Gas Density: d = PM/RT and Molar Mass from Density

By Aniket Bhardwaj · 26 October 2026 · Updated 25 October 2026 · Class 11-12 Gas Laws

“Find the density of oxygen at 25°C” or “a gas has density 1.78 g/L at STP, find its molar mass” are favourite short questions in Class 11 and Class 12. Both come from one equation, which is the ideal gas law rearranged. Once you see where it comes from, you do not need to remember it as a separate formula. You also stop making the classic mistake of using the liquid density idea for a gas.

A bar chart compares CO₂ at 1.799 g/L with O₂ at 1.308 g/L and shows their density ratio of 1.375 at the same temperature and pressure. — At the same temperature and pressure, CO₂ is 1.375 times as dense as O₂.

The formula

d = m/V = PM/(RT)
M = dRT/P

d is the density of the gas in g/L. P is the pressure and T is the absolute temperature in kelvin (°C + 273.15). M is the molar mass in g/mol. R must match the pressure unit: R = 0.082057 L·atm/(mol·K) when P is in atm, and R = 8.314 J/(mol·K) when P is in pascal and you want M in kg/mol.

Where it comes from

Start with PV = nRT. The number of moles is n = m/M, where m is the mass of the gas. So PV = (m/M)RT. Move V to the right side and M to the left: PM = (m/V)RT. Since m/V is the density, PM = dRT. That is all. Notice what the equation tells you: at the same T and P, the density of a gas is directly proportional to its molar mass. This is why CO₂ sinks below air and hydrogen rises, and why Avogadro’s idea (equal volumes, equal numbers of molecules) leads straight to the molar mass.

Worked example 1: oxygen at 273.15 K and 1 atm

Question. Find the density of O₂ at 0°C and 1.000 atm.

Step 1. M = 2 × 15.999 = 31.998 g/mol. T = 273.15 K.

Step 2. d = PM/(RT) = (1.000 × 31.998)/(0.082057 × 273.15) = 31.998/22.414 = 1.428 g/L.

Check by another route. One mole of an ideal gas occupies 22.414 L at 273.15 K and 1 atm (see the note below on 22.4 vs 22.7). Density = mass of 1 mol ÷ volume = 31.998 ÷ 22.414 = 1.428 g/L. Same answer.

Worked example 2: CO₂ compared with O₂ at room temperature

Question. Find the density of CO₂ and of O₂ at 25°C and 1.00 atm. How many times denser is CO₂?

Step 1. T = 25 + 273.15 = 298.15 K, so RT = 0.082057 × 298.15 = 24.466 L·atm/mol.

Step 2. M(CO₂) = 12.011 + 2(15.999) = 44.009 g/mol. d = 44.009 ÷ 24.466 = 1.799 g/L.

Step 3. M(O₂) = 31.998 g/mol. d = 31.998 ÷ 24.466 = 1.308 g/L.

Step 4. Ratio = 1.799 ÷ 1.308 = 1.375. This is exactly 44.009 ÷ 31.998, the ratio of molar masses, as the equation predicts. At the same T and P you do not even need R.

Worked example 3: molar mass from a measured density

Question. A gas has density 1.78 g/L at 0°C and 1.00 atm. Find its molar mass.

Step 1. M = dRT/P = 1.78 × 0.082057 × 273.15 ÷ 1.00.

Step 2. RT = 22.414, so M = 1.78 × 22.414 = 39.9 g/mol.

Answer. About 39.9 g/mol. This matches the molar mass of argon, a monatomic gas. The match is a good sign that the working is right, but the data alone cannot prove the gas is argon, because other gases could have a similar molar mass.

Worked example 4: molar mass of a vapour above 100°C

Question. A volatile liquid sample of mass 0.560 g is completely vaporised. The vapour occupies 250 mL at 100°C and 1.00 atm. Find the molar mass.

Step 1. V = 0.250 L, T = 100 + 273.15 = 373.15 K, so density d = 0.560 ÷ 0.250 = 2.24 g/L.

Step 2. M = dRT/P = 2.24 × 0.082057 × 373.15 ÷ 1.00 = 68.6 g/mol.

Check using PV = nRT. n = PV/RT = (1.00 × 0.250)/(0.082057 × 373.15) = 0.008165 mol. M = 0.560 ÷ 0.008165 = 68.6 g/mol. Same.

This is the idea of the Victor Meyer experiment taught in some syllabuses. If your book gives pressure in mmHg, divide by 760 first to get atm.

Using SI units

If P is in pascal, keep M in kg/mol and R = 8.314 J/(mol·K). The density then comes out in kg/m³, which numerically equals g/L. As a check, CO₂ at 101325 Pa and 298.15 K: d = (101325 × 0.044009)/(8.314 × 298.15) = 1.799 kg/m³, which is 1.799 g/L. If you mix kPa with R = 8.314, the numbers still work because kPa × g/mol gives g/L directly: 101.325 × 44.009 ÷ (8.314 × 298.15) = 1.799.

A note on 22.4 vs 22.7 L/mol. Many Indian textbooks use STP as 0°C and 1 atm, where the molar volume is 22.4 L (more exactly 22.414 L). The IUPAC standard since 1982 uses 1 bar (100 kPa), which gives 22.711 L at 0°C. Always use the value your syllabus or question states. Using d = PM/RT with the actual P and T avoids the question entirely.

Common mistakes

Using °C in the formula. T must be in kelvin. 25°C is 298.15 K, not 25.
Mixing units of R and P. R = 0.082057 goes with atm and litres. With mmHg or kPa, convert first or change R.
Using atomic mass for a molecular gas. Oxygen gas is O₂, so M = 32, not 16. The same applies to H₂, N₂ and Cl₂.
Quoting the density at STP for room conditions. Gas density changes with both T and P. A gas is about 8% less dense at 25°C than at 0°C at the same pressure.
Confusing density with vapour density. Vapour density is a ratio (gas compared with hydrogen), so it has no unit. Molar mass = 2 × vapour density.

Exam relevance

Question type	Use
Density at given T and P	d = PM/(RT)
Molar mass from density	M = dRT/P
Which gas is denser	Compare molar masses at the same T and P
Mass and volume of vapour given	d = m/V, then M = dRT/P
Density at STP	M ÷ 22.4 (or the molar volume in your book)

Check the limits of your syllabus in the current official notification of your board or exam. The method above is the same everywhere.

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