IB Chemistry SL
Topic 1 — Counting Particles by Mass
Paper 1 & 2
Core idea
~9 min read
Avogadro’s Law
For gases, volume behaves just like moles. Avogadro’s Law says equal volumes of any gases (at the same temperature and pressure) hold equal numbers of molecules — so you can read mole ratios straight off gas volumes.
📘 What you need to know
- Avogadro’s Law: equal volumes of gases at the same temperature and pressure contain the same number of molecules.
- So the volume ratio of gases equals their mole ratio in a balanced equation.
- At STP (0 °C / 273 K and 100 kPa), one mole of any gas occupies 22.7 dm3.
- These relationships only hold when all gases are at the same temperature and pressure.
- If reactant gases aren’t in the equation’s ratio, use limiting reactant reasoning.
Equal volumes, equal molecules
In 1811 Amedeo Avogadro proposed that equal volumes of gases — measured at the same temperature and pressure — contain the same number of molecules. This is Avogadro’s Law, and it’s incredibly useful: it lets us treat gas volumes exactly like moles.
At standard temperature and pressure (STP) — defined as 0 °C (273 K) and 100 kPa — one mole of any gas takes up 22.7 dm3 (units: dm3 mol-1).
The key insight: for gases you can skip converting to moles entirely. If the volumes are measured under the same conditions, the volume ratio is the mole ratio. That saves a lot of work.
Volume ratios from equations
Because volume tracks moles, the coefficients in a balanced equation give you the volume ratio of the gases directly.
Take the combustion of propane:
Combustion of propane
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)
The gas mole ratio is 1 : 5 : 3 (propane : oxygen : CO2 — water is liquid, so it’s left out). If you burn 50 cm3 of propane:
- O2 needed = 5 × 50 = 250 cm3
- CO2 formed = 3 × 50 = 150 cm3
Note: only count gases. State symbols matter — H2O here is a liquid (l), so it doesn’t appear in the gas volume ratio.
When gases aren’t in the right ratio
If the reactant gases aren’t supplied in the equation’s exact ratio, one runs out first — the limiting reactant. A quick trick: divide each gas volume by its coefficient; the lowest result is the limiting reactant.
WORKED EXAMPLEWhat total volume of gas remains when 70 cm3 of ammonia reacts fully with 50 cm3 of oxygen? 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(l)
Step 1 — gas mole ratio (water is liquid)
NH₃ : O₂ : NO = 4 : 5 : 4
Step 2 — find the limiting reactant
NH₃: 70 ÷ 4 = 17.5 O₂: 50 ÷ 5 = 10 → O₂ limits
50 cm³ O₂ needs (4/5 × 50) = 40 cm³ NH₃ to react.
Step 3 — NO produced
40 cm³ NH₃ reacting makes 40 cm³ NO.
Step 4 — leftover NH₃ + product
NH₃ left = 70 − 40 = 30 cm³; total = 40 + 30
= 70 cm³ of gas
💡 Exam tip
- The “lowest is limiting” trick — divide each gas volume by its coefficient — quickly tells you which reactant runs out first.
- Gas-volume relationships only work if all gases share the same temperature and pressure. Watch for liquids or solids, which don’t count.
That completes Counting Particles by Mass — the Mole. Next you’ll move on to The Behaviour of Ideal Gases, applying these ideas to how gases respond to changes in pressure, volume and temperature.
Want this explained one-to-one?
Book a free session with an experienced IB Chemistry tutor and get your trickiest topics made simple.
Book a Free Session →