Relative masses of atoms and molecules Exam Questions
4 past-paper questions on this unit. Four of them are below. Answer on the page: each one is marked the moment you pick, the correct option is shown whether or not you found it, and the full explanation opens either way.
CIE 0620 ChemistryPaper 1 and Paper 2 MCQsFree account
Relative masses of atoms and molecules: four questions to try now
Real past-paper questions, the answer key from the mark scheme, and the explanation that goes with it. No account needed to answer them.
Question 1
The relative atomic mass of chlorine is 35.5. When calculating relative atomic mass, which particle is the mass of a chlorine atom compared to?
Answer: C.
Relative atomic mass compares the average mass of an atom of an element with one twelfth of the mass of a carbon-12 atom, which is the agreed international standard. Hydrogen-1 was the historic standard in the nineteenth century, so it is the most plausible of the wrong answers, but it was replaced because carbon-12 can be measured far more precisely by mass spectrometry. A proton and a neutron each weigh close to one atomic mass unit, so either would give roughly similar numbers, yet neither is a defined standard and neither represents a whole atom. The value 35.5 for chlorine is itself a reminder that this scale averages over isotopes.
Question 2
What is the definition of relative molecular mass, Mr?
Answer: C.
Relative molecular mass is found by adding the relative atomic masses of every atom shown in the formula, which is why water comes to 18 and carbon dioxide to 44. Adding atomic numbers instead would give the total proton count, a different quantity that ignores neutrons and would make water 10 rather than 18. Counting the atoms in a compound gives 3 for water, a number that carries no mass information at all. The description involving an average over isotopes belongs to the relative atomic mass of a single element, such as chlorine at 35.5, rather than to a whole compound.
Question 3
What is the relative molecular mass, Mr, of sulfuric acid, H2SO4?
Answer: D.
Adding the relative atomic masses gives 2 × 1 for the hydrogen, 32 for the sulfur and 4 × 16 = 64 for the oxygen, so the total is 2 + 32 + 64 = 98. The value 97 drops one hydrogen atom, the easiest atom to overlook because it contributes so little. The value 82 counts only three oxygen atoms, 2 + 32 + 48. The value 81 combines both of those errors at once, losing a hydrogen and an oxygen together. Writing out each element with its subscript before adding is what prevents these single atom slips.
Question 4
What is the relative molecular mass, Mr, of sulfur dioxide?
Answer: D.
Relative molecular mass is the sum of the relative atomic masses of every atom in the formula, so sulfur dioxide needs 32 for the sulfur plus 16 for each of the two oxygen atoms, giving 32 + 32 = 64. The value 48 counts only one oxygen atom, which is what happens whenever a subscript is read past. The value 32 is sulfur on its own and leaves the oxygen out entirely. The value 24 corresponds to no part of this molecule at all, being closer to the relative atomic mass of magnesium than to anything in SO2.
These questions are drawn from past CIE 0620 Chemistry papers and filtered to relative masses of atoms and molecules. You answer, you find out immediately whether you were right, and you get the reasoning for the correct option and for each distractor. Wrong answers go to a mistakes locker so you can come back to exactly those.
Practice is free. You need an account only so your progress and your mistakes are still there next time.
These are the errors that cost marks on relative masses of atoms and molecules, taken from our own topic notes. Read them before you practise and you will recognise the traps in the questions.
Giving relative atomic mass or relative molecular mass a unit. Both are ratios.
Calling the Mᵣ of an ionic compound a relative molecular mass.
Ignoring the number outside a bracket, or applying it to only part of the group.
Forgetting the water of crystallisation in a hydrated salt.
Using the Mᵣ of one substance with the balancing number of another.
Leaving out the balancing number altogether, so 2MgO is treated as 40 rather than 80.
Adding masses of substances that are not in the same equation.