⬢ Higher Tier Only
SPEC 4.3.2.1 • QUANTITATIVE CHEMISTRY • PAPER 1
Moles — The Chemist's Counting Unit
One number — 6.02 × 10²³ — connects every quantitative chemistry calculation in the rest of the course. This lesson is the foundation; lessons 3.04, 3.05, and 3.07 all build on it.
What Is a Mole?
A mole is just a very specific number. Just as a dozen means 12, a mole means 6.02 × 10²³.
Mole
The mole (symbol mol) is the unit chemists use to measure amount of substance. One mole of any substance contains 6.02 × 10²³ particles — atoms, molecules, ions, electrons, or formula units, depending on the substance.
Why such a strange number?
Atoms are tiny. To measure them out in countable quantities, chemists chose a number big enough that one mole of a substance has a mass we can weigh on a school balance — typically a few grams to a few hundred grams. 6.02 × 10²³ is the number that makes the maths work for grams.
Memory hook
"A mole of X" means 6.02 × 10²³ of whatever the particles of X are. One mole of carbon = 6.02 × 10²³ C atoms. One mole of water = 6.02 × 10²³ H₂O molecules. One mole of NaCl = 6.02 × 10²³ NaCl formula units.
Avogadro's Constant
The number 6.02 × 10²³ per mole has its own name and symbol — Avogadro's constant. You will be given this value in any question that needs it.
Avogadro's constant = 6.02 × 10²³ per mole
What it means in the spec
The spec gives a striking example: in one mole of carbon (C), the number of atoms equals the number of molecules in one mole of carbon dioxide (CO₂). Both are 6.02 × 10²³ particles — even though the particles are different (atoms vs molecules).
This is the whole point: a mole is a count, regardless of what's being counted.
⚠️ When to use Avogadro's constant: ONLY when the question asks for a number of particles ("how many atoms / molecules / ions"). For straightforward moles ↔ mass conversion, you do not need it.
The Mole Formula
The single formula that connects moles to mass — and the one you will use in every Module 3 calculation from here on.
moles = mass (g) ÷ M_r
Why it works
The mass of one mole of a substance, in grams, is numerically equal to its M_r (or A_r for an element counted as single atoms).
1 mol of C atoms has mass 12 g (A_r = 12)
1 mol of H₂O molecules has mass 18 g (M_r = 18)
1 mol of CO₂ molecules has mass 44 g (M_r = 44)
1 mol of NaCl formula units has mass 58.5 g (M_r = 58.5)
1 mol of O₂ molecules has mass 32 g (M_r = 2 × 16 = 32)
⚠️ Diatomic gas trap: several elements exist as diatomic molecules when free — H₂, N₂, O₂, F₂, Cl₂ (gases at room temperature), plus Br₂ (liquid) and I₂ (solid). When the question is about the molecular form, use M_r of the molecule (= 2 × A_r), not A_r of the atom. So 32 g of oxygen gas = 32 ÷ 32 = 1 mol of O₂, not 32 ÷ 16 = 2 mol. This trap appears in nearly every Module 3 calculation involving gases.
Use M_r or A_r?
Match the entity you're counting:
Counting molecules (H₂O, CO₂, O₂, H₂) → use M_r
Counting formula units (NaCl, CaCO₃) → use M_r
Counting individual atoms (C in graphite, Mg metal, Fe metal — anything written as a bare element symbol in an equation) → use A_r
Three Rearrangements
The mole formula has three forms — one for each missing quantity. You'll need all three.
Find moles
moles = mass ÷ M_r
Use when given mass and M_r.
Find mass
mass = moles × M_r
Use when given moles and M_r.
Find M_r
M_r = mass ÷ moles
Use when given mass and moles.
Triangle Memory Trick
Cover the variable you want, and the triangle gives you the formula. Cover mass → moles × M_r (multiply across the bottom). Cover moles → mass ÷ M_r (top divided by remaining bottom). Cover M_r → mass ÷ moles.
Worked Example 1 — Mass to Moles
Calculate the number of moles of water in 36 g of water. (M_r(H₂O) = 18.)
1
Identify the formula. Given mass and M_r; want moles. → Use moles = mass ÷ M_r
2
Substitute and label.
moles of H₂O = 36 ÷ 18
3
Calculate.
moles of H₂O = 2 mol
moles = 2 mol
Sanity check: 2 mol × 18 g/mol = 36 g ✓
Worked Example 2 — Moles to Mass
Calculate the mass of 0.5 mol of calcium carbonate (CaCO₃). [A_r: Ca = 40, C = 12, O = 16]
1
Calculate M_r. The question gives A_r values, not M_r directly — you must compute M_r first.
M_r(CaCO₃) = 40 + 12 + 3(16) = 100
2
Identify the formula. Given moles and M_r; want mass. → Use mass = moles × M_r
3
Substitute and label.
mass of CaCO₃ = 0.5 × 100 = 50 g
mass = 50 g
Sanity check: 50 g ÷ 100 g/mol = 0.5 mol ✓ (and 0.5 mol is half of 1 mol, so should be half of 100 g)
Worked Example 3 — Number of Particles
Calculate the number of molecules in 0.2 mol of CO₂. (Avogadro's constant = 6.02 × 10²³ per mole.)
The Trigger
The question asks for "number of molecules" — this is the cue to use Avogadro's constant. Multiply moles by 6.02 × 10²³.
1
Identify the formula. Number of particles = moles × Avogadro's constant.
number of molecules = 0.2 × (6.02 × 10²³)
2
Calculate.
= 0.2 × 6.02 × 10²³ = 1.204 × 10²³
≈ 1.20 × 10²³ molecules
Calculator Tip
Use the EXP or ×10ˣ button on your calculator. Type: 0.2 × 6.02 EXP 23 = . Don't write out 6.02 × 10 × 10 × 10 ... — that's a recipe for mistakes.
Significant Figures (MS 2a)
The data here is given to 3 significant figures (0.2 has 1 sig fig if literal, but exam convention treats it as exact; 6.02 × 10²³ has 3 sig figs). Quote your final answer to the same number of sig figs as the data — usually 2 or 3. Writing "1.204 × 10²³" claims precision the data doesn't support; "1.20 × 10²³" matches.
Counting Atoms Within Molecules
The spec gives a striking example: "in one mole of carbon (C) the number of atoms is the same as the number of molecules in one mole of carbon dioxide (CO₂)." Both equal 6.02 × 10²³. But what about the atoms inside the CO₂ molecules?
The Distinction
Each CO₂ molecule contains 1 carbon atom and 2 oxygen atoms — 3 atoms in total. So in 1 mole of CO₂:
6.02 × 10²³ CO₂ molecules
6.02 × 10²³ C atoms (1 per molecule × 6.02 × 10²³)
1.204 × 10²⁴ O atoms (2 per molecule × 6.02 × 10²³)
1.806 × 10²⁴ total atoms (3 per molecule × 6.02 × 10²³)
⚠️ Read the question carefully: "How many molecules in 0.5 mol CO₂?" → 0.5 × 6.02 × 10²³. But "how many oxygen atoms in 0.5 mol CO₂?" → 0.5 × 2 × 6.02 × 10²³ (because each molecule contains 2 O atoms).
The Rule
To find the number of atoms of a specific element in a sample: multiply by the number of those atoms in one formula unit / molecule. The subscripts in the formula tell you the multiplier.
Multi-Step Calculation: Mass → Particles
The trickier exam question: combine moles ↔ mass and Avogadro's constant in two steps.
Question
Calculate the number of molecules in 9.0 g of water. (M_r(H₂O) = 18; Avogadro's constant = 6.02 × 10²³.)
1
Mass → moles.
moles of H₂O = mass ÷ M_r = 9.0 ÷ 18 = 0.5 mol
2
Moles → number of molecules.
number of molecules = 0.5 × (6.02 × 10²³) = 3.01 × 10²³ molecules
3.01 × 10²³ molecules of H₂O
⚠️ Pattern to spot: "Number of molecules" / "number of atoms" + a mass given → 2-step calculation. Mass to moles first, then moles to particles. Never try to go straight from grams to particles.
Label Every Step
The single most-penalised error in Module 3 — across every Higher Tier paper, every year. Examiners can't credit a method they can't see.
⚠️ EXAMINER ALERT (Cross-cutting theme): Examiners describe the typical poor answer as "a page full of additions and subtractions with no words to explain what these represented." Numbers without labels can't earn the method mark — even when the chemistry is correct.
UNLABELLED — loses marks
5.00 ÷ 100 = 0.05 × 6.02 × 10²³ = 3.01 × 10²²Examiner: "What is 0.05? What is 100? I cannot give credit for working I cannot interpret."
LABELLED — full marks
M_r(CaCO₃) = 100 moles = 5.00 ÷ 100 = 0.05 mol molecules = 0.05 × 6.02 × 10²³ = 3.01 × 10²² moleculesEvery step has a label and a unit — full marks even if the arithmetic slips.
HT Calculator Hygiene — The % Button
One specific calculator error appears across HT papers. It's not unique to moles, but the same mistake will cost you marks throughout Module 3 — particularly in lesson 3.06 (% yield, atom economy) and any question involving percentages.
⚠️ EXAMINER ALERT (P1 Jun24 Q07.5 — a percentage-by-mass question): students used both the % button on the calculator AND manually divided by 100. The % button already divides by 100. Doing both gives an answer 100× too small. (The cited question is from spec 4.3.1.2, not 4.3.2.1 — but the calculator habit is one to fix now, before it costs you marks elsewhere.)
The Rule
Pick one method and stick to it:
EITHER use the % button (which automatically divides by 100)
OR type the percentage as a decimal yourself (e.g. 25% = 0.25)
OR divide by 100 manually after typing the percentage value
NEVER combine two methods. If you've used the % key, you have already divided by 100 — don't divide again.
Safer Option
Most chemistry teachers recommend converting percentages to decimals manually (25% → 0.25; 7.5% → 0.075) and never touching the % key at all. It's one less thing to go wrong under exam pressure.
Practice Question 1
Calculate the number of moles in 11.0 g of carbon dioxide (CO₂).
A_r: C = 12, O = 16. 2 marks
1
Calculate M_r.
M_r(CO₂) = 12 + 2(16) = 441 mark — method
2
Apply the formula.
moles of CO₂ = mass ÷ M_r = 11.0 ÷ 44 = 0.25 mol1 mark — answer
Practice Question 2
A student weighs 0.10 mol of an unknown compound and finds it has a mass of 4.6 g.
Calculate the relative formula mass (M_r) of the compound. 2 marks
1
Identify the formula. Given mass and moles; want M_r. → Use M_r = mass ÷ moles1 mark — method
2
Substitute and calculate.
M_r = 4.6 ÷ 0.10 = 461 mark — answer
(For interest: 46 is the M_r of NO₂ or ethanol C₂H₅OH — but the question doesn't require you to identify the compound.)
Practice Question 3
Calculate the number of molecules in 4.4 g of carbon dioxide (CO₂).
A_r: C = 12, O = 16. Avogadro's constant = 6.02 × 10²³ per mole. 3 marks
1
M_r. M_r(CO₂) = 12 + 2(16) = 44
2
Mass → moles.
moles = 4.4 ÷ 44 = 0.10 mol1 mark
3
Moles → molecules.
molecules = 0.10 × (6.02 × 10²³) = 6.02 × 10²² molecules2 marks (1 method, 1 answer)
Common Errors — Mark-Scheme Verbatim
The error patterns examiners flag most often on mole calculations. Right-hand column is what scores.
Students often writeCorrect versionWhy it matters
moles = M_r ÷ mass (inverted)moles = mass ÷ M_rInversion gives an answer wrong by a factor of M_r²
Numbers scattered with no labelsLabel every step: "M_r = ", "moles = ", "molecules = "No method mark without visible working (cross-cutting theme)
Used % button AND divided by 100Pick one method — the % key already divides by 100Answer 100× too small. (Calculator habit; cross-cutting across HT — surfaced in P1 Jun24 Q07.5, a % by mass question.)
Multiplying mass by Avogadro's constant directlyAlways convert to moles first , then × 6.02 × 10²³Avogadro's constant relates moles ↔ particles, not mass ↔ particles
Forgetting to compute M_r firstIf A_r values are given (not M_r), compute M_r in a labelled step before the moles formulaHidden step → no method mark
Using A_r(O) = 16 for "moles of oxygen gas"Oxygen gas = O₂; use M_r(O₂) = 32Diatomic gas trap — answer wrong by factor of 2 (also N₂, H₂, Cl₂)
Why This Matters for the Rest of Module 3
The mole formula is the foundation of every Higher Tier calculation in Module 3. Lesson 3.04 builds directly on it.
What's coming next
3.04: Using moles to find masses of reactants and products from balanced equations
3.05: Limiting reactants and concentration in mol/dm³
3.06: Percentage yield and atom economy
3.07: Volumes of gases at room temperature and pressure
Strategy
If you make any error in this lesson — inverted formula, missing M_r, lost label — the same mistake will cost you marks across five lessons. Master moles → mass and mass → moles now, with labelling, and the rest of Module 3 becomes substitution into bigger calculations.
Summary
Key Definitions
Mole (mol): the unit for amount of substance; one mole contains 6.02 × 10²³ particles
Avogadro's constant: 6.02 × 10²³ per mole — the number of particles in one mole of any substance
The Three Formulae
moles = mass ÷ M_r
mass = moles × M_r
M_r = mass ÷ moles
Avogadro's Constant
Use ONLY for "number of particles" questions:
number of particles = moles × 6.02 × 10²³
M_r first
If A_r values are given (not M_r), compute M_r in a labelled step before applying the moles formula.
Label everything
Every step must have a label: "M_r = ", "moles = ", "molecules = ". Cross-cutting theme — affects every Module 3 calculation.
Exam Tip
The moles formula is the most-used equation in Higher Tier Chemistry. Memorise the triangle. Compute M_r first. Label every step. Use Avogadro's constant only when asked for a number of particles.
Ready for 3.04 [HT only]
Next up: Amounts of Substances in Equations and Using Moles to Balance Equations — applying moles = mass ÷ M_r across a balanced equation to find unknown masses, and using moles to deduce coefficients in equations from experimental data.
Next: What Is a Mole?