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Air is not a single gas, yet every density, buoyancy and vapour-density question treats it as if it had one molar mass. That number is an average, and the way you average decides whether you get it right. This post shows the one rule, works it for dry air, humid air, a mass-based mixture and a dissociating vapour, and says where the topic sits in the current syllabus.
Where is this in the syllabus?
This needs care. The NCERT Class 11 Chemistry book no longer has a chapter on the gaseous state, and the JEE (Main) 2026 and NEET (UG) 2026 chemistry syllabi have no gas-laws unit. Mixtures of gases now appear in physics: the NCERT Class 11 Physics chapter Kinetic Theory (2026-27 reprint) derives PV = (μ1 + μ2 + …)RT for a mixture and Dalton's law of partial pressures, and both exams list the kinetic theory of gases. Molar mass and mole fraction are in the first chemistry unit, Some Basic Concepts of Chemistry, and partial pressures return in equilibrium (Kp). The JEE (Advanced) 2026 brochure still lists "States of Matter: Gases and Liquids", including the "Law of partial pressures". Check your year's syllabus.
What is the rule?
For a mixture, the average molar mass is:
M_avg = total mass ÷ total moles = x1M1 + x2M2 + …
where x is each gas's mole fraction. For gases, volume percentage equals mole percentage (equal volumes at the same T and p hold equal moles), so a composition "by volume" can be used directly.
It follows from the physics. The mixture obeys pV = n_total RT. Divide the total mass by n_total and you have a single M that works in pV = (m/M)RT and in ρ = pM/RT.
Worked example 1: dry air
The NASA Earth fact sheet gives dry air as 78.08% N2, 20.95% O2, 9340 ppm Ar and 420 ppm CO2 by volume. With molar masses from the IUPAC standard atomic weights: N2 28.014, O2 31.998, Ar 39.95, CO2 44.009 g/mol.
N2: 0.7808 × 28.014 = 21.873
O2: 0.2095 × 31.998 = 6.704
Ar: 0.00934 × 39.95 = 0.373
CO2: 0.00042 × 44.009 = 0.018
Sum: 28.97 g/mol
NASA lists a mean molecular weight of 28.97, so the calculation agrees. The quick classroom version, 0.79 × 28 + 0.21 × 32 = 28.84, is close enough for most problems, but it puts argon in with nitrogen.
Worked example 2: a mixture given by mass
A flask holds 2.0 g of helium and 16.0 g of oxygen (sample numbers). Find M_avg.
n(He) = 2.0/4.003 = 0.50 mol; n(O2) = 16.0/31.998 = 0.50 mol.
M_avg = 18.0 g ÷ 1.00 mol = 18.0 g/mol.
The wrong method weights by mass: oxygen is 89% of the mass, so 0.11 × 4 + 0.89 × 32 ≈ 28.9 g/mol. That is the molar mass of air, not of a mixture that is half helium by moles.
Worked example 3: is humid air heavier or lighter?
Take air with 2.0% water vapour by moles (a sample value; NASA notes water "typically makes up about 1%" and is highly variable). M(H2O) = 18.02 g/mol.
M_avg = 0.98 × 28.97 + 0.02 × 18.02 = 28.75 g/mol.
Density at 1.00 bar and 298 K, from ρ = pM/RT: dry air 10^5 × 0.02897/(8.314 × 298) = 1.169 kg/m³; humid air 1.160 kg/m³.
Humid air is lighter, because a water molecule replaces a heavier N2 or O2 molecule. Students often guess the opposite, because "wet" sounds heavy.
Worked example 4: vapour density and dissociation
Vapour density (VD) is the density of a gas relative to hydrogen at the same T and p, so M ≈ 2 × VD. Suppose N2O4 (M = 92.01) is partly dissociated, N2O4 ⇌ 2NO2, and the measured vapour density of the mixture is 38.3 (a sample value).
M_avg = 2 × 38.3 = 76.6 g/mol.
Start with 1 mol N2O4. If a fraction α dissociates, total moles = 1 + α, while the total mass stays 92.01 g.
So 92.01/(1 + α) = 76.6, giving 1 + α = 1.201 and α ≈ 0.20.
Twenty percent of the N2O4 has dissociated. This is the same average-molar-mass rule, used backwards. Our post on equilibrium constant calculations takes such a mixture on to Kp.
Common mistakes
Weighting by mass fractions. Convert masses to moles first.
Treating M_avg as a property of the gases alone. It changes with composition, so it changes when a reaction shifts the equilibrium.
Using 22.4 L without checking conditions. ρ = pM/RT works at any T and p; our post on ideal gas problems covers the unit traps.
Forgetting that VD is half of M, not equal to it.
A practice routine
Day 1: Recalculate dry air with and without argon and note the difference.
Day 2: Three mass-based mixtures: H2 with N2, He with CO2, and CH4 with O2. Find M_avg and the density at 1 bar and 300 K.
Day 3: Two vapour-density problems: one for N2O4 and one for PCl5 ⇌ PCl3 + Cl2.
Day 4: Explain in three sentences why a balloon filled with humid air would rise very slightly in dry air.
If you keep mixing up moles and mass in these problems, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Today, find M_avg for a mixture of 4.0 g of H2 and 56 g of N2, and check that the answer is about 15 g/mol.
