Relative Masses Exam Questions
Practice IGCSE Chemistry exam questions on relative masses. Covers relative atomic mass, relative molecular mass, relative formula mass, and calculations using Ar and Mr with mark schemes and examiner notes.
Published by IGCSEChemistry.com.my
Chemistry teaching team: K. S. Tan (15+ years teaching IGCSE Chemistry) and Ms Yash (10+ years teaching IGCSE Chemistry) and Ms Kartini (15+ years teaching IGCSE Chemistry).
Mapped to Cambridge IGCSE Chemistry 0620 (2026–2028). Last updated 2026-08-19.
These questions cover all aspects of relative masses tested in the 0620 exam. Write each answer fully before checking.
Question 1 (2 marks, Core)
Define relative atomic mass (Ar).
Mark scheme
- The average mass of naturally occurring atoms of an element [1]
- on a scale where an atom of carbon-12 has a mass of exactly 12 [1]
Examiner note: Both parts are needed. “The mass of an atom” without reference to the carbon-12 scale scores only 1 mark. The word “average” is important because elements have isotopes.
Question 2 (3 marks, Core)
Calculate the relative formula mass (Mr) of the following compounds. (Ar: H = 1, C = 12, N = 14, O = 16, S = 32, Ca = 40)
(a) H2O (1)
(b) CaCO3 (1)
(c) (NH4)2SO4 (1)
Mark scheme
(a) Mr = (2 x 1) + 16 = 18 [1]
(b) Mr = 40 + 12 + (3 x 16) = 100 [1]
(c) Mr = 2(14 + 4x1) + 32 + (4 x 16) = 2(18) + 32 + 64 = 36 + 32 + 64 = 132 [1]
Examiner note: For (c), the bracket means there are two NH4 groups. Work out the mass inside the bracket first: N + 4H = 14 + 4 = 18, then multiply by 2. Add SO4: 32 + 64 = 96. Total = 36 + 96 = 132.
Question 3 (3 marks, Core)
Calculate the percentage by mass of iron in iron(III) oxide, Fe2O3. (Ar: Fe = 56, O = 16)
Mark scheme
- Mr of Fe2O3 = (2 x 56) + (3 x 16) = 112 + 48 = 160 [1]
- Mass of Fe = 2 x 56 = 112 [1]
- Percentage of Fe = (112 / 160) x 100 = 70.0% [1]
Examiner note: Show all three steps clearly. Even if you make an arithmetic error, correct method earns 2 of the 3 marks. The formula is: % = (total Ar of element / Mr of compound) x 100.
Question 4 (3 marks, Supplement)
A fertiliser has the formula NH4NO3. Calculate the percentage of nitrogen by mass. (Ar: H = 1, N = 14, O = 16)
Mark scheme
- Mr of NH4NO3 = 14 + (4 x 1) + 14 + (3 x 16) = 14 + 4 + 14 + 48 = 80 [1]
- Total mass of N = 14 + 14 = 28 (there are 2 nitrogen atoms) [1]
- Percentage of N = (28 / 80) x 100 = 35.0% [1]
Examiner note: A common error is counting only one nitrogen atom. NH4NO3 contains two nitrogen atoms — one in the NH4 group and one in the NO3 group. Always count carefully from the formula.
Question 5 (2 marks, Core)
Define relative molecular mass (Mr).
Mark scheme
- The sum of the relative atomic masses of all atoms in a molecule [1]
- on a scale where a carbon-12 atom has a mass of exactly 12 [1]
Examiner note: “Mr is the mass of a molecule” is insufficient. You must state it is the sum of the Ar values of all atoms shown in the formula. The term “relative formula mass” is used for ionic compounds (which are not molecules), but the calculation method is identical.
Question 6 (4 marks, Supplement)
Two compounds of carbon and oxygen are carbon monoxide (CO) and carbon dioxide (CO2).
(a) Calculate the Mr of each compound. (2)
(b) Calculate the percentage by mass of carbon in each compound. (2)
(Ar: C = 12, O = 16)
Mark scheme
(a) Mr of CO = 12 + 16 = 28 [1] Mr of CO2 = 12 + (2 x 16) = 44 [1]
(b) % C in CO = (12/28) x 100 = 42.9% [1] % C in CO2 = (12/44) x 100 = 27.3% [1]
Examiner note: Although both contain carbon and oxygen, they have different percentage compositions because the ratio of atoms differs. CO has a higher percentage of carbon because there is less oxygen per carbon atom.
Question 7 (3 marks, Core)
A student needs to calculate the Mr of hydrated copper sulfate, CuSO4.5H2O. (Ar: H = 1, O = 16, S = 32, Cu = 64)
Calculate the Mr and state the percentage by mass of water in this compound.
Mark scheme
- Mr = 64 + 32 + (4 x 16) + 5(2 x 1 + 16) = 64 + 32 + 64 + 5(18) = 64 + 32 + 64 + 90 = 250 [1]
- Mass of water = 5 x 18 = 90 [1]
- Percentage of water = (90/250) x 100 = 36.0% [1]
Examiner note: The “dot” in CuSO4.5H2O means 5 molecules of water are chemically associated with each formula unit. Include the water of crystallisation in the total Mr. This is a common exam calculation.
Question 8 (2 marks, Supplement)
Two hydrocarbons have the formulae C2H6 and C3H8.
(a) Calculate the Mr of each hydrocarbon. (1)
(b) Predict which hydrocarbon has the lower boiling point. Explain your answer. (1)
(Ar: H = 1, C = 12)
Mark scheme
(a) Mr of C2H6 = (2 x 12) + (6 x 1) = 30; Mr of C3H8 = (3 x 12) + (8 x 1) = 44 [1]
(b) C2H6 has the lower boiling point [1] because it has a smaller Mr / fewer electrons / weaker intermolecular forces between molecules
Examiner note: In a homologous series, boiling point increases with Mr because larger molecules have stronger intermolecular forces (more surface area for van der Waals interactions). This links stoichiometry to organic chemistry and bonding.
What to revise if you scored below 6
If Mr calculations caused errors, practise breaking down complex formulae step by step — always deal with brackets first. If percentage composition was tricky (Questions 3, 4, 6), drill the formula: % = (mass of element / Mr) x 100. Revisit the relative masses notes.
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