Relative formula mass - AQA GCSE Chemistry Exam Questions

Practise 19 exam-style questions on Relative formula mass 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.

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1–5 marks each
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19 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

Calculate the relative formula mass (Mr) of calcium hydroxide, Ca(OH)₂.

Relative atomic masses (Ar): H = 1, O = 16, Ca = 40

Model answer

Mr = 40 + 2 × (16 + 1) = 40 + 34 = 74

Explanation

The subscript 2 after the bracket multiplies both the oxygen and the hydrogen, so there are 2 O and 2 H atoms. A common mistake is 40 + 16 + 1 = 57, which ignores the bracket.

Mark scheme

1 mark
40 + 2 × (16 + 1)
allow 40 + 32 + 2
1 mark
74
correct answer with no working scores 2 marks; ignore units

Question 2

2 marks

A student calculated the relative formula mass (Mr) of magnesium hydroxide, Mg(OH)₂, and got the answer 41.

Explain the mistake the student made and give the correct Mr.

Relative atomic masses (Ar): H = 1, O = 16, Mg = 24

Model answer

The student added 24 + 16 + 1 = 41, counting only one oxygen and one hydrogen atom. The 2 outside the bracket means there are two OH groups, so the correct Mr is 24 + 2 × (16 + 1) = 58.

Explanation

A subscript after a bracket multiplies every atom inside the bracket. Writing out the number of each atom (1 Mg, 2 O, 2 H) before adding helps avoid this mistake.

Mark scheme

1 mark
the student did not multiply O and H by 2 / ignored the subscript 2 outside the bracket
allow added only one O and one H
1 mark
58

Question 3

3 marks

A carbonate, X, reacts with hydrochloric acid:

X + 2HCl → MgCl₂ + H₂O + CO₂

Use relative formula masses to calculate the Mr of X.

Relative atomic masses (Ar): H = 1, C = 12, O = 16, Mg = 24, Cl = 35.5

Model answer

Products: MgCl₂ = 24 + 71 = 95; H₂O = 18; CO₂ = 44
Total for products = 95 + 18 + 44 = 157
Reactants: 2HCl = 2 × 36.5 = 73
Mr of X = 157 - 73 = 84

Explanation

The sum of the relative formula masses of the reactants equals the sum for the products, so the unknown Mr is found by subtraction. Remember 2HCl means two lots of 36.5. (X is magnesium carbonate, MgCO₃, which does have Mr = 84.)

Mark scheme

1 mark
total Mr of products = 95 + 18 + 44 = 157
1 mark
Mr of 2HCl = 73
1 mark
Mr of X = 157 - 73 = 84
allow ecf; allow 84 found by deducing that X is MgCO₃ and adding its Ar values (scores 3 marks)

Question 4

4 marks

Iron is extracted from two ores: haematite contains Fe₂O₃ and magnetite contains Fe₃O₄.

Which compound has the higher percentage by mass of iron? Show your working.

Relative atomic masses (Ar): O = 16, Fe = 56

Model answer

Fe₂O₃: Mr = (2 × 56) + (3 × 16) = 160
Percentage of iron = (112 ÷ 160) × 100 = 70.0%
Fe₃O₄: Mr = (3 × 56) + (4 × 16) = 232
Percentage of iron = (168 ÷ 232) × 100 = 72.4%
Fe₃O₄ (magnetite) has the higher percentage by mass of iron.

Explanation

Use the mass of all the iron atoms in the formula (2 × 56 or 3 × 56), not just one. The ratio of iron to oxygen atoms is slightly higher in Fe₃O₄ (3 : 4) than in Fe₂O₃ (2 : 3), which is why its percentage of iron is higher.

Mark scheme

1 mark
Mr of Fe₂O₃ = 160 and Mr of Fe₃O₄ = 232
1 mark
percentage of Fe in Fe₂O₃ = (112 ÷ 160) × 100 = 70 (%)
allow ecf from Mr
1 mark
percentage of Fe in Fe₃O₄ = (168 ÷ 232) × 100 = 72.4 (%)
allow 72%; allow ecf from Mr
1 mark
Fe₃O₄ / magnetite
must be supported by calculated values

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