IB Chemistry HLTopic 5 — How Much? Quantifying Chemical ChangePaper 1 & 2Core idea~11 min read
Avogadro’s Law and Molar Gas Volume
Weighing a gas is awkward. Measuring its volume is easy. Avogadro spotted something in 1811 that makes this a gift: at the same temperature and pressure, equal volumes of any gases contain equal numbers of particles. That means for gases, volumes behave exactly like moles.
📚 What you need to know
Avogadro’s law: equal volumes of gases at the same temperature and pressure contain equal numbers of particles.
So for gases, the volume ratio is the same as the mole ratio straight from the equation.
At STP, one mole of any gas occupies 22.7 dm3.
STP means 273 K (0 °C) and 100 kPa. Learn both numbers.
n = V / 22.7 and V = n × 22.7, with V in dm3.
Divide cm3 by 1000 to get dm3 before using the molar volume.
Only species labelled (g) count towards a gas volume — liquids and solids take up almost none.
Why the law is so useful
A carbon dioxide molecule is about 22 times heavier than a hydrogen molecule, so you would expect a box of it to be very different. But gas particles are tiny compared with the space between them. What fills the container is mostly empty space, and the size of the particle barely matters.
So a fixed volume at a fixed temperature and pressure holds a fixed number of particles, whatever gas you choose.
Both boxes hold eight molecules here. The carbon dioxide box weighs far more, but that is irrelevant — volume counts particles, not mass.
The molar gas volume
At STP: 273 K and 100 kPaV (dm3) = n × 22.7 n = V (dm3) ÷ 22.7
One mole of any gas at STP fills 22.7 dm3 — that is about the volume of a large bucket. Hydrogen, carbon dioxide, ammonia, it makes no difference. The molar mass never enters the calculation.
22.7, not 24. Some textbooks and older courses use 24 dm3 mol−1, which is the value at room temperature. The IB defines STP as 273 K and 100 kPa, giving 22.7 dm3 mol−1. Use 22.7 unless a question tells you otherwise.
Reading volume ratios straight off the equation
Here is where Avogadro saves you real time. If a question is entirely about gases at the same conditions, you do not need moles at all — just scale the volumes by the coefficients.
No molar masses, no moles, no 22.7. When every substance you care about is a gas at the same conditions, the coefficients are the answer.
Watch the state symbol on water. If the question says H2O(l), it contributes nothing to the final gas volume. If it says H2O(g), it counts fully. Same reaction, very different answer.
Quick reference
You have
You want
Do this
Moles of gas
Volume in dm3 at STP
× 22.7
Volume in dm3 at STP
Moles of gas
÷ 22.7
Volume in cm3
Volume in dm3
÷ 1000
Volume in dm3
Volume in cm3
× 1000
Mass of a gas
Volume at STP
÷ M, then × 22.7
Worked examples
WORKED EXAMPLE
(a) What volume does 0.250 mol of carbon dioxide occupy at STP? (b) How many moles are in 480 cm3 of methane at STP?
(a) moles to volume: multiplyV = 0.250 × 22.7 = 5.675 dm³V = 5.68 dm³ (3 s.f.)(b) convert the volume first480 ÷ 1000 = 0.480 dm³then divide by the molar volumen = 0.480 ÷ 22.7 = 0.02115 moln = 2.11 × 10⁻² molNeither part needed the molar mass. That is the whole point of Avogadro’s law.
WORKED EXAMPLE
5.00 g of calcium carbonate reacts with excess hydrochloric acid: CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g). What volume of gas is collected at STP?
Step 1: moles of the solidn = 5.00 ÷ 100.09 = 0.04996 molStep 2: mole ratio is 1 : 1n(CO₂) = 0.04996 molStep 3: moles to gas volumeV = 0.04996 × 22.7 = 1.134 dm³1.13 dm³, which is 1130 cm³“Excess acid” tells you the carbonate is the limiting reactant, so all of your moles come from the 5.00 g.
WORKED EXAMPLE
40 cm3 of ethene is burned completely in 200 cm3 of oxygen: C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(l). What total volume of gas remains, measured at the same conditions?
Step 1: how much oxygen is actually used?
Ratio is 1 : 3, so 40 × 3 = 120 cm³ of O₂ reactsStep 2: oxygen left over200 − 120 = 80 cm³ of O₂ unreactedStep 3: carbon dioxide made
Ratio is 1 : 2, so 40 × 2 = 80 cm³ of CO₂Step 4: add up the gases only
Water is (l), so it counts as zero: 80 + 80 = 160 cm³160 cm³ of gas remainsEthene ran out first, so the leftover oxygen has to be included in the final volume. Forgetting the excess gas is the most common slip in this question type.
💡 Exam tip
Convert cm3 to dm3 on its own line. Dividing by 1000 inside a bigger calculation is where the factor of 1000 gets lost.
Ask whether you need 22.7 at all. If the question only involves gas volumes at the same conditions, you do not.
Underline the state symbols before you start. They tell you what to count.
For “total volume remaining”, include unreacted gas as well as the gaseous products.
Quote STP properly if asked: 273 K and 100 kPa.
Check the size of your answer. A fraction of a mole should give a few dm3, not hundreds.
⚠ Common mix-up
Using 24 instead of 22.7. That is the room temperature value, not the IB’s STP value.
Multiplying by 22.7 when you should divide. Moles are small, volumes are big — use that to check.
Counting liquid water as a gas in a volume total.
Applying the molar volume to a solid or a solution. 22.7 dm3 mol−1 is for gases only.
Assuming equal volumes have equal masses. Equal numbers of particles, not equal mass.
Forgetting that Avogadro’s law needs the same temperature and pressure for both gases.
Solids you weigh, gases you measure by volume — and solutions you measure by concentration. That is the last of the three, and it brings titrations with it. Up next: Calculating Concentration.
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