Percentage yield (chemistry only) - AQA GCSE Chemistry Exam Questions

Practise 27 exam-style questions on Percentage yield (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.

27 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

In a reaction, the maximum theoretical mass of product is 9.0 g. A student actually made 6.3 g of the product.

Calculate the percentage yield.

Model answer

percentage yield = (mass of product actually made ÷ maximum theoretical mass) × 100
= (6.3 ÷ 9.0) × 100 = 70%

Explanation

Always divide the actual mass by the theoretical mass, not the other way round. A percentage yield must be 100% or less, which is a quick check that you have divided correctly.

Mark scheme

1 mark
(6.3 ÷ 9.0) × 100
1 mark
70 (%)
correct answer with no working scores 2 marks

Question 2

2 marks

No atoms are gained or lost in a chemical reaction.

Explain why the mass of product obtained is usually less than the maximum theoretical mass.

Model answer

The atoms are not destroyed, but not all of them end up in the product that is collected. Some stay in unreacted reactants because the reaction does not go to completion, some end up in other products from side reactions, and some product is left behind on the apparatus during separation.

Explanation

Conservation of mass still applies: the 'missing' mass is somewhere else, not destroyed. Percentage yield measures how much ends up in the product you actually collect.

Mark scheme

1 mark
some atoms remain in unreacted reactants (the reaction does not go to completion / is reversible)
max 2
1 mark
some atoms end up in other products (side reactions)
1 mark
some product is lost / left behind during separation

Question 3

3 marks

Zinc reacts with hydrochloric acid:

Zn + 2HCl → ZnCl₂ + H₂

Calculate the maximum theoretical mass of zinc chloride that can be made from 6.5 g of zinc with excess acid.

Relative atomic masses (Ar): Cl = 35.5, Zn = 65

Model answer

moles of Zn = 6.5 ÷ 65 = 0.10 mol
Ratio Zn : ZnCl₂ is 1 : 1, so moles of ZnCl₂ = 0.10 mol
Mr of ZnCl₂ = 65 + (2 × 35.5) = 136
theoretical mass = 0.10 × 136 = 13.6 g

Explanation

The theoretical mass is calculated from the limiting reactant, zinc, because the acid is in excess. This is the maximum mass; any real experiment would give less.

Mark scheme

1 mark
moles of Zn = 6.5 ÷ 65 = 0.10 (mol)
1 mark
moles of ZnCl₂ = 0.10 (mol) and Mr of ZnCl₂ = 136
allow ecf
1 mark
theoretical mass = 0.10 × 136 = 13.6 g
allow ecf; correct answer with no working scores 3 marks

Question 4

5 marks

Ammonia is made in the Haber process:

N₂ + 3H₂ ⇌ 2NH₃

In one pass through the reactor, 2.8 tonnes of nitrogen produced 0.51 tonnes of ammonia.

Calculate the percentage yield. Explain why the yield is low and why this is not a big problem for the manufacturer.

Relative atomic masses (Ar): H = 1, N = 14
1 tonne = 1 000 000 g

Model answer

moles of N₂ = 2 800 000 ÷ 28 = 100 000 mol
Ratio N₂ : NH₃ is 1 : 2, so moles of NH₃ = 200 000 mol
theoretical mass of NH₃ = 200 000 × 17 = 3 400 000 g = 3.4 tonnes
percentage yield = (0.51 ÷ 3.4) × 100 = 15%
The yield is low because the reaction is reversible, so it does not go to completion: some ammonia breaks down back into nitrogen and hydrogen. This is not a big problem because the unreacted nitrogen and hydrogen are recycled through the reactor again.

Explanation

A reversible reaction can never convert all the reactants into product, which is one of the main reasons for a low yield. Industry gets round this by recycling unreacted gases, so very little is actually wasted.

Mark scheme

1 mark
moles of N₂ = 2 800 000 ÷ 28 = 100 000 (mol), so moles of NH₃ = 200 000 (mol)
allow ratio method: 28 tonnes N₂ gives 34 tonnes NH₃
1 mark
theoretical mass of NH₃ = 200 000 × 17 = 3 400 000 g = 3.4 tonnes
allow ecf
1 mark
percentage yield = (0.51 ÷ 3.4) × 100 = 15 (%)
allow ecf
1 mark
the reaction is reversible so it does not go to completion
allow ammonia breaks down back into nitrogen and hydrogen / an equilibrium is reached
1 mark
unreacted nitrogen and hydrogen are recycled (passed through the reactor again)

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