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

This guide teaches 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 shared by AS and A-level. In the specification’s assessment tables, section 3.1.2 sits inside the physical chemistry listed for Paper 1 and Paper 2 of each qualification; the A-level’s third paper covers any content.

Two companion pages go with this one: the condensed revision notes and an exam-style question set with worked answers. Both sections lean on moles, Ar and Mr; if those are rusty, go back to the relative mass and the mole study guide and its revision notes. Formula work returns in introduction to organic chemistry. Track your progress against the AQA A-Level Chemistry printable checklist, find every page for the course on the AQA A-Level Chemistry hub, and take one of the free diagnostics if you are unsure where your gaps lie.

What this unit covers

Spec point	You should be able to	Status
3.1.2.3	Use pV = nRT with every variable in SI units; rearrange it to find p, V, n or T	AS and A-level
3.1.2.3 (skills)	Convert units before substituting; find the Mr of a volatile liquid from gas data	AS and A-level
3.1.2.4	Define empirical formula and molecular formula, and state how they are related	AS and A-level
3.1.2.4	Calculate an empirical formula from composition by mass or percentage by mass	AS and A-level
3.1.2.4	Calculate a molecular formula from the empirical formula and Mr	AS and A-level
3.1.2.4 (skills)	Find the empirical formula of a metal oxide by experiment; judge errors and uncertainty	AS and A-level

Neither section is linked to a required practical activity. The specification suggests two practical opportunities instead, the Mr of a volatile liquid and the empirical formula of a metal oxide; both are taught below.

3.1.2.3 The ideal gas equation

The equation and its units

pV = nRT

Symbol	Quantity	SI unit
p	pressure	pascal, Pa
V	volume	cubic metre, m³
n	amount of gas	mol
R	gas constant	J K⁻¹ mol⁻¹
T	temperature	kelvin, K

You do not need to memorise R; the specification says you will not be expected to recall it. The examples on this page use R = 8.31 J K⁻¹ mol⁻¹.

The equation only works if every value is in SI units. Convert first, then substitute.

You are given	Convert to SI
kPa	multiply by 10³ to get Pa
MPa	multiply by 10⁶ to get Pa
dm³	multiply by 10⁻³ to get m³
cm³	multiply by 10⁻⁶ to get m³
°C	add 273 to get K

A useful check: 1 J = 1 Pa m³, so the units cancel to leave mol.

Rearranging

From pV = nRT you can make any variable the subject:

• n = pV / RT
• V = nRT / p
• p = nRT / V
• T = pV / nR

Combine it with n = m / M when a mass is involved. This gives M = mRT / pV, the form used for a volatile liquid.

Worked example 1: amount and mass of a gas

A sealed glass bulb of volume 0.750 dm³ contains chlorine, Cl₂, at 18 °C and 96.0 kPa. Calculate the mass of chlorine in the bulb. R = 8.31 J K⁻¹ mol⁻¹; Mr of Cl₂ = 71.0.

1. Convert: p = 96.0 × 10³ Pa; V = 0.750 × 10⁻³ m³; T = 18 + 273 = 291 K.
2. n = pV / RT = (96.0 × 10³ × 0.750 × 10⁻³) / (8.31 × 291) = 72.0 / 2418 = 0.02977 mol.
3. Mass = n × Mr = 0.02977 × 71.0 = 2.11 g (3 significant figures, matching the data).

Worked example 2: finding a temperature

0.0400 mol of nitrogen is held in a 1.20 dm³ container at 75.0 kPa. Calculate the temperature of the gas.

1. Convert: p = 75.0 × 10³ Pa; V = 1.20 × 10⁻³ m³.
2. T = pV / nR = (75.0 × 10³ × 1.20 × 10⁻³) / (0.0400 × 8.31) = 90.0 / 0.3324.
3. T = 271 K, which is about −2 °C.

Worked example 3: a volume, converted back

Calculate the volume, in dm³, occupied by 4.40 g of propane, C₃H₈ (Mr = 44.0), at 310 K and 2.00 × 10⁵ Pa.

1. n = 4.40 / 44.0 = 0.100 mol.
2. V = nRT / p = (0.100 × 8.31 × 310) / (2.00 × 10⁵) = 1.288 × 10⁻³ m³.
3. Convert back: 1.288 × 10⁻³ m³ × 10³ = 1.29 dm³.

The equation returns m³; convert at the end if needed.

Practical: the Mr of a volatile liquid

The method links AT a (measuring mass, volume and temperature), AT b (heating) and AT k (safe handling of flammable liquids).

1. Heat a gas syringe in an oven or steam jacket to a steady temperature above the liquid’s boiling point. Record the starting volume reading.
2. Weigh a small hypodermic syringe containing the liquid. Inject a few drops through the self-sealing cap into the gas syringe. Reweigh the small syringe at once. The mass injected is the difference.
3. The liquid vaporises. Record the new volume reading, the temperature and the room pressure.
4. Use M = mRT / pV with the volume of vapour (final reading minus starting reading).

Worked example 4: processing the data

0.188 g of a liquid gives 73.5 cm³ of vapour at 120 °C and 98.5 kPa. R = 8.31 J K⁻¹ mol⁻¹.

1. Convert: T = 393 K; p = 98.5 × 10³ Pa; V = 73.5 × 10⁻⁶ m³.
2. M = mRT / pV = (0.188 × 8.31 × 393) / (98.5 × 10³ × 73.5 × 10⁻⁶) = 84.8 g mol⁻¹.
3. Mr = 84.8 (no units).

Uncertainty (PS 3.3). If the balance reads to ±0.001 g, the mass by difference uses two readings, so its uncertainty is ±0.002 g, which is 1.1% of 0.188 g. A volume read to ±0.5 cm³ is 0.7% of 73.5 cm³. The combined percentage uncertainty is about 1.7%, so the Mr lies roughly between 83.3 and 86.3.

Errors that push the answer the wrong way:

• Liquid that does not fully vaporise gives a smaller volume, so the calculated Mr is too high.
• Liquid that evaporates from the needle before reweighing makes the recorded mass too large, so Mr is too high.
• Vapour escaping past a leaky cap gives a smaller volume, so Mr is again too high.
• Using the syringe reading without subtracting the starting volume gives a larger volume, so Mr is too low.

3.1.2.4 Empirical and molecular formula

Definitions

• Empirical formula: the ratio of atoms of each element in a compound, written in its simplest whole-number form.
• Molecular formula: how many atoms of each element are actually present in the compound’s formula unit, such as one molecule.

You reach the molecular formula by scaling every subscript of the empirical formula by the same whole number. For a molecule:

multiplier = Mr ÷ empirical formula mass

The multiplier can be 1. Then the two formulae are the same.

Method in steps

1. Write the mass (or percentage) of each element. Find any missing element by difference.
2. Turn each into moles of atoms using its Ar.
3. Scale the mole values so the smallest becomes 1.
4. Clear fractions: × 2 for a .5; × 3 for .33 or .67; × 4 for .25 or .75.
5. Write the formula with the whole numbers.

Keep at least three significant figures in step 2. Rounding early can turn 1.33 into 1.3 and hide the true ratio.

Worked example 5: percentage composition to molecular formula

A compound contains 40.68% carbon, 5.08% hydrogen and 54.24% oxygen by mass. Its Mr is 118. Find its molecular formula. (Ar values to use: O = 16.0, C = 12.0, H = 1.0)

	C	H	O
% by mass	40.68	5.08	54.24
÷ Ar	3.39	5.08	3.39
÷ smallest	1.00	1.50	1.00
× 2	2	3	2

Empirical formula = C₂H₃O₂. Empirical formula mass = 24.0 + 3.0 + 32.0 = 59.0. Multiplier = 118 ÷ 59.0 = 2. Molecular formula = C₄H₆O₄.

Linking the two sections: the liquid in worked example 4 is 14.1% carbon, 2.4% hydrogen and 83.5% chlorine by mass (Cl = 35.5). Moles are 1.175 : 2.4 : 2.352, which scale to 1 : 2 : 2, so the empirical formula is CH₂Cl₂ (mass 85.0). The gas data gave Mr = 84.8, so the multiplier is 1 and the molecular formula is also CH₂Cl₂.

Practical: the empirical formula of a metal oxide

Two common routes (AT a and AT k):

• Make the oxide. Heat a weighed metal in a crucible in air until the mass stops changing, then weigh the oxide. Raise the lid briefly from time to time so oxygen can enter while little solid escapes.
• Reduce the oxide. Heat a weighed oxide in a stream of a reducing gas such as hydrogen until only the metal is left, then weigh the metal. Let it cool in the gas stream so it does not reoxidise.

In both, heat to constant mass: heat, cool, weigh, and repeat until two readings agree. This shows the reaction is complete.

Worked example 6: an oxide of lead

A lead oxide is reduced by heating in hydrogen. Lead compounds are toxic, so the work is done in a fume cupboard. Results:

Reading	Mass / g
empty tube and boat	30.412
tube, boat and oxide	32.469
tube, boat and lead (constant mass)	32.277

(Ar: Pb = 207.2, O = 16.0)

1. Mass of lead = 32.277 − 30.412 = 1.865 g. Mass of oxygen = 32.469 − 32.277 = 0.192 g.
2. Moles: Pb = 1.865 / 207.2 = 0.009001; O = 0.192 / 16.0 = 0.01200.
3. Ratio Pb : O = 1 : 1.333. Multiply by 3: 3 : 4.
4. Empirical formula = Pb₃O₄.

Evaluation (PS 2.3). The oxygen mass comes from two readings of ±0.001 g, so ±0.002 g out of 0.192 g is about 1.0%. If the oxide is not fully reduced, the final mass is too high, the oxygen mass too low, and the ratio of O to Pb comes out too small.

Common errors

• Putting kPa, cm³ or °C straight into pV = nRT. Each one gives an answer out by a factor of 10³, 10⁶ or more.
• Forgetting to change m³ back into the cm³ or dm³ the question wanted.
• Rounding 1.33 or 1.50 to 1 in an empirical formula ratio.
• Using the oxide’s mass as the oxygen’s mass.
• Giving only the empirical formula when the question wants the molecular formula.
• Giving Mr with units of g mol⁻¹. Mr has no units; molar mass M does.
• Quoting more figures than the weakest measurement justifies (MS 1.1).

Next steps

• Recall drill: ideal gas and formula revision notes
• Exam-style work: ideal gas and formula practice questions
• Formula skills in context: organic analysis
• Course pages: AQA A-Level Chemistry hub; progress: printable checklist
• Find your weak spots: free diagnostics

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.