Practise 30 real questions on Atoms/molecules/stoichiometry from 2021 to 2026. See the answer and explanation as soon as you answer.
Atoms/molecules/stoichiometry is one of the 18 main topics in our A Level Chemistry (Cambridge (CAIE) 9701) question bank, and about 5% of its real questions come from it. Tayyari Karo has 30 real questions on this topic from 2021 to 2026, from Paper 1, Paper 2; 29 are multiple choice and marked instantly, and the rest are structured questions with a model answer. By difficulty: 19 medium, 8 hard, 3 easy.
Work through the questions below, then open the explanation for each one. Redo the ones you missed a few days later.
These are real past-paper questions. Think of your answer first, then open the explanation.
Which statement is correct?
Answer: C. The relative formula mass of magnesium chloride is 95.3.
MgCl2 = 24.3 + 2(35.5) = 95.3. NaCl is a giant lattice, so it has relative formula mass (not Mr), and it is 58.5.
ICl3 reacts with water in a redox reaction: vICl3 + wH2O → xHCl + yHI + zHIO3 + zHClO3. What are the numbers v, x and z in the correctly balanced equation? (given as v, x, z)
Answer: C. v=3, x=8, z=1
Balancing I, Cl, H and O atoms across the equation gives v=3, x=8, z=1.
A sample of magnesium contains the isotopes 24Mg, 25Mg and 26Mg only. The percentage abundance of 25Mg and 26Mg is the same. The relative atomic mass of magnesium in the sample is 24.3. What is the percentage abundance of 24Mg?
Answer: D. 80%
Let 24Mg = p%, so 25Mg = 26Mg = (100−p)/2 % each. 24p + 51(100−p)/2 = 2430 (×100 form) ⇒ 48p + 51(100−p) = 4860 ⇒ 48p + 5100 − 51p = 4860 ⇒ −3p = −240 ⇒ p = 80%.
A 1.7 g sample of Mg reacts with 50.0 cm3 of 2.2 mol dm⁻3 HCl at 303 K and 110 400 Pa (assume ideal gas). Which volume of gas is produced, measured under these conditions?
Answer: A. 1.3 dm3
mol Mg = 1.7/24.3 = 0.0700; mol HCl = 0.050×2.2 = 0.110 (limiting, since Mg + 2HCl → MgCl2 + H2 needs 0.140 mol HCl for all Mg); mol H2 = 0.110/2 = 0.055; V = nRT/p = 0.055×8.31×303/110400 ≈ 1.26 dm3 ≈ 1.3 dm3.
A 3.0 g sample of Na2CO3 powder is stirred into 50 cm3 of 1.0 mol dm⁻3 HCl. The volume of CO2 produced is 600 cm3. Na2CO3(s) + 2HCl(aq) → 2NaCl(aq) + CO2(g) + H2O(l). [Mr: Na2CO3, 106.0]. Which volume of CO2 is produced if 1.0 g of Na2CO3 powder is stirred into 50 cm3 of 1.0 mol dm⁻3 HCl under the same conditions?
Answer: C. 226 cm3
3.0 g Na2CO3 = 0.0283 mol (excess relative to 0.05 mol HCl, so HCl limiting, giving 600 cm3 CO2 from 0.025 mol HCl→0.0125 mol CO2). With 1.0 g Na2CO3 = 0.00943 mol, now Na2CO3 is limiting (needs 0.01887 mol HCl < 0.05 mol available), producing 0.00943 mol CO2, i.e. 600 × (0.00943/0.0125) ≈ 226 cm3.
What is the minimum mass of oxygen required to ensure the complete combustion of 12 dm³ of propane measured under room conditions?
Answer: B. 80 g
C3H8 + 5O2 → 3CO2 + 4H2O. 12 dm³ propane ÷ 24 dm³ mol⁻¹ = 0.5 mol; O2 needed = 0.5 × 5 = 2.5 mol × 32 g mol⁻¹ = 80 g.
| Session | Paper 1 | Paper 2 |
|---|---|---|
| May/June 2026 | 4 | – |
| May/June 2025 | 2 | – |
| October/November 2025 | 4 | – |
| October/November 2024 | 3 | – |
| May/June 2023 | 3 | – |
| October/November 2023 | 2 | – |
| May/June 2022 | 3 | – |
| October/November 2022 | 2 | – |
| May/June 2021 | 3 | 1 |
| October/November 2021 | 3 | – |
30 real A Level Chemistry (9701) questions on Atoms/molecules/stoichiometry from 2021 to 2026, from Paper 1, Paper 2. Each one shows the year, session and paper it came from.
Yes. Every question shows the correct answer and a short explanation the moment you answer.
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