If 5.6L of gas at STP has mass 11g — find molecular mass

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Question

5.6 L of a gas at STP has a mass of 11 g. Find the molecular mass of the gas.

Solution — Step by Step

At STP (Standard Temperature and Pressure: 0°C and 1 atm), one mole of any ideal gas occupies 22.4 L.

This is a fundamental result from the ideal gas law. It is the same for ALL gases regardless of molecular mass — only the number of molecules (moles) matters.

Key fact: 22.4 L at STP = 1 mole of gas.

n=Volume at STP22.4 L/mol=5.622.4=0.25 moln = \frac{\text{Volume at STP}}{22.4 \text{ L/mol}} = \frac{5.6}{22.4} = 0.25 \text{ mol}

So 5.6 L of gas at STP = 0.25 moles.

M=Massn=11 g0.25 mol=44 g/molM = \frac{\text{Mass}}{n} = \frac{11 \text{ g}}{0.25 \text{ mol}} = 44 \text{ g/mol}

Molecular mass = 44 g/mol.

This matches: CO2\text{CO}_2 (12+16×2=4412 + 16 \times 2 = 44), N2O\text{N}_2\text{O} (44), or C3H8\text{C}_3\text{H}_8 (44).

Without additional information, we conclude the molecular mass is 44 g/mol.

Why This Works

The molar volume at STP (22.4 L/mol) comes from the ideal gas law PV=nRTPV = nRT at T=273T = 273 K and P=1P = 1 atm:

V=nRTP=1×0.0821×273122.4 LV = \frac{nRT}{P} = \frac{1 \times 0.0821 \times 273}{1} \approx 22.4 \text{ L}

This is independent of the identity of the gas — all ideal gases have the same volume per mole at the same conditions. This is essentially Avogadro’s hypothesis: equal volumes of gases at the same T and P contain equal numbers of molecules.

The molecular mass is then simply mass per mole — total mass divided by number of moles.

Alternative Method — Direct Proportion

At STP: 22.4 L has mass = 1 molecular mass (M) grams.

So 5.6 L has mass = 5.622.4×M=M4\frac{5.6}{22.4} \times M = \frac{M}{4} grams.

Given mass = 11 g: M4=11M=44\frac{M}{4} = 11 \Rightarrow M = 44 g/mol.

Same answer, one-step proportion.

Common Mistake

Using 22.4 L for conditions other than STP. The molar volume 22.4 L/mol applies specifically at STP (0°C, 1 atm). At different temperatures or pressures, the molar volume changes. Some problems specify “at 27°C and 1 atm” — then use Vm=RT/P=0.0821×300/1=24.63V_m = RT/P = 0.0821 \times 300 / 1 = 24.63 L/mol. Always check whether the problem says STP before using 22.4.

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