Vapour Density and Molecular Mass — Formula and Worked Examples
Long before mass spectrometers existed, chemists found molecular masses by weighing gases. The method survives in the syllabus because it is quick, it is examinable, and it leads straight into one of the most useful ideas in equilibrium: when the measured molecular mass disagrees with the formula, the molecule must be splitting up or joining together. This guide covers vapour density, the two routes from gas density to molar mass, and abnormal vapour density with full working.
What vapour density means
Vapour density (VD) is a ratio: the density of a gas compared with the density of hydrogen gas measured at the same temperature and pressure.
Because equal volumes of gases at the same temperature and pressure contain equal numbers of molecules (Avogadro's law), that ratio of densities is also the ratio of molecular masses:
Two things follow immediately, and both are asked in one-mark questions:
- Vapour density has no unit. It is a pure number, because it is a ratio of two densities. Density in g/L is a completely different quantity.
- Strictly M(H₂) = 2.016 g/mol, so M = 2.016 × VD. School practice rounds this to 2, and the error is about 0.8% — negligible for the kind of numbers these questions use.
Worked example 1 — the simplest kind
Problem: A gas has a vapour density of 22. What is its molecular mass, and which gas could it be?
M = 2 × 22 = 44 g/mol
Candidates: CO₂ (44.01), N₂O (44.01) and propane C₃H₈ (44.10) all fit.
This is the honest limit of the method: it gives a mass, not an identity. To decide between the candidates you need chemical tests or a spectrum. An exam answer that names one gas without saying "or" is over-claiming.
Route 1 — molar mass from density at STP
One mole of any ideal gas occupies the same molar volume at a given temperature and pressure, so:
The STP conventions genuinely differ, and you must know which one your question uses:
| Convention | Conditions | Molar volume | Where you meet it |
|---|---|---|---|
| Older STP | 0 °C (273.15 K) and 1 atm (101.325 kPa) | 22.4 L/mol | Most Indian school textbooks and older papers |
| Current IUPAC STP | 0 °C (273.15 K) and 1 bar (100 kPa) | 22.7 L/mol | Newer syllabus material and IUPAC documents |
Neither is wrong — they are different reference pressures. What is wrong is mixing them, because the density of the gas also depends on the pressure it was measured at.
Problem: A gas has a density of 1.25 g/L at 0 °C and 1 atm. Find its molar mass.
M = 1.25 × 22.4 = 28.0 g/mol — consistent with N₂ (28.02) or CO (28.01).
The same gas at 1 bar. Lower pressure means lower density: the density
becomes 1.234 g/L. Using the matching molar volume,
M = 1.234 × 22.7 = 28.0 g/mol
Both conventions give the same molar mass, provided the density and the molar volume come from the same conditions. Pairing a 1 atm density with 22.7 L/mol would have given 28.4 — wrong by 1.3%, and wrong for no reason other than carelessness.
Vapour density here = M ÷ 2 = 14.0.
Route 2 — the ideal gas equation, which works at any conditions
You do not need STP at all. Combining PV = nRT with n = w ÷ M:
Problem: 0.550 g of a gas occupies 250 mL at 27 °C and 1.00 atm. Find its molar mass and vapour density. Take R = 0.0821 L atm K⁻¹ mol⁻¹.
Step 1 — convert units. V = 250 mL = 0.250 L; T = 27 + 273 = 300 K
Step 2 — moles.
n = PV ÷ RT = (1.00 × 0.250) ÷ (0.0821 × 300)
0.0821 × 300 = 24.63
n = 0.250 ÷ 24.63 = 0.010150 mol
Step 3 — molar mass.
M = 0.550 ÷ 0.010150 = 54.2 g/mol
Step 4 — vapour density. VD = 54.2 ÷ 2 = 27.1
54 g/mol is consistent with butadiene, C₄H₆ (54.09 g/mol) — again, one candidate among several possible formulas.
Abnormal vapour density — when the answer disagrees with the formula
Sometimes the measured molecular mass is well below the value calculated from the formula. That is not experimental error: it means the molecules are breaking into more, smaller particles, so the same mass of substance occupies more volume. For a molecule that dissociates into n particles, with α the fraction dissociated:
or equivalently, in vapour densities: α = (D − d) ÷ [(n − 1) × d]
Problem: Phosphorus pentachloride dissociates on heating, PCl₅ ⇌ PCl₃ + Cl₂. At a certain temperature the observed vapour density is 70.2. Find the degree of dissociation.
Step 1 — normal molecular mass. P = 30.974, Cl = 35.45
M(PCl₅) = 30.974 + 5 × 35.45 = 30.974 + 177.25 = 208.22 g/mol
Normal vapour density D = 208.22 ÷ 2 = 104.11
Step 2 — observed molecular mass. Mobserved = 2 × 70.2 = 140.4 g/mol
Step 3 — apply the formula. One molecule gives two, so n = 2 and
(n − 1) = 1.
α = (208.22 − 140.4) ÷ (1 × 140.4) = 67.82 ÷ 140.4 = 0.483, i.e. 48.3%
Cross-check using vapour densities directly:
α = (104.11 − 70.2) ÷ (1 × 70.2) = 33.91 ÷ 70.2 = 0.483 ✓
The two routes must agree, because vapour density is just molecular mass divided by two. If they do not, you have made an arithmetic slip somewhere.
The opposite case — association
Some substances do the reverse: their molecules join into dimers, so the observed molecular mass comes out higher than the formula predicts. Carboxylic acids in the vapour phase or in a non-polar solvent are the classic example, because two hydrogen bonds hold two acid molecules together. For n molecules associating into one:
Problem: Acetic acid, CH₃COOH, shows an observed molecular mass of 100 g/mol under conditions where it dimerises. Find the degree of association.
Step 1 — normal molecular mass.
2 × 12.011 + 4 × 1.008 + 2 × 15.999 = 24.022 + 4.032 + 31.998 = 60.05 g/mol
Step 2 — substitute with n = 2.
Mnormal ÷ Mobserved = 60.05 ÷ 100 = 0.6005
α = (1 − 0.6005) ÷ (1 − 0.5) = 0.3995 ÷ 0.5 = 0.799, i.e. about 80%
Check. Start with 1 mol. If 0.799 mol associates into 0.3995 mol of dimer, the total particles = (1 − 0.799) + 0.3995 = 0.6005 mol, so the observed mass per particle = 60.05 ÷ 0.6005 = 100.0 g/mol ✓
Common mistakes that cost marks
- Giving vapour density a unit. It is a ratio and has none. Gas density in g/L does have a unit, and the two are not interchangeable.
- Writing VD = 2 × M. It is the other way round: M = 2 × VD.
- Mixing the two STP conventions. Pair a 1 atm density with 22.4 L/mol, or a 1 bar density with 22.7 L/mol — never one with the other.
- Using °C in PV = nRT. Temperature must be in kelvin.
- Leaving the volume in millilitres when R is in L atm K⁻¹ mol⁻¹.
- Assuming (n − 1) = 1 for every dissociation. n is the number of particles produced by one molecule, read off the balanced equation. PCl₅ ⇌ PCl₃ + Cl₂ and N₂O₄ ⇌ 2NO₂ both give n = 2, so (n − 1) = 1 — but that is not automatic. Where one molecule breaks into three particles, (n − 1) = 2, and using 1 doubles your answer.
- Naming one compound as "the" answer. A molecular mass narrows the field; it does not identify a substance on its own.
Where this appears in exams
| Exam | Typical question |
|---|---|
| CBSE/ICSE Class 11 | Molecular mass from vapour density or from gas density at STP |
| ICSE Class 10 | Vapour density definition and simple M = 2 × VD calculations |
| JEE/NEET | Degree of dissociation from abnormal vapour density; equilibrium linkage |
| IIT-JAM / CUET-PG | Association and dissociation, Kp from α |
| GATE / CSIR-NET | Real-gas corrections to molar-mass determinations |
Route 2 is a PV = nRT problem in disguise. Feed in the pressure, the volume and the temperature in kelvin and the Ideal Gas Law tool returns the number of moles; divide your sample mass by that and you have the molar mass, then halve it for the vapour density. It handles the R-value and unit matching that cause most of the errors above.
Open the Ideal Gas Law (PV = nRT) Calculator →Mole concept and gaseous state are the two chapters that decide how comfortable Class 11 feels for the rest of the year. ABC Chemistry teaches them at the Gurugram centre and through online classes across India: abcchemistry.in.