Gas Volumes and Amounts of Substance

Higher tier only: This whole lesson is Higher tier. Foundation candidates do not need the molar gas volume, but should be confident converting between mass and moles from Topic C3.

Many reactions give off a gas — a metal fizzing in acid, a carbonate decomposing — and chemists often need to know the volume of gas produced. There is a beautifully simple rule for this: at room temperature and pressure, one mole of any gas takes up the same volume, $24 \text{ dm}^3$24 dm3. This lets you turn moles into a gas volume in a single step. This lesson, part of Topic C5 of OCR Gateway Science A, introduces the molar gas volume and uses it to calculate gas volumes from moles and from reacting masses.

By the end of this lesson you should be able to state the molar gas volume at rtp, calculate the volume of a gas from its number of moles, calculate gas volumes from reacting masses, and use volume ratios from equations.

This lesson builds directly on the mole work from Topic C3: once you can find the moles of a gas, a single multiplication turns that into a volume. That makes gas-volume questions some of the most predictable calculations in the whole course.

---

The Molar Gas Volume

At room temperature and pressure (rtp) — about $20 \text{ °C}$20 °C and normal atmospheric pressure — one mole of any gas occupies $24 \text{ dm}^3$24 dm3 (which is the same as $24{,}000 \text{ cm}^3$24,000 cm3). This is the molar gas volume, and remarkably it is the same for every gas: one mole of hydrogen, one mole of carbon dioxide and one mole of oxygen each take up $24 \text{ dm}^3$24 dm3 at rtp, even though their masses are very different.

The relationship is:

$\text{volume of gas (dm}^3) = \text{moles} \times 24$volume of gas (dm3)=moles×24

which rearranges to:

$\text{moles} = \dfrac{\text{volume of gas (dm}^3)}{24}$moles=24volume of gas (dm3)​

Exam Tip: The molar gas volume is $24 \text{ dm}^3$24 dm3 ($24{,}000 \text{ cm}^3$24,000 cm3) per mole at rtp, and it is the same for all gases. Learn $\text{volume} = \text{moles} \times 24$volume=moles×24 and its rearrangement — it appears in almost every gas calculation in C5.

Why the volume is the same for every gas

It can seem surprising that one mole of a heavy gas such as carbon dioxide takes up exactly the same volume as one mole of a light gas such as hydrogen. The reason is that in a gas the particles are very far apart compared with their own size — the space a gas fills is mostly empty space between the particles, not the particles themselves. So the volume depends only on how many particles there are (and the temperature and pressure), not on how big or heavy each particle is. One mole of any gas contains the same number of particles (the Avogadro constant, $6.02 \times 10^{23}$6.02×1023, from C3), so at the same temperature and pressure they all spread out to fill the same volume. This is why a single value — $24 \text{ dm}^3$24 dm3 per mole at rtp — works for every gas, which makes gas calculations refreshingly simple.

It also means the rule only applies to gases. Solids and liquids have their particles packed closely together, so their volumes are tiny and depend strongly on the substance; you can never use $24 \text{ dm}^3$24 dm3 for a solid or a liquid. For those you go back to $\text{moles} = \dfrac{\text{mass}}{M_r}$moles=Mr​mass​.

---

Calculating a Gas Volume from Moles

Worked Example 1 — Moles to volume

What is the volume, at rtp, of $0.5 \text{ mol}$0.5 mol of carbon dioxide?

Step 1 — write the equation: $\text{volume} = \text{moles} \times 24$volume=moles×24.

Step 2 — substitute: $\text{volume} = 0.5 \times 24$volume=0.5×24.

Step 3 — calculate: $\text{volume} = 12 \text{ dm}^3$volume=12 dm3.

Answer: $12 \text{ dm}^3$12 dm3 (which is $12{,}000 \text{ cm}^3$12,000 cm3).

Worked Example 2 — Volume to moles

How many moles of gas are there in $6 \text{ dm}^3$6 dm3 of oxygen at rtp?

Step 1 — rearrange: $\text{moles} = \dfrac{\text{volume}}{24}$moles=24volume​.

Step 2 — substitute: $\text{moles} = \dfrac{6}{24}$moles=246​.

Step 3 — calculate: $\text{moles} = 0.25 \text{ mol}$moles=0.25 mol.

Answer: $0.25 \text{ mol}$0.25 mol.

Exam Tip: If a volume is given in cm³, convert it to dm³ first (÷1000) before dividing by 24 — or use $24{,}000 \text{ cm}^3$24,000 cm3 as the molar volume. Keep your volume unit consistent throughout the calculation.

---

Gas Volumes from Reacting Masses

The real power comes from combining the molar gas volume with $\text{moles} = \dfrac{\text{mass}}{M_r}$moles=Mr​mass​ (from C3). This lets you work from the mass of a solid reactant all the way to the volume of gas it produces, using the mole ratio from the balanced equation.

The route is: mass → moles of solid → mole ratio → moles of gas → volume of gas.

Worked Example 3 — Mass of carbonate giving a gas volume

What mass of calcium carbonate must be heated to produce $6 \text{ dm}^3$6 dm3 of carbon dioxide at rtp? Equation: $\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$CaCO3​→CaO+CO2​. ($M_r$Mr​ of $\text{CaCO}_3 = 100$CaCO3​=100.)

Step 1 — moles of $\text{CO}_2$CO2​ from the volume: $\dfrac{6}{24} = 0.25 \text{ mol}$246​=0.25 mol.

Step 2 — mole ratio $\text{CaCO}_3 : \text{CO}_2 = 1 : 1$CaCO3​:CO2​=1:1, so moles of $\text{CaCO}_3 = 0.25 \text{ mol}$CaCO3​=0.25 mol.

Step 3 — mass of $\text{CaCO}_3$CaCO3​: $\text{mass} = \text{moles} \times M_r = 0.25 \times 100 = 25 \text{ g}$mass=moles×Mr​=0.25×100=25 g.

Answer: $25 \text{ g}$25 g of calcium carbonate.

Worked Example 4 — Volume of gas from a mass of metal

What volume of hydrogen, at rtp, is produced when $2.4 \text{ g}$2.4 g of magnesium reacts with excess hydrochloric acid? Equation: $\text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2$Mg+2HCl→MgCl2​+H2​. ($A_r$Ar​: Mg = 24.)

Step 1 — moles of Mg: $\dfrac{2.4}{24} = 0.1 \text{ mol}$242.4​=0.1 mol.

Step 2 — mole ratio $\text{Mg} : \text{H}_2 = 1 : 1$Mg:H2​=1:1, so moles of $\text{H}_2 = 0.1 \text{ mol}$H2​=0.1 mol.

Step 3 — volume of $\text{H}_2$H2​: $\text{volume} = \text{moles} \times 24 = 0.1 \times 24 = 2.4 \text{ dm}^3$volume=moles×24=0.1×24=2.4 dm3.

Answer: $2.4 \text{ dm}^3$2.4 dm3 (which is $2{,}400 \text{ cm}^3$2,400 cm3) of hydrogen.

Exam Tip: For a mass-to-gas-volume calculation, follow the chain mass → moles → ratio → moles of gas → volume. The two key conversions are $\text{moles} = \dfrac{\text{mass}}{M_r}$moles=Mr​mass​ (for the solid) and $\text{volume} = \text{moles} \times 24$volume=moles×24 (for the gas).

---

Volume Ratios from Equations

For reactions where both the reactant and product are gases, the mole ratio is also the volume ratio (because equal numbers of moles of any gas have equal volumes at the same temperature and pressure). For example, in $\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$N2​+3H2​→2NH3​, $1 \text{ dm}^3$1 dm3 of nitrogen reacts with $3 \text{ dm}^3$3 dm3 of hydrogen to give $2 \text{ dm}^3$2 dm3 of ammonia.

Worked Example 5 — A volume ratio

What volume of oxygen reacts with $50 \text{ cm}^3$50 cm3 of methane in complete combustion, and what volume of carbon dioxide forms (all gases at the same conditions)? Equation: $\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$CH4​+2O2​→CO2​+2H2​O.