Vapour density

Weigh a flask of carbon dioxide and an identical flask of hydrogen, same temperature, same pressure — the carbon dioxide is 22 times heavier. That single number, 22, is the vapour density of carbon dioxide, and it is one of the cheapest tricks in chemistry for pinning down a molecule's mass without ever counting atoms directly. Hydrogen is chosen as the yardstick because it is the lightest gas known, so almost every other gas comes out a clean, convenient multiple of it. Vapour density matters in the Mole Concept chapter because it links a simple relative-weighing experiment to the relative molecular mass scale used everywhere else in stoichiometry — turn the ratio around and molecular mass drops straight out, doubled. This article compresses the definition, the reasoning behind it, and the standard exam calculations into one tight review.

The core insight

Vapour density is the mass of a gas compared with the mass of an equal volume of hydrogen under the same conditions, and because equal volumes of gases hold equal numbers of molecules, doubling it gives the gas's relative molecular mass.

The core idea

Definition. The vapour density (V.D.) of a gas is the ratio of the mass of a certain volume of that gas to the mass of an equal volume of hydrogen gas, both measured at the same temperature and pressure:

\text{V.D.} = \dfrac{\text{mass of a certain volume of gas}}{\text{mass of the same volume of hydrogen}}

where: the numerator and denominator are masses of equal volumes of the two gases, taken under matched T and P; V.D. carries no units, since it is a ratio of two masses.

Why hydrogen, and why it links to molecular mass. By Avogadro's Law, equal volumes of any gases at the same temperature and pressure contain equal numbers of molecules. So a mass ratio of equal volumes is automatically the same as a ratio of the mass of one molecule of the gas to the mass of one molecule of hydrogen. Since hydrogen is diatomic (H₂) with relative molecular mass 2, this collapses to a single clean relation:

\text{Vapour density} = \dfrac{\text{Relative molecular mass}}{2}

where: relative molecular mass is measured on the usual scale, referenced (like relative atomic mass) to one-twelfth the mass of a carbon-12 atom. Turned around, this is the form used constantly in numericals:

\text{Molecular mass} = 2 \times \text{Vapour density}

Because hydrogen is the lightest known gas, nearly every other gas has a molecular mass bigger than 2, so vapour densities almost always come out as convenient numbers greater than 1. In the lab, V.D. can be measured directly — weigh equal volumes of the unknown gas and of hydrogen under identical conditions — without needing to know beforehand how many atoms the gas molecule contains; the molecular formula is worked out afterward, from the mass the ratio hands you.

Figure

Two identical flasks holding equal volumes of hydrogen and carbon dioxide at the same temperature and pressure on a balance, with the carbon dioxide flask twenty two times heavier, illustrating the definition of vapour density

Derivation AWhy molecular mass equals twice the vapour density

The goal is to turn the bare definition of vapour density, a ratio of two masses of equal gas volumes, into the working numerical formula molecular mass equals two times vapour density, using nothing beyond Avogadro's Law and the definition of relative molecular mass, and to show precisely where the factor of 2 in that formula comes from.

Step 1. Start from the definition: V.D. is the mass of a volume of gas divided by the mass of the same volume of hydrogen, both at the same temperature and pressure. Written with molecule counts, if a volume holds $N$ molecules of the gas (mass per molecule $m_{\text{gas}}$ ) and the same volume holds $N$ molecules of hydrogen (mass per molecule $m_{\text{H}_2}$ ), then V.D. $= \dfrac{N \cdot m_{\text{gas}}}{N \cdot m_{\text{H}_2}} = \dfrac{m_{\text{gas}}}{m_{\text{H}_2}}$ — the molecule count $N$ is the same in both because Avogadro's Law says equal volumes at equal T and P hold equal numbers of molecules, so it cancels cleanly.

Step 2. Recognise that a ratio of single-molecule masses is exactly what "relative molecular mass" means, on the same scale for every substance in this chapter. So V.D. equals the relative molecular mass of the gas divided by the relative molecular mass of hydrogen.

Step 3. Substitute the known molecular mass of hydrogen, H₂, which is $2 \times 1 = 2$ on that scale. This gives $\text{V.D.} = \dfrac{\text{Molecular mass}}{2}$ , and rearranging for molecular mass gives the working formula, $\text{Molecular mass} = 2 \times \text{V.D.}$ — the factor of 2 is simply hydrogen's own molecular mass, nothing more mysterious than that.

Worked example

Problem: The vapour density of a gas is 22. Find its relative molecular mass, and suggest a possible identity for the gas if it is known to contain only carbon and oxygen. Then find the vapour density of ammonia, NH₃. (RAM: C = 12, O = 16, N = 14, H = 1)

Step 1. Apply the relation to the unknown gas: $\text{Molecular mass} = 2 \times \text{V.D.} = 2 \times 22 = 44$ .

Step 2. Test a candidate formula built only from carbon and oxygen. Carbon dioxide, CO₂, has $M(\text{CO}_2) = 12 + 2(16) = 44$ , matching exactly.

Step 3. Now go the other way for ammonia: find its molecular mass first, $M(\text{NH}_3) = 14 + 3(1) = 17$ , then its vapour density, $\text{V.D.} = \dfrac{17}{2} = 8.5$ .

Answer: the gas of vapour density 22 has molecular mass 44 and is consistent with CO₂; ammonia, NH₃, has vapour density 8.5, meaning a given volume of ammonia is 8.5 times as heavy as the same volume of hydrogen under identical conditions.

Competitive stretch

Problem: Two different gases, X and Y, both have vapour density 22. Gas X is carbon dioxide, CO₂ (shown above). Show that propane, C₃H₈, is also consistent with this vapour density, and explain what this tells you about identifying a gas from vapour density alone. (RAM: C = 12, H = 1)

Step 1. Find the molecular mass propane would need to have: $\text{Molecular mass} = 2 \times \text{V.D.} = 2 \times 22 = 44$ , the same target as before.

Step 2. Compute propane's actual molecular mass and compare: $M(\text{C}_3\text{H}_8) = 3(12) + 8(1) = 36 + 8 = 44$ . It matches exactly, alongside carbon dioxide.

Answer: both CO₂ and C₃H₈ have relative molecular mass 44 and so both give vapour density 22 — vapour density fixes only the total molecular mass, not the identity of the gas, so a full identification needs extra evidence such as elemental composition or other chemical tests.

Common traps

Forgetting the factor of 2 and writing molecular mass = vapour density instead of $2 \times$ V.D. — the 2 comes from hydrogen's own molecular mass.
Giving vapour density a unit (like g/dm³); it is a pure number, being a ratio of two masses of equal volumes.
Confusing vapour density with ordinary gas density (mass per unit volume); vapour density is always relative to hydrogen, never an absolute value.
Applying the vapour density relation to a solid or liquid, where equal-volume, equal-molecule-count reasoning does not hold.

Ask the tutor: Want to practice converting between vapour density and molecular mass with a few more gases, or see why the factor of 2 traces back to hydrogen being diatomic? Ask the AI tutor for extra worked numbers.

Why is hydrogen used as the reference gas instead of, say, oxygen or air?

Hydrogen is the lightest known gas, so almost every other gas ends up heavier than it, giving vapour densities that are convenient positive numbers greater than 1. Historically it was also the gas whose molecular mass (2) was firmly established early, making it a stable, simple reference. Using a heavier reference like oxygen would just rescale every vapour density by a different constant — the physics would be the same, but the numbers less convenient.

Why does doubling the vapour density give the molecular mass, and not tripling or halving it?

Because the reference gas, hydrogen, is diatomic with a relative molecular mass of exactly 2. Vapour density is defined as molecular mass of the gas divided by molecular mass of hydrogen, so molecular mass equals vapour density multiplied by 2, the reference gas's own molecular mass. If a different reference gas were used, the multiplying factor would change to that gas's molecular mass instead.

Do I need to know how vapour density is measured experimentally, like Victor Meyer's method?

No — for this syllabus you only need the definition, the relation to molecular mass, and how to use it in numericals such as finding molecular mass from vapour density or vice versa. The experimental determination method is not part of this topic's requirement; focus on applying $\text{Molecular mass} = 2 \times \text{V.D.}$ confidently and interpreting what the ratio means physically.

Vapour density

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads