IB Chemistry SLTopic 5 — Quantifying Chemical ChangePaper 1 & 2Core skill~12 min read
Avogadro’s Law and Molar Gas Volume
Gases barely notice what they are made of. A cubic decimetre of hydrogen and a cubic decimetre of carbon dioxide hold the same number of molecules, and that one fact turns awkward mass calculations into arithmetic you can do in your head.
📚 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 mole ratio, read straight off the coefficients.
At STP (0 °C = 273 K and 100 kPa), one mole of any gas occupies 22.7 dm3.
V = n × 22.7 and n = V / 22.7, with V in dm3.
Convert cm3 to dm3 by dividing by 1000.
Only gases count. Solids and liquids contribute no gas volume at all.
Avogadro’s law
The molecules differ enormously in size and mass, yet the same box holds the same number. In a gas, the particles are so far apart that their own size is irrelevant.
That last point is the physical reason the law works. In a gas the molecules are separated by distances vastly larger than the molecules themselves, so what fills the container is mostly empty space. Swapping a small molecule for a large one changes the mass in the box but not how many will fit.
Avogadro’s law
equal volumes of gases at the same temperature and pressure contain equal numbers of particles
The practical payoff is large: for gaseous species, the coefficients in a balanced equation can be read as volume ratios directly. No molar masses, no weighing.
Notice the condition. Equal volumes contain equal numbers only at the same temperature and pressure. If a question quietly changes conditions between two measurements, the shortcut is no longer available.
Molar gas volume
At STP: 273 K and 100 kPa
V = n × 22.7 n = V ÷ 22.7
V in dm3, molar volume 22.7 dm3 mol–1
Two conventions exist and they are easy to confuse. STP is 273 K and 100 kPa, giving 22.7 dm3 mol–1 — this is the IB value and it is in the data booklet. The older 22.4 dm3 mol–1 belongs to a different pressure (101.3 kPa) and is not the one to use.
WORKED EXAMPLE
(a) What volume does 0.250 mol of carbon dioxide occupy at STP? (b) How many moles are there in 500 cm3 of oxygen at STP?
(a) moles to volumeV = 0.250 × 22.7 = 5.675V = 5.68 dm³(b) convert the units first500 ÷ 1000 = 0.500 dm³n = 0.500 ÷ 22.7 = 0.02203n = 0.0220 molThe identity of the gas never entered either calculation. That is the whole point of a molar volume.
Volume ratios straight from the equation
When every substance you care about is a gas, you can work entirely in volumes. Divide each volume by its coefficient: the smallest answer is the reactant that runs out first, exactly as with moles.
Total gas volume falls from 140 cm3 to 60 cm3, because three gas molecules become one and the water condenses.
WORKED EXAMPLE
40 cm3 of methane is burnt in 100 cm3 of oxygen. Calculate the total volume of gas remaining, measured at the same temperature and pressure, at which water is a liquid. CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
Step 1 — which runs out firstCH₄: 40 ÷ 1 = 40 O₂: 100 ÷ 2 = 50Methane gives the smaller number, so methane is limiting.Step 2 — oxygen used and left40 × 2 = 80 cm³ used, so 100 − 80 = 20 cm³ leftStep 3 — carbon dioxide made40 × 1 = 40 cm³Step 4 — add up the gasesThe water is a liquid at this temperature, so it contributes nothing.40 + 20 = 6060 cm³ of gas remainsLeftover reactant still counts as gas in the container. It is the single most missed part of these questions.
WORKED EXAMPLE
Calculate the volume of carbon dioxide, measured at STP, released when 25.0 g of calcium carbonate decomposes completely. CaCO3(s) → CaO(s) + CO2(g)
Step 1 — moles of the solidn = 25.0 ÷ 100.09 = 0.2498 molStep 2 — ratio 1 : 1n(CO₂) = 0.2498 molStep 3 — moles to gas volumeV = 0.2498 × 22.7 = 5.670V = 5.67 dm³Mass in, volume out — the two halves of the mole map joined in one question. Only CO₂ gets the 22.7; the solids never do.
💡 Exam tip
Check the units before anything else. cm3 must become dm3, and dividing by 1000 is the most commonly forgotten line.
Use 22.7, not 22.4, and only at STP.
In volume-ratio questions, ignore anything that is not a gas — check the state symbols.
Remember to add unreacted excess gas when asked for the total volume remaining.
“Same temperature and pressure” in a question is your signal that volumes behave like moles.
⚠️ Common mix-up
Applying 22.7 to a solid or a liquid. It is a molar gas volume.
Forgetting water is often liquid and counting it in a final gas volume.
Leaving the volume in cm3 and getting an answer 1000 times too big.
Assuming the limiting gas is the one with the smaller volume. Divide by the coefficient first.
Using 22.7 at non-standard conditions. Different temperature or pressure, different molar volume.
Up next: Calculating Concentration — the third way of measuring an amount, and the one that makes titration the most precise technique in a school laboratory.
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