8 min readConcept sheetPart II (2 marks: state the hypothesis), Part III (4 marks: RMM = 2 × VD) or part of a Part IV answer (7 marks)

Avogadro's hypothesis and its applications

Equal volumes of all gases at the same temperature and pressure hold equal numbers of molecules. This gives the key result: RMM is twice the vapour density.

Avogadro's hypothesisEqual volumes of all gases at the same temperature and pressure contain equal numbers of molecules

Vapour density (VD)Mass of a volume of gas ÷ mass of the same volume of hydrogen, at the same temperature and pressure

Key relationRelative molecular mass = 2 × vapour density

AtomicityAtoms in one molecule = molecular mass ÷ atomic mass (O₃: 48 ÷ 16 = 3)

Applications of the hypothesis

It explains Gay-Lussac's law of combining volumes of gases.
It helps to find the atomicity of gases (the number of atoms in one molecule): He 1, O₂ 2, O₃ 3.
It helps to work out the molecular formula of gases.
It gives the relation RMM = 2 × vapour density.
It shows that one mole of every gas has the same volume, 22.4 L at STP (the gram molar volume).

Derivation: RMM = 2 × vapour density

1
VD = mass of a certain volume of the gas ÷ mass of the same volume of hydrogen (same temperature and pressure).
2
Let that volume contain n molecules. By Avogadro's hypothesis, the same volume of hydrogen also contains n molecules.
3
VD = mass of n molecules of gas ÷ mass of n molecules of hydrogen = mass of 1 molecule of gas ÷ mass of 1 molecule of hydrogen.
4
Hydrogen is diatomic, so mass of 1 molecule of H₂ = 2 × mass of 1 hydrogen atom.
5
VD = mass of 1 molecule of gas ÷ (2 × mass of 1 hydrogen atom).
6
RMM = mass of 1 molecule of gas ÷ mass of 1 hydrogen atom, so VD = RMM ÷ 2.
7★
Therefore RMM = 2 × VD.

★ key step

Using the hypothesis to show that hydrogen is diatomic

H₂(g) + Cl₂(g) → 2HCl(g). Experiment: 1 volume of hydrogen + 1 volume of chlorine → 2 volumes of hydrogen chloride.

1
By Avogadro's hypothesis: n molecules of hydrogen + n molecules of chlorine → 2n molecules of hydrogen chloride.
2
Divide by 2n: 1/2 molecule of hydrogen + 1/2 molecule of chlorine → 1 molecule of hydrogen chloride.
3
One molecule of HCl has at least one hydrogen atom, so half a hydrogen molecule holds one atom.
4
So one molecule of hydrogen has two atoms: hydrogen is diatomic (H₂). The same reasoning shows chlorine is Cl₂.

Worked example: from vapour density to volume

Problem

The vapour density of a gas is 8. Find (i) its relative molecular mass and (ii) the volume of 32 g of the gas at STP.

Working

Formula: RMM = 2 × VD
RMM = 2 × 8 = 16, so the molar mass is 16 g mol⁻¹
Number of moles = mass ÷ molar mass = 32 g ÷ 16 g mol⁻¹ = 2 mol
Volume at STP = moles × 22.4 L = 2 × 22.4 L = 44.8 L

Answer

RMM = 16. Volume of 32 g at STP = 44.8 L. (Methane, CH₄, fits these values.)

In real life

When you blow up a balloon for a birthday or a festival, each breath adds more air molecules. At the same temperature and pressure, more molecules need more space, so the balloon grows. That is Avogadro's idea at work.

Exam angle

2 marks: “State Avogadro's hypothesis.” Equal volumes of all gases under the same conditions of temperature and pressure contain equal numbers of molecules.

4 marks: “Using Avogadro's hypothesis, show that the relative molecular mass of a gas is twice its vapour density.” Define VD. Apply Avogadro's hypothesis to cancel n. Use mass of H₂ molecule = 2 × mass of H atom. Conclude RMM = 2 × VD.

7 marks: “(a) State Avogadro's hypothesis and give its applications. (b) The vapour density of a gas is 22. Find its relative molecular mass.” (a) Statement and five applications. (b) RMM = 2 × VD = 2 × 22 = 44 (carbon dioxide fits this).

Remember it

Same T, same P, same V: same number of molecules. Then RMM is double the VD.