AQA A-Level Chemistry: The Ideal Gas Equation and Empirical and Molecular Formula (7405) – 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 sections 3.1.2.3 The ideal gas equation and 3.1.2.4 Empirical and molecular formula of the AQA AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for exams from June 2016 onwards. Both sections are common to AS and A-level, so every question suits both courses. Questions 9 to 11 combine the two sections with practical skills.

If any method is unfamiliar, read the ideal gas and formula study guide or skim the ideal gas and formula revision notes before you start. Mark your progress on the AQA A-Level Chemistry printable checklist, see every page for the course on the AQA A-Level Chemistry hub, and try the free diagnostics to find other gaps.

Use R = 8.31 J K⁻¹ mol⁻¹ throughout. Relative atomic masses for this set: hydrogen 1.0, carbon 12.0, nitrogen 14.0, oxygen 16.0, fluorine 19.0, silicon 28.1, sulfur 32.1, vanadium 50.9, tin 118.7.

Questions

1. State the ideal gas equation and give the SI unit of pressure, volume and temperature used in it. [3]

2. Define the terms empirical formula and molecular formula. [2]

3. A gas sample has a volume of 350 cm³ at 85.0 kPa and 27 °C. Convert each of these values to the units needed for the ideal gas equation. [3]

4. Calculate the volume, in cm³, occupied by 1.28 g of sulfur dioxide, SO₂ (Mr = 64.1), at 45 °C and 112 kPa. Give your answer to 3 significant figures. [4]

5. 2.04 g of vanadium is heated in oxygen until the mass is constant. The oxide formed has a mass of 3.64 g. Calculate the empirical formula of the oxide. [3]

6. A gaseous hydride of silicon contains 90.4% silicon and 9.6% hydrogen by mass. Its Mr is 62.2. Calculate its empirical formula and deduce its molecular formula. [4]

7. A 5.00 dm³ steel cylinder holds 0.180 mol of helium at a pressure of 92.4 kPa. Calculate the temperature of the helium in kelvin. [3]

8. An evacuated glass bulb of volume 400 cm³ is filled with an unknown gas at 20 °C and 101 kPa. The mass of the bulb increases by 0.712 g. Calculate the Mr of the gas. [3]

9. A student finds the empirical formula of an oxide of tin. Tin is warmed with concentrated nitric acid in a crucible, and the residue is then heated strongly. Results:

Reading	Mass / g
empty crucible	21.346
crucible and tin	22.536
crucible and tin oxide (constant mass)	22.857

(a) Use the results to work out the empirical formula of the tin oxide. [3] (b) Give the reason for heating the residue to constant mass. [1] (c) The balance reads to ±0.001 g. Calculate the percentage uncertainty in the mass of oxygen. [1] (d) Suggest why the reaction with nitric acid is done in a fume cupboard. [1]

10. A volatile liquid X contains 54.5% carbon, 9.1% hydrogen and 36.4% oxygen by mass. In an experiment, 0.186 g of X is injected into a gas syringe held in an oven at 110 °C. It vaporises completely, giving 67.9 cm³ of vapour at 99.0 kPa.

(a) Calculate the empirical formula of X. [3] (b) Calculate the Mr of X from the gas data. [3] (c) Deduce the molecular formula of X. [1] (d) State why the oven must be above the boiling point of X, and explain the effect on the calculated Mr if some vapour leaked out of the syringe. [2]

11. A 0.520 g sample of a gaseous fluoride of nitrogen contains 0.140 g of nitrogen and 0.380 g of fluorine. The sample occupies 125 cm³ at 30 °C and 101 kPa.

(a) Calculate the empirical formula of the compound. [2] (b) Calculate the amount, in mol, of gas in the sample. [2] (c) Calculate the Mr of the compound and deduce its molecular formula. [3] (d) The container is sealed and heated to 150 °C. Calculate the new pressure in kPa. [2]

Answers

1. pV = nRT [1]. Pressure in Pa and volume in m³ [1]. Temperature in K [1]. [3] Examiner insight: Give the units as named SI units; writing “kPa” or “dm³” here would contradict the requirement that every variable is in SI units.

2. Empirical formula: the ratio of atoms of each element in the compound, in its simplest whole-number form [1]. Molecular formula: the true number of atoms of each element present in the compound’s formula [1]. [2] Examiner insight: “Simplest ratio” without “whole number” is incomplete; include both words to make the definition precise.

3. V = 350 × 10⁻⁶ = 3.50 × 10⁻⁴ m³ [1]. p = 85.0 × 10³ = 8.50 × 10⁴ Pa [1]. T = 27 + 273 = 300 K [1]. [3] Examiner insight: Write each conversion out on its own line so every step can be credited; dividing cm³ by 1000 gives dm³, not m³.

4. n = 1.28 / 64.1 = 0.01997 mol [1]. Convert: p = 112 × 10³ Pa, T = 318 K [1]. V = nRT / p = (0.01997 × 8.31 × 318) / (112 × 10³) = 4.71 × 10⁻⁴ m³ [1]. V = 471 cm³ [1]. [4] Examiner insight: Converting m³ back to cm³ is part of the answer; an answer left as 4.71 × 10⁻⁴ does not answer the question as set.

5. Mass of oxygen = 3.64 − 2.04 = 1.60 g [1]. Moles: V = 2.04 / 50.9 = 0.0401; O = 1.60 / 16.0 = 0.100 [1]. Ratio 1 : 2.5 = 2 : 5, so V₂O₅ [1]. [3] Examiner insight: Using 3.64 g as the mass of oxygen is the usual slip; show the subtraction so the oxygen mass is clear.

6. Si: 90.4 / 28.1 = 3.217; H: 9.6 / 1.0 = 9.6 [1]. Ratio 1 : 2.98 ≈ 1 : 3, empirical formula SiH₃ [1]. Empirical formula mass = 28.1 + 3.0 = 31.1; 62.2 ÷ 31.1 = 2 [1]. Molecular formula Si₂H₆ [1]. [4] Examiner insight: Show the division 62.2 ÷ 31.1 = 2; a molecular formula stated with no link to Mr leaves the reasoning unseen.

7. p = 92.4 × 10³ Pa and V = 5.00 × 10⁻³ m³ [1]. T = pV / nR = (92.4 × 10³ × 5.00 × 10⁻³) / (0.180 × 8.31) [1]. T = 309 K [1]. [3] Examiner insight: An answer near 0.309 K signals kPa left unconverted; check that a gas temperature is physically sensible before moving on.

8. Convert: V = 4.00 × 10⁻⁴ m³, p = 101 × 10³ Pa, T = 293 K [1]. M = mRT / pV = (0.712 × 8.31 × 293) / (101 × 10³ × 4.00 × 10⁻⁴) [1]. Mr = 42.9 [1]. [3] Examiner insight: Quote Mr without units; if you give a molar mass instead, the unit must be g mol⁻¹.

9. (a) Mass of tin = 22.536 − 21.346 = 1.190 g; mass of oxygen = 22.857 − 22.536 = 0.321 g [1]. Moles: Sn = 1.190 / 118.7 = 0.01003; O = 0.321 / 16.0 = 0.02006 [1]. Ratio 1 : 2.00, so SnO₂ [1]. (b) To make sure the reaction is complete and all water and other volatile matter has been driven off [1]. (c) Two readings give ±0.002 g; (0.002 / 0.321) × 100 = 0.62% [1]. (d) Toxic gas (nitrogen dioxide) is given off [1]. [6] Examiner insight: In (c), a mass found by difference uses two readings, so the uncertainty is doubled before it is divided by the mass.

10. (a) C: 54.5 / 12.0 = 4.54; H: 9.1 / 1.0 = 9.1; O: 36.4 / 16.0 = 2.275 [1]. Divide by 2.275: 2.00 : 4.00 : 1 [1]. Empirical formula C₂H₄O [1]. (b) Convert: T = 383 K, p = 99.0 × 10³ Pa, V = 67.9 × 10⁻⁶ m³ [1]. M = mRT / pV = (0.186 × 8.31 × 383) / (99.0 × 10³ × 67.9 × 10⁻⁶) [1]. Mr = 88.1 [1]. (c) C₂H₄O = 44.0; 88.1 ÷ 44.0 = 2, so C₄H₈O₂ [1]. (d) So that all of X is a gas and its whole mass contributes to the vapour volume [1]. A leak makes the measured volume too small, so the calculated Mr is too high [1]. [9] Examiner insight: For (d), a direction (“too high”) without the reason (V is smaller and sits on the bottom of mRT / pV) is unlikely to earn the explanation mark.

11. (a) N: 0.140 / 14.0 = 0.0100; F: 0.380 / 19.0 = 0.0200 [1]. Ratio 1 : 2, empirical formula NF₂ [1]. (b) p = 101 × 10³ Pa, V = 1.25 × 10⁻⁴ m³, T = 303 K [1]. n = pV / RT = (101 × 10³ × 1.25 × 10⁻⁴) / (8.31 × 303) = 5.01 × 10⁻³ mol [1]. (c) Mr = 0.520 ÷ (5.01 × 10⁻³) = 104 [1]. NF₂ = 14.0 + 38.0 = 52.0 [1]. 104 ÷ 52.0 = 2, so N₂F₄ [1]. (d) n and V are fixed, so p is proportional to T (in kelvin) [1]. p = 101 × 423 / 303 = 141 kPa [1]. [9] Examiner insight: In (d), scaling by 150 / 30 in °C gives 505 kPa; both temperatures must be in kelvin before any ratio is taken.

Where marks are usually lost

• Putting kPa, cm³ or °C straight into pV = nRT.
• Dividing cm³ by 1000 and calling the result m³.
• Leaving a final volume in m³ when cm³ or dm³ was asked for.
• Taking ratios of Celsius temperatures instead of kelvin temperatures.
• Rounding a 1 : 2.5 or 1 : 1.33 ratio to the nearest whole number instead of multiplying through.
• Using the mass of the oxide as the mass of oxygen.
• Ending with the empirical formula when the molecular formula is the target.
• Forgetting that a mass by difference has two readings, so twice the reading uncertainty.
• Stating that an error makes Mr “inaccurate” without saying whether it is too high or too low, and why.

Next steps

• Ideal gas and formula revision notes
• Ideal gas and formula study guide
• Mole practice questions
• Organic analysis practice
• Course pages: AQA A-Level Chemistry hub
• Progress tracker: printable checklist
• Other gaps: all free 10-minute diagnostics
• Prefer live teaching? Book a free trial class

Official syllabus

Source document: AQA, AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams June 2016 onwards: section 3.1.2 Amount of substance, sub-sections 3.1.2.3 The ideal gas equation and 3.1.2.4 Empirical and molecular formula, with their opportunities for skills development.