IB Chemistry HLTopic 1 — Counting Particles by MassPaper 1 & 2Core idea~10 min read
Avogadro’s Law
For gases there is a shortcut that does not exist for anything else: you can count particles just by measuring volume. No balance, no molar mass, no conversion. Understand why that works and a whole family of exam questions becomes a few seconds of arithmetic.
📘 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, volume ratio = mole ratio — the balancing numbers in an equation can be read straight as volumes.
At STP (0 °C, or 273 K, and 100 kPa) one mole of any gas occupies 22.7 dm3.
n = V ÷ 22.7, with V in dm3. This is the molar volume, Vm.
This works for gases only, and only when everything is compared at the same temperature and pressure.
Liquids and solids in an equation are ignored when you are adding up gas volumes.
Why volume can count particles
In a gas, the particles are tiny compared with the space between them — the molecules themselves take up around a thousandth of the container. So the volume of a gas is basically a measure of how much elbow room the particles have claimed, not of how big the particles are.
That is the key idea. A CO2 molecule is 22 times heavier than an H2 molecule, but at the same temperature and pressure it commands the same amount of space. Swap them and the volume does not change. Volume depends on how many, not on which.
Ten molecules are drawn in each box. In reality it would be around 1022, but the point stands: the count is set by the box, not by the gas.
Molar volume and STP
If one mole of any gas takes up the same volume, that volume is worth learning. The IB uses STP: standard temperature and pressure.
Molar volume at STPVm = 22.7 dm3 mol−1 at 273 K and 100 kPa
So a mole of any gas at STP would fill a cube roughly 28 cm along each edge — about the size of a football. That is the same box for hydrogen, chlorine or carbon dioxide.
Amount of gas from volumen = V ÷ 22.7 and V = n × 22.7
Watch the conditions. 22.7 dm3 mol−1 only applies at STP. If a question gives a different temperature or pressure, you cannot use it — you need the ideal gas equation instead.
Reading volumes straight off an equation
Here is the part that saves time. Because equal volumes contain equal numbers of particles, the coefficients in a balanced equation are also the volume ratio — provided everything you are comparing is a gas at the same conditions.
Notice the total volume falls from 4 units to 2 units. Gas reactions can lose or gain volume, which is exactly how equilibrium questions later use pressure.
Because the ratio is all you need, you can often answer a gas volume question with a single multiplication and no molar mass at all. If you find yourself converting to grams, stop and check whether the shortcut applies.
🧩 Gas volume questions: the method
Balance the equation and cross out anything that is not a gas — liquids and solids take up almost no volume.
Check whether a reactant runs out. Divide each given volume by its coefficient; the smallest answer is the limiting reactant.
Scale from the limiting reactant using the coefficient ratio to find each product volume.
Work out what is left over of the reactant in excess, if the question asks for a total.
Add up only the gases at the stated conditions. Water is often liquid — check the state symbol.
Worked examples
WORKED EXAMPLE
Mass of gas to volume at STP
Calculate the volume occupied by 3.20 g of oxygen gas, O2, at STP.
Step 1: mass to molesM(O₂) = 2 × 16.00 = 32.00 g mol⁻¹n = 3.20 ÷ 32.00 = 0.100 molStep 2: moles to volume using the molar volumeV = 0.100 × 22.7 = 2.27V = 2.27 dm³use 32.00, not 16.00 — oxygen gas is diatomic
WORKED EXAMPLE
Volumes straight from the equation
100 cm3 of propane is burned completely in excess oxygen. Calculate the volume of oxygen used and the volume of carbon dioxide formed, all measured at the same temperature and pressure.
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)
Step 1: read the ratio off the equationC₃H₈ : O₂ : CO₂ = 1 : 5 : 3Step 2: scale from the propaneO₂ = 5 × 100 = 500 cm³CO₂ = 3 × 100 = 300 cm³500 cm³ O₂, 300 cm³ CO₂the water is liquid, so it contributes no gas volume
WORKED EXAMPLE
Limiting reactant with gas volumes
60 cm3 of methane is mixed with 180 cm3 of oxygen and ignited. Calculate the total volume of gas remaining, measured at room temperature.
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
Step 1: which one runs out? Divide by the coefficientsCH₄: 60 ÷ 1 = 60O₂: 180 ÷ 2 = 9060 is smaller, so methane is limitingStep 2: oxygen actually used2 × 60 = 120 cm³, so 180 − 120 = 60 cm³ left overStep 3: carbon dioxide formed1 × 60 = 60 cm³Step 4: add up the gases only60 (excess O₂) + 60 (CO₂) = 120120 cm³ of gas remainingat room temperature the water has condensed — do not count it
WORKED EXAMPLE
Identifying a gas from its density
An unknown gas has a density of 1.25 g dm−3 at STP. Calculate its molar mass and suggest an identity.
Step 1: 1 mol occupies 22.7 dm³, so weigh that muchM = 1.25 × 22.7 = 28.375M ≈ 28.4 g mol⁻¹Step 2: what has that molar mass?N₂ is 28.02 and CO is 28.01 — either fits.density × molar volume = molar mass, because g dm⁻³ × dm³ mol⁻¹ leaves g mol⁻¹
💡 Exam tip
Check the state symbols before you add volumes. Water is very often (l), and liquids contribute essentially nothing to gas volume.
If every substance in the comparison is a gas, do not convert to moles at all — work directly in volumes.
Use the divide-by-the-coefficient trick to find the limiting reactant. The smallest answer wins, every time.
22.7 dm3 mol−1 is for STP only. If the question says room temperature or gives a pressure in the question, use the ideal gas equation instead.
Keep an eye on cm3 versus dm3. Ratios are safe in either, but the molar volume needs dm3.
For “total volume remaining”, remember to include unreacted excess reactant as well as the products.
⚠ Common mix-up
Counting liquid water as a gas. If the equation says H2O(l), it takes no part in the volume total.
Using 22.7 dm3 at the wrong conditions. It is an STP value, not a universal one, and older textbooks quote 22.4 dm3 for a different standard.
Applying Avogadro’s law to solids or solutions. Equal volumes of two liquids contain nothing like equal numbers of particles.
Forgetting the excess reactant when adding up the gas left at the end.
Assuming the reactant with the smaller volume is limiting. You must divide by the coefficient first — 180 cm3 of O2 can still be in excess.
Using Ar instead of Mr for diatomic gases. Oxygen gas is 32.00, not 16.00.
Up next: The Ideal Gas Equation — what to do when the gas is not at STP, and how pressure, volume and temperature all tie back to the same number of moles.
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