IB Chemistry SL Topic 1 — Counting Particles by Mass Paper 1 & 2 Core skill ~9 min read

Molar Mass

You can’t count atoms directly, but you can weigh them. Molar mass is the bridge: it links the mass you measure on a balance to the number of moles — and from there to the actual number of particles.

📘 What you need to know

Why we use moles

Atoms are so tiny that even a pinch of a substance contains an astronomical number of them. Those numbers are impossible to work with directly, so we bundle particles into moles — a practical counting unit. Two simple formula triangles handle all the conversions.

Linking moles and particles

To go between a number of moles and the actual number of particles, use the Avogadro constant (L):

PARTICLES MOLES L × ÷ ÷
Cover the quantity you want: particles = moles × L, and moles = particles ÷ L. (L = 6.02 × 10²³ mol⁻¹.)
WORKED EXAMPLE

How many hydrogen atoms are in 0.010 moles of CH3CHO?

Count H atoms per molecule CH₃CHO has 4 H atoms, so 0.010 mol of molecules = 0.040 mol of H atoms. atoms = moles × L 0.040 × (6.02 × 10²³) = 2.4 × 10²² H atoms

Linking moles and mass

The molar mass (M) is the relative mass in grams — units g mol-1. A second formula triangle links moles, mass and molar mass:

MASS MOLES M × ÷ ÷
Cover the quantity you want: mass = moles × M, and moles = mass ÷ M. (M in g mol⁻¹.)
The key relationship moles = mass (g) ÷ molar mass (g mol-1)
WORKED EXAMPLE

(a) What is the mass of 0.250 moles of zinc (Ar = 65.38)? (b) How many moles are in 2.64 g of sucrose, C12H11O22 (Mr = 342.3)?

(a) mass = moles × M 0.250 × 65.38 = 16.3 g (b) moles = mass ÷ M 2.64 ÷ 342.3 = 7.71 × 10⁻³ mol 16.3 g · 7.71 × 10⁻³ mol

💡 Exam tip

Up next: Empirical Formulae — working out the simplest whole-number ratio of atoms in a compound from mass or percentage data.

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