Carboxylic acids (chemistry only) - AQA GCSE Chemistry Exam Questions

Practise 30 exam-style questions on Carboxylic acids (chemistry only) for AQA GCSE Chemistry. Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.

30 questions
1–6 marks each
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20 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

Name the carboxylic acids that have these formulae:
CH₃CH₂COOH
HCOOH

Model answer

CH₃CH₂COOH is propanoic acid.
HCOOH is methanoic acid.

Explanation

Count all the carbon atoms, including the one in the –COOH group: CH₃CH₂COOH has three carbon atoms, so it is propanoic acid, and HCOOH has one, so it is methanoic acid. The ending -oic acid shows a carboxylic acid.

Mark scheme

1 mark
CH₃CH₂COOH: propanoic acid
do not accept propanol or propanoate
1 mark
HCOOH: methanoic acid
do not accept methanol

Question 2

2 marks

Ethyl ethanoate is made by this reaction:
CH₃COOH + C₂H₅OH → CH₃COOC₂H₅ + H₂O
Relative formula masses (Mr): CH₃COOH = 60, C₂H₅OH = 46, CH₃COOC₂H₅ = 88

Calculate the percentage atom economy for making ethyl ethanoate. Give the answer to 3 significant figures.

Model answer

atom economy = Mr of desired product ÷ total Mr of reactants × 100
= 88 ÷ (60 + 46) × 100
= 88 ÷ 106 × 100 = 83.0%

Explanation

The only waste product is water, so the atom economy is high but not 100%. Use the total Mr of all the reactants on the bottom of the fraction. 83.02% rounds to 83.0% to 3 significant figures - the final zero is needed.

Mark scheme

1 mark
atom economy = 88 ÷ (60 + 46) × 100
allow 88 ÷ 106 × 100
1 mark
83.0(%)
must be 3 significant figures; correct answer with no working scores 2 marks

Question 3

3 marks

The diagram shows the displayed formulae of the substances in a reaction.

Name compounds A, B and C.

Model answer

A is ethanoic acid, B is ethanol and C is ethyl ethanoate.

Explanation

A contains the –COOH group with two carbon atoms, so it is ethanoic acid. B has an –OH group on a two-carbon chain, so it is ethanol. They react to form the ester ethyl ethanoate, which contains the –COO– link, and water.

Mark scheme

1 mark
A: ethanoic acid
1 mark
B: ethanol
1 mark
C: ethyl ethanoate
'an ester' alone does not score; do not accept ethyl ethanol

Question 4

4 marks

Ethyl ethanoate is made by this reaction:
CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O

A chemist reacted 12.0 g of ethanoic acid with excess ethanol and obtained 11.0 g of ethyl ethanoate.

Calculate the percentage yield of ethyl ethanoate.
Relative atomic masses (Ar): H = 1, C = 12, O = 16

Model answer

Mr of CH₃COOH = 60
moles of ethanoic acid = 12.0 ÷ 60 = 0.200 mol
Mr of CH₃COOC₂H₅ = (4 × 12) + (8 × 1) + (2 × 16) = 88
maximum mass of ester = 0.200 × 88 = 17.6 g
percentage yield = (11.0 ÷ 17.6) × 100 = 62.5%

Explanation

The ethanol is in excess, so the ethanoic acid decides the maximum amount of ester. Percentage yield = actual mass ÷ maximum mass × 100. The yield is well below 100% because the reaction is reversible and some ester is lost when it is separated.

Mark scheme

1 mark
moles of ethanoic acid = 12.0 ÷ 60 = 0.200
allow Mr = 60 for this mark
1 mark
Mr of ethyl ethanoate = 88
1 mark
maximum mass of ethyl ethanoate = 0.200 × 88 = 17.6 (g)
allow ecf; allow ratio method: 60 g gives 88 g, so 12.0 g gives 17.6 g
1 mark
percentage yield = (11.0 ÷ 17.6) × 100 = 62.5(%)
allow ecf; correct answer with no working scores 4 marks

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