OxfordAQA A-Level Chemistry: Amount of Substance (9620) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section 3.1.2 (Amount of substance) of the OxfordAQA International AS and A-level Chemistry 9620 specification, Version 5.3. Constants where needed: Avogadro constant L = 6.02 × 10²³ mol⁻¹; gas constant R = 8.31 J K⁻¹ mol⁻¹. Ar values are given in the questions that need them. Work through the revision notes or the study guide first if a method is unfamiliar. The course hub and checklist show what else to cover.

Questions

1. Define (a) relative atomic mass and (b) relative molecular mass, in terms of carbon-12. [2] [2]

2. Ammonium iron(II) sulfate crystals have the formula (NH₄)₂Fe(SO₄)₂·6H₂O. (Ar: H 1.0, N 14.0, O 16.0, S 32.1, Fe 55.8)

(a) Calculate the relative formula mass. [3] (b) Explain why the term relative formula mass is used for this compound. [1]

3. Calculate the number of hydrogen atoms in 3.40 g of ammonia, NH₃ (Mr 17.0). [4]

4. A technician needs 250 cm³ of sodium hydroxide solution of concentration 0.150 mol dm⁻³ (Mr of NaOH 40.0).

(a) Calculate the mass of NaOH required. [2] (b) 25.0 cm³ of this solution is diluted with water to 500 cm³. Calculate the new concentration. [2]

5. Describe how you would make up 250 cm³ of a standard solution from a weighed solid sample, naming the apparatus you use. [4]

6. A sample of gas of mass 0.350 g occupies 215 cm³ at 330 K and 101 kPa. Calculate the relative molecular mass of the gas and suggest one hydrocarbon it could be. [5]

7. A compound of phosphorus and oxygen contains 4.96 g of phosphorus and 6.40 g of oxygen (Ar: P 31.0, O 16.0). Its relative molecular mass is 284. Determine the empirical formula and the molecular formula. [5]

8. (a) Balance the equation: Al + Fe₃O₄ → Al₂O₃ + Fe [2] (b) Aqueous barium chloride reacts with aqueous sodium sulfate to form a precipitate. Write the ionic equation, including state symbols. [2] (c) Name the spectator ions in part (b). [1]

9. Calcium carbonate dissolves in hydrochloric acid according to CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O. A 2.50 g sample of CaCO₃ (Ar: Ca 40.1, C 12.0, O 16.0) reacts completely. Calculate the volume of 1.00 mol dm⁻³ hydrochloric acid needed, and the volume of carbon dioxide released at 298 K and 101 kPa. [6]

10. Urea can be made by this reaction: 2NH₃ + CO₂ → (NH₂)₂CO + H₂O (Mr: NH₃ 17.0, CO₂ 44.0, (NH₂)₂CO 60.0).

(a) Calculate the percentage atom economy for making urea. [3] (b) 51.0 g of ammonia reacts with excess carbon dioxide and 72.9 g of urea is isolated. Calculate the percentage yield. [4] (c) Give two advantages of developing a process with a high atom economy. [2]

11. A 1.00 g sample of impure solid sodium hydroxide is dissolved and made up to 250 cm³. In a titration, 25.0 cm³ portions of this solution need a mean of 22.95 cm³ of 0.100 mol dm⁻³ hydrochloric acid.

(a) Give the ionic equation for the neutralisation. [1] (b) Calculate the amount of HCl used. [1] (c) State the amount of NaOH in each 25.0 cm³ portion. [1] (d) Calculate the amount of NaOH in the 250 cm³ solution. [1] (e) Calculate the mass of NaOH, then the percentage purity of the sample (Mr NaOH 40.0). [2] (f) Suggest why a pipette is used for the 25.0 cm³ portions instead of a measuring cylinder. [1]

12. A hydrocarbon of mass 0.700 g burns completely to give 2.20 g of carbon dioxide and 0.900 g of water (Ar: H 1.0, C 12.0; Mr: CO₂ 44.0, H₂O 18.0). A separate 0.700 g sample, as vapour, occupies 364 cm³ at 350 K and 100 kPa.

(a) Calculate the amounts of carbon and hydrogen atoms in the sample. [2] (b) Deduce the empirical formula. [2] (c) Calculate the relative molecular mass from the gas data. [3] (d) Deduce the molecular formula. [2] (e) Write a balanced equation for the complete combustion of this hydrocarbon. [1]

Answers

1. (a) Relative atomic mass is the weighted mean mass of an atom of the element, allowing for its isotopes, [1] divided by one twelfth of the mass of an atom of carbon-12. [1] (b) Relative molecular mass is the mean mass of a molecule [1] divided by one twelfth of the mass of an atom of carbon-12. [1] Examiner insight: A definition usually needs both the averaging idea and the one-twelfth comparison; saying only “compared with carbon-12” is often too vague.

2. (a) (NH₄)₂ = 2 × (14.0 + 4.0) = 36.0 [1]; Fe + 2 SO₄ = 55.8 + 2 × (32.1 + 64.0) = 248.0 [1]; 6H₂O = 6 × 18.0 = 108.0, giving a total of 392.0 [1] (b) The compound is ionic, so it exists as a lattice of ions and not as separate molecules. [1] Examiner insight: Showing the bracket multipliers one group at a time makes slips easy to spot and usually keeps method credit available when the total is wrong.

3. n(NH₃) = 3.40 / 17.0 = 0.200 mol [1]; n(H atoms) = 3 × 0.200 = 0.600 mol [1]; number = 0.600 × 6.02 × 10²³ [1] = 3.61 × 10²³ atoms [1] Examiner insight: The question asks for atoms of hydrogen, not molecules. Answers usually lose credit by stopping at 1.20 × 10²³ molecules.

4. (a) n = 0.150 × 0.250 = 0.0375 mol [1]; m = 0.0375 × 40.0 = 1.50 g [1] (b) New concentration = 0.150 × 25.0 / 500 [1] = 0.00750 mol dm⁻³ [1] Examiner insight: Convert cm³ to dm³ in the first line. Mixing cm³ and dm³ is the most common source of a factor-of-1000 error.

5. Weigh the solid accurately (for example by difference, using a weighing bottle). [1] Dissolve it in a beaker in less than 250 cm³ of distilled water, stirring. [1] Transfer to a 250 cm³ volumetric flask, rinsing the beaker and funnel into the flask. [1] Fill so the meniscus bottom sits on the mark, then stopper and invert to mix. [1] Examiner insight: The rinsing step and the choice of a volumetric flask (not a beaker or measuring cylinder) are the details that usually separate full and partial credit.

6. Convert: V = 215 × 10⁻⁶ = 2.15 × 10⁻⁴ m³ and p = 101 000 Pa [1]; n = pV / RT = (101 000 × 2.15 × 10⁻⁴) / (8.31 × 330) [1] = 0.00792 mol [1]; Mr = 0.350 / 0.00792 = 44.2 [1]; propane, C₃H₈ (Mr 44.0) [1] Examiner insight: A line showing the conversions before substitution is what examiners can credit; a bare 44.2 with unconverted units shows no evidence of method.

7. n(P) = 4.96 / 31.0 = 0.160 mol [1]; n(O) = 6.40 / 16.0 = 0.400 mol [1]; ratio 1 : 2.5, so multiply by 2 to give 2 : 5, so the empirical formula is P₂O₅ [1]; empirical mass = 142 [1]; 284 / 142 = 2, so the molecular formula is P₄O₁₀ [1] Examiner insight: A ratio of 1 : 2.5 should be doubled. Rounding it to 1 : 2 or 1 : 3 is a typical way of losing the final marks.

8. (a) 8Al + 3Fe₃O₄ → 4Al₂O₃ + 9Fe (correct formulae balanced; [1] for the Al₂O₃ and Fe coefficients, [1] for the rest) (b) Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s) [1] for species, [1] for state symbols (c) Na⁺ and Cl⁻ [1] Examiner insight: An ionic equation is expected to balance for charge as well as atoms, and spectator ions on either side usually cost credit.

9. n(CaCO₃) = 2.50 / 100.1 = 0.02498 mol [1]; n(HCl) = 2 × 0.02498 = 0.04995 mol [1]; V(HCl) = 0.04995 / 1.00 = 0.0500 dm³ = 50.0 cm³ [1]; n(CO₂) = 0.02498 mol [1]; V = nRT / p = (0.02498 × 8.31 × 298) / 101 000 [1] = 6.12 × 10⁻⁴ m³ = 612 cm³ [1] Examiner insight: Mole ratios come from the balanced equation. Using 1 : 1 for HCl is the usual slip, and a clear ratio line helps to show the method.

10. (a) Sum of reactant masses = 2 × 17.0 + 44.0 = 78.0 [1]; (60.0 / 78.0) × 100 [1] = 76.9% [1] (b) n(NH₃) = 51.0 / 17.0 = 3.00 mol [1]; n(urea) = 1.50 mol [1]; theoretical mass = 1.50 × 60.0 = 90.0 g [1]; yield = (72.9 / 90.0) × 100 = 81.0% [1] (c) Any two from: less raw material wasted, which lowers costs; less waste to dispose of or treat; conserves finite resources; less environmental harm. [1] [1] Examiner insight: Atom economy uses the equation alone while yield uses the experiment. Answers to (c) tend to earn credit when they name a consequence for industry or society, not just “less waste”.

11. (a) H⁺ + OH⁻ → H₂O [1] (b) n(HCl) = 0.02295 × 0.100 = 2.295 × 10⁻³ mol [1] (c) 1 : 1 ratio, so n(NaOH) = 2.295 × 10⁻³ mol [1] (d) × 10 for the whole 250 cm³: 0.02295 mol [1] (e) Mass = 0.02295 × 40.0 = 0.918 g [1]; purity = 0.918 / 1.00 × 100 = 91.8% [1] (f) A pipette measures a fixed volume with much smaller uncertainty. [1] Examiner insight: Scaling up from the portion to the whole solution is a separate step to show; skipping it is a typical way to get a purity that is ten times too small.

12. (a) n(C) = n(CO₂) = 2.20 / 44.0 = 0.0500 mol [1]; n(H) = 2 × 0.900 / 18.0 = 0.100 mol [1] (b) Ratio C : H = 0.0500 : 0.100 = 1 : 2 [1], so the empirical formula is CH₂ [1] (c) Convert to SI: V = 3.64 × 10⁻⁴ m³ [1]; n = (100 000 × 3.64 × 10⁻⁴) / (8.31 × 350) = 0.01252 mol [1]; Mr = 0.700 / 0.01252 = 55.9 [1] (d) Empirical mass = 14.0; 55.9 / 14.0 = 4 [1], so the molecular formula is C₄H₈ [1] (e) C₄H₈ + 6O₂ → 4CO₂ + 4H₂O [1] Examiner insight: A multi-part question rewards a clear chain of results. Giving each intermediate value with its unit lets later parts be credited even if an earlier slip has occurred.

Where marks are usually lost

• Putting cm³ or kPa directly into pV = nRT.
• Forgetting that each H₂O or NH₃ molecule contains more than one atom of the element being counted.
• Skipping the scaling step between a titration portion and the whole solution.
• Taking a 1 : 1 mole ratio where the equation gives 1 : 2.
• Rounding empirical-formula ratios such as 2.5 or 1.33 instead of multiplying to whole numbers.
• Mixing up percentage yield (compares masses made) and atom economy (compares masses in the equation).
• Describing high atom economy as “good for the environment” without saying why.
• Leaving out state symbols or including spectator ions when an ionic equation is requested.

Next steps

• Re-read the revision notes and the study guide.
• Tick off what you have covered on the printable checklist and browse the course hub.
• Try the free diagnostics to find your next weak spot.
• Book a free trial class if you want to work through these with a teacher.

Frequently asked questions

What does OxfordAQA A-level Chemistry 3.1.2 Amount of substance cover? Section 3.1.2 of OxfordAQA International AS and A-level Chemistry (9620) covers relative masses, the mole and the Avogadro constant, solution concentrations, the ideal gas equation, empirical and molecular formulae, and calculations from balanced equations including yield and atom economy.

What are common mistakes in amount of substance calculations? Frequent errors include putting cm³ or kPa directly into pV = nRT, forgetting that a molecule such as NH₃ contains more than one atom of the element being counted, skipping the scaling step from a titration portion to the whole solution, and using a 1 : 1 ratio where the equation gives 1 : 2.

What units are used in pV = nRT? Use pressure in Pa, volume in m³ and temperature in K. Putting cm³ or kPa directly into the equation is one of the most common mistakes in this section.

What is the difference between percentage yield and atom economy? Percentage yield compares the mass of product actually made with the theoretical mass, so it comes from the experiment. Atom economy compares the mass of desired product with the total mass of reactants in the balanced equation.

Official syllabus

OxfordAQA, International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards, section 3.1.2 Amount of substance.