be awarded indicative content and 1 mark for partial structure and some

for indicative content. linkages and lines of reasoning).

Number of indicative Number of marks awarded

marking points seen in for indicative marking If there are no linkages between points, the same five

answer points indicative marking points would yield an overall score of 3

64 marks (3 marks for indicative content and no marks for

5-4 3 linkages).

3-2 2

11 In general it would be expected that 5 or 6 indicative points

00 would get 2 reasoning marks, and 3 or 4 indicative points

would get 1 mark for reasoning, and 0, 1 or 2 indicative points

The following table shows how the marks should be awarded would score zero marks for reasoning.

for structure and lines of reasoning.

Number of marks If there is any incorrect chemistry, deduct mark(s) from the

awarded for structure reasoning. If no reasoning mark(s) awarded do not deduct

and sustained lines of mark(s).

reasoning

Answer shows a coherent and

logical structure with linkages 2

and fully sustained lines of

reasoning demonstrated

throughout.

Answer is partially structured

with some linkages and lines of 1

reasoning.

Answer has no linkages between

points and is unstructured. 0

Indicative content: Ignore states in equations even if incorrect

• IP1 potassium chloride (and bromide) produces Allow white fumes

misty / steamy fumes (of hydrogen halide) Ignore identification of the fumes using ammonia

Do not award white smoke for misty fumes

• IP2 equation for reaction between potassium chloride KCl + H2SO4 → KHSO4 + HCl

+ concentrated sulfuric acid Allow 2KCl + H2SO4 → K2SO4 + 2HCl

Allow ions given in equation for KCl

• IP3 brown fumes of bromine Allow orange / orange-brown fumes of bromine

Allow orange/ brown liquid of bromine

Do not award yellow fumes

Do not award reference to ‘eggy smell’ / yellow solid of

sulfur

• IP4 equation for HBr producing SO2 and Br2 2HBr + H2SO4 → Br2 + SO2 + 2H2O

2KBr + 2H2SO4 → Br2 + SO2 + 2H2O + K2SO4

Allow ions given in equation for KBr or HBr

• IP5 no change in oxidation numbers of (potassium)

chloride / sulfur

• IP6 with (potassium) bromide the sulfur is reduced to

+4 (therefore the stronger reducing agent)

Ignore any explanations or justifications, even if incorrect