AQA A-Level Chemistry: The Ideal Gas Equation and Empirical and Molecular Formula (7405) – Revision Notes

These notes condense 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. Everything here is explained step by step, with more examples, in the ideal gas and formula study guide.

Once these feel familiar, test them on the ideal gas and formula question set. Mole basics are in the mole revision notes. Tick off each point on the AQA A-Level Chemistry printable checklist, browse the rest of the course from the AQA A-Level Chemistry hub, and use the free diagnostics to spot weak topics.

Key definitions

• Ideal gas equation: pV = nRT, used with every variable in SI units.
• Empirical formula: the ratio of atoms of each element in a compound, in its simplest whole-number form.
• Molecular formula: how many atoms of each element are really present in the compound’s formula.
• Relationship: molecular formula = empirical formula × a whole number, where the whole number is Mr ÷ empirical formula mass.

3.1.2.3 The ideal gas equation

Formula and units

Quantity	Symbol	SI unit	Common conversion
pressure	p	Pa	kPa × 10³; MPa × 10⁶
volume	V	m³	dm³ × 10⁻³; cm³ × 10⁻⁶
amount	n	mol	n = m / M
gas constant	R	J K⁻¹ mol⁻¹	value given in the question
temperature	T	K	°C + 273

You are not expected to recall R (the specification says so). These notes use R = 8.31 J K⁻¹ mol⁻¹.

Rearrangements to know

Find	Use
n	pV / RT
V	nRT / p
p	nRT / V
T	pV / nR
M of a gas or vapour	mRT / pV

Method in steps

1. List p, V, n (or m) and T from the question.
2. Convert each to SI units before substituting.
3. Rearrange for the unknown.
4. Substitute and calculate.
5. Convert the answer to the unit asked for (m³ to dm³: × 10³; m³ to cm³: × 10⁶).
6. Round to the number of significant figures in the least precise data.

Worked reminder

0.0250 mol of argon is held in 600 cm³ at 300 K. Find the pressure in kPa.

• V = 600 × 10⁻⁶ m³ = 6.00 × 10⁻⁴ m³
• p = nRT / V = (0.0250 × 8.31 × 300) / (6.00 × 10⁻⁴) = 1.04 × 10⁵ Pa = 104 kPa

Mr of a volatile liquid in brief

• Inject a weighed mass of liquid into a heated gas syringe held above the liquid’s boiling point.
• Mass injected = mass of small syringe before − after.
• Volume of vapour = final reading − starting reading.
• M = mRT / pV with T the syringe temperature and p the room pressure.
• Errors that make Mr too high: liquid not fully vaporised, vapour leaking, liquid lost from the needle before reweighing.

3.1.2.4 Empirical and molecular formula

Method in steps

1. Masses or percentages of each element (find oxygen or a missing element by difference).
2. Divide each by Ar.
3. Divide all by the smallest.
4. Clear fractions: × 2 for a .5; × 3 for .33 or .67; × 4 for .25 or .75.
5. Write the empirical formula.
6. For the molecular formula: multiplier = Mr ÷ empirical formula mass; multiply every subscript.

Worked reminder

A carbide contains 75.0% aluminium and 25.0% carbon by mass. (Use C = 12.0 and Al = 27.0.)

	Al	C
÷ Ar	75.0 / 27.0 = 2.778	25.0 / 12.0 = 2.083
÷ smallest	1.333	1.000
× 3	4	3

Empirical formula Al₄C₃. Rounding 1.333 to 1 would give AlC, which is wrong.

Metal oxide practical in brief

• Make the oxide by heating a weighed metal in a crucible, or reduce a weighed oxide in a stream of hydrogen.
• Heat to constant mass so you know the reaction is complete.
• Mass of oxygen = mass of oxide − mass of metal.
• Each mass by difference uses two balance readings, so double the reading uncertainty.

Linking the two sections

Longer problems often join pV = nRT to formula work. The pattern is always the same: the gas data give Mr, the composition gives the empirical formula, and Mr ÷ empirical formula mass gives the multiplier.

Worked reminder

A gaseous fluorocarbon is 24.0% carbon and 76.0% fluorine by mass. A 0.450 g sample occupies 110 cm³ at 300 K and 102 kPa. Find its molecular formula. (Use C = 12.0 and F = 19.0.)

1. Moles of gas: n = pV / RT = (102 × 10³ × 110 × 10⁻⁶) / (8.31 × 300) = 4.50 × 10⁻³ mol.
2. Mr = 0.450 ÷ (4.50 × 10⁻³) = 100.
3. Ratio: C 24.0 / 12.0 = 2.00; F 76.0 / 19.0 = 4.00; scale to 1 : 2. Empirical formula CF₂ (mass 50.0).
4. Multiplier = 100 ÷ 50.0 = 2. Molecular formula C₂F₄.

Note the order: composition alone cannot give the molecular formula, because CF₂, C₂F₄ and C₃F₆ share the same percentages. Only the Mr tells them apart.

Significant figures and uncertainty

• Round the final answer to match the least precise piece of data (MS 1.1). In the reminder above the data are to 3 significant figures, so the amount is written as 4.50 × 10⁻³ mol, keeping the final zero.
• Carry unrounded values through between steps. Round only once, at the end.
• A percentage uncertainty is (uncertainty ÷ measured value) × 100. For a 0.450 g mass found from two readings of ±0.001 g, that is (0.002 ÷ 0.450) × 100 = 0.44%.
• To combine uncertainties in a product or quotient such as mRT / pV, add the percentage uncertainties of the measured quantities.

Must-know distinctions

• Empirical vs molecular formula. One is a simplest ratio, the other the true atom count; they coincide when the multiplier is 1.
• Mr vs M. Mr has no units. Molar mass M is in g mol⁻¹. M = mRT / pV gives M; quote Mr as the same number without units.
• Pa vs kPa. pV = nRT needs Pa. A value like “101” is almost certainly kPa.
• m³ vs dm³ vs cm³. 1 m³ = 10³ dm³ = 10⁶ cm³.
• °C vs K. Never put a Celsius temperature into pV = nRT.
• Composition by mass vs percentage by mass. Same method; with percentages you treat the values as masses in a 100 g sample.

Quick self-test

1. Convert 2.5 MPa to Pa.
2. Convert 45.0 cm³ to m³.
3. Convert −23 °C to K.
4. Calculate the amount of gas in 2.00 dm³ at 100 kPa and 400 K.
5. Calculate the volume, in dm³, of 0.500 mol of gas at 500 K and 200 kPa.
6. 0.375 g of a gas occupies 210 cm³ at 300 K and 101 kPa. Calculate its Mr.
7. 0.640 g of sulfur combines with 0.960 g of oxygen. Find the empirical formula. (S = 32.1, O = 16.0)
8. A compound is 30.4% nitrogen and 69.6% oxygen by mass, with Mr = 92.0. Find its molecular formula. (N = 14.0, O = 16.0)
9. A hydrocarbon has empirical formula C₃H₇ and Mr = 86.0. Find its molecular formula.
10. A mole ratio comes out as C 1.00 : H 2.67. Write the empirical formula.
11. In a volatile-liquid experiment, some liquid stays as droplets. Is the calculated Mr too high or too low? Explain.

Answers

1. 2.5 × 10⁶ Pa.
2. 4.50 × 10⁻⁵ m³.
3. 250 K.
4. n = (100 × 10³ × 2.00 × 10⁻³) / (8.31 × 400) = 0.0602 mol.
5. V = (0.500 × 8.31 × 500) / (200 × 10³) = 1.04 × 10⁻² m³ = 10.4 dm³.
6. M = (0.375 × 8.31 × 300) / (101 × 10³ × 210 × 10⁻⁶) = 44.1.
7. S: 0.640 / 32.1 = 0.01994; O: 0.960 / 16.0 = 0.0600; ratio 1 : 3.01, so SO₃.
8. N: 2.171; O: 4.350; ratio 1 : 2.00, so NO₂ (46.0). 92.0 ÷ 46.0 = 2, so N₂O₄.
9. C₃H₇ = 43.0; 86.0 ÷ 43.0 = 2, so C₆H₁₄.
10. Multiply by 3: 3 : 8, so C₃H₈.
11. Too high. The vapour volume is too small, and M = mRT / pV has V on the bottom.

Where marks are usually lost

• Substituting kPa, cm³ or °C into pV = nRT without converting.
• Converting cm³ to m³ with ÷ 1000 (that gives dm³).
• Forgetting to change m³ back into the cm³ or dm³ the question wanted.
• Rounding mole values to one or two figures before dividing, so a 1.33 ratio is missed.
• Rounding 1.5 down to 1 instead of multiplying the whole ratio by 2.
• Using the mass of oxide in place of the mass of oxygen.
• Writing down only the empirical formula when the molecular one is required.
• Showing no division of Mr by the empirical formula mass, so the multiplier is not justified.
• Giving units with Mr, or quoting more significant figures than the data support.
• Saying “heat it for longer” instead of “heat to constant mass” when asked how to make sure a reaction is complete.

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.