Amounts of Substance and Gas Volumes

(Separate Chemistry, Higher Tier only) The whole of section 4.3.5 is assessed only in AQA GCSE Chemistry and only at Higher tier.

At the same temperature and pressure, equal numbers of moles of any gases occupy equal volumes. This is true whatever the gas: one mole of hydrogen and one mole of carbon dioxide take up the same space, even though a CO₂ molecule is far heavier than an H₂ molecule. Gas molecules are so far apart that the size of the molecule barely affects the volume.

At room temperature and pressure (rtp), defined by AQA as 20°C and 1 atmosphere, one mole of any gas occupies 24 dm³. This value is the molar gas volume.

Learn the two forms of the same figure: the molar gas volume is 24 dm³, which is 24,000 cm³. Watch which unit the question uses, because 1 dm³ = 1000 cm³.

The molar gas volume links moles and volume for any gas at rtp.

${\text{volume (dm}}^3)=\text{moles}\times 24$

Rearranging lets you find moles from a measured volume:

$\text{moles}=\frac{{\text{volume (dm}}^3)}{24}$

Worked example. How many moles are there in 6 dm³ of carbon dioxide at rtp?

$\text{moles}=\frac{6}{24}=0.25\text{ mol}$

If a volume is given in cm³, either divide by 24,000 or convert to dm³ first by dividing by 1000. For instance, 480 cm³ = 0.480 dm³, which is $0.480÷24=0.020$ mol.

To go from a mass to a gas volume, chain two steps: mass to moles using $M_r$, then moles to volume using 24 dm³.

Worked example. Calculate the volume of 8 g of oxygen gas, O₂, at rtp. Use $A_r(\text{O})=16$.

Step 1 — relative formula mass of O₂:

$M_r({\text{O}}_2)=2\times 16=32$

Step 2 — moles of O₂:

$\text{moles}=\frac{\text{mass}}{M_r}=\frac{8}{32}=0.25\text{ mol}$

Step 3 — volume at rtp:

$\text{volume}=\text{moles}\times 24=0.25\times 24=6{\text{ dm}}^3$

The 8 g of oxygen occupies 6 dm³. The same three steps in reverse turn a volume back into a mass, as the next slide shows.

If you are given a gas volume and asked for a mass, reverse the chain: volume to moles, then moles to mass.

Worked example. What mass of carbon dioxide is present in 48 dm³ of CO₂ at rtp? Use $A_r$: C = 12, O = 16.

Step 1 — moles from volume:

$\text{moles}=\frac{48}{24}=2.0\text{ mol}$

Step 2 — relative formula mass of CO₂:

$M_r({\text{CO}}_2)=12+(2\times 16)=44$

Step 3 — mass from moles:

$\text{mass}=\text{moles}\times M_r=2.0\times 44=88\text{ g}$

The 48 dm³ of carbon dioxide has a mass of 88 g. Deciding which direction to work in is the whole skill: read whether the unknown is a volume or a mass, and set up the chain to end on it.

Because equal moles of gases occupy equal volumes, the balancing numbers in an equation give the ratio of gas volumes directly, with no need to work out moles at all when every substance involved is a gas.

Worked example. In the Haber process nitrogen reacts with hydrogen:

${\text{N}}_2+3{\text{H}}_2\to 2{\text{NH}}_3$

The mole ratio is 1 : 3 : 2, so the volume ratio is also 1 : 3 : 2. If 50 cm³ of nitrogen reacts completely:

Gas	Ratio	Volume
N₂	1	50 cm³
H₂	3	150 cm³
NH₃	2	100 cm³

So 50 cm³ of nitrogen reacts with 150 cm³ of hydrogen to make 100 cm³ of ammonia. Notice the total gas volume falls from 200 cm³ of reactants to 100 cm³ of product, because there are fewer gas molecules on the right.

When a reactant is a solid or liquid but a product is a gas, use $M_r$ to get moles of the known substance, apply the equation ratio, then convert to gas volume.

Worked example. Magnesium reacts with excess hydrochloric acid:

$\text{Mg}+2\text{HCl}\to {\text{MgCl}}_2+{\text{H}}_2$

Calculate the volume of hydrogen produced at rtp from 0.12 g of magnesium. Use $A_r(\text{Mg})=24$.

Step 1 — moles of magnesium:

$\text{moles}=\frac{0.12}{24}=0.0050\text{ mol}$

Step 2 — the ratio of Mg to H₂ is 1 : 1, so moles of H₂ = 0.0050 mol.

Step 3 — volume of hydrogen at rtp:

$\text{volume}=0.0050\times 24=0.12{\text{ dm}}^3=120{\text{ cm}}^3$

The reaction produces 0.12 dm³ (120 cm³) of hydrogen. The acid being in excess confirms that magnesium is the limiting reactant, so all of it reacts.

1. Muddling dm³ and cm³

The molar gas volume is 24 dm³, which is 24,000 cm³. Dividing a cm³ volume by 24 instead of 24,000 gives an answer 1000 times too large. Convert to a single unit before calculating.

2. Forgetting to balance the formula mass of diatomic gases

Oxygen, hydrogen, nitrogen and chlorine exist as diatomic molecules, so $M_r({\text{O}}_2)=32$, not 16. Using the atomic mass instead of the molecular mass halves your moles.

3. Skipping the equation ratio

Moles of the known substance are not always equal to moles of the gas you want. Apply the balancing-number ratio before converting to volume, as in the 1 : 3 : 2 Haber ratio.

4. Using the wrong direction of the chain

Volume to mass runs volume → moles → mass; mass to volume runs mass → moles → volume. Decide what the unknown is first, then build the chain to finish on it.

5. Assuming 24 dm³ applies to solids or liquids

The molar gas volume only applies to gases at rtp. It has no meaning for a solid or a liquid, so never multiply a mole quantity of a solid by 24 to get a "volume".