Overview
What you’ll learn
- State what a mole represents and convert masses to moles and back with mol = mass ÷ Mr.
- Read balanced equations as mole recipes and use the ratios to predict reacting and produced masses.
- Decide which reactant is limiting and calculate the product from it.
- Convert between moles of gas and volumes using 24 dm³ per mole (Chemistry only, Higher).
Specification checklist (AQA 8462 · Edexcel 1CH0 · OCR Gateway J248)
Statements are paraphrased from the shared DfE subject content and grouped by tier. Foundation content is studied by every student; Higher statements are examined on Higher-tier papers only, and statements badged Chemistry only are examined in separate (triple) Chemistry, not Combined Science. QuizQuip is not affiliated with AQA, Pearson Edexcel or OCR.
Higher — Higher tier only
- 6.1Recall that one mole of any substance contains 6.02 × 10²³ particles (the Avogadro constant), and that the mass of one mole of a substance in grams equals its relative formula mass.
- 6.2Use the relationship moles = mass ÷ Mr to convert between the mass of a substance and the amount in moles.
- 6.3Interpret balanced symbol equations as mole ratios: the coefficients state how many moles of each substance react and form.
- 6.4Calculate the mass of a product from a given mass of reactant (and vice versa) using moles and the equation’s mole ratio.
- 6.5Identify the limiting reactant from the moles of each reactant and the equation ratio, and explain that the amount of product is set by the limiting reactant while the excess is left over.
- 6.6Use the molar gas volume: one mole of any gas occupies 24 dm³ (24 000 cm³) at room temperature and pressure, so volume of gas = moles × 24 dm³.Chemistry only
- 6.7Explain why the mole concept underpins conservation of mass: equal numbers of particles are rearranged according to the equation, and no atoms appear or disappear.
Key points
- 1One mole = 6.02 × 10²³ particles (the Avogadro constant, Nₐ).
- 2The mass of one mole in grams is the Mr: 1 mol of water = 18 g, 1 mol of CO₂ = 44 g.
- 3moles = mass ÷ Mr and mass = moles × Mr — the two working relationships of the whole topic.
- 4A balanced equation is a mole recipe: 2Mg + O₂ → 2MgO means 2 mol : 1 mol : 2 mol.
- 5Reacting masses (Higher): convert the known mass to moles, multiply by the ratio, convert back.
- 6The limiting reactant (Higher) runs out first and fixes the product amount; anything extra stays unreacted.
- 7One mole of any gas occupies 24 dm³ at room temperature and pressure (Chemistry only, Higher): volume = mol × 24 dm³.
Notes · 1 of 5
The mole: chemistry’s counting unit
Atoms are too small to count one by one — so we count in enormous, weighable batches.
A mole is simply a very large number of particles: 6.02 × 10²³ of them — the Avogadro constant. One mole of carbon atoms, one mole of water molecules, one mole of electrons: each packet holds the same count. Why that strange number? Because it makes the packet weigh nicely: the mass of one mole of a substance in grams equals its relative formula mass. One mole of carbon: 12 g. One mole of water (Mr 18): 18 g. One mole of carbon dioxide (Mr 44): 44 g. The chemist’s scale connects straight to the atom count.
| Substance | Formula | Mr | Mass of 1 mol | Particles in 1 mol |
|---|---|---|---|---|
| Carbon | C | 12 | 12 g | 6.02 × 10²³ atoms |
| Water | H₂O | 18 | 18 g | 6.02 × 10²³ molecules |
| Carbon dioxide | CO₂ | 44 | 44 g | 6.02 × 10²³ molecules |
| Magnesium oxide | MgO | 40 | 40 g | 6.02 × 10²³ ion pairs |
Fractions of moles scale the same way: 0.5 mol of water weighs 9 g and holds 3.01 × 10²³ molecules; 2 mol of CO₂ weighs 88 g. The mole is a proportion, not a prescription.
Notes · 2 of 5
Equations as mole recipes
The big numbers in an equation are a ratio — the recipe linking reactants to products.
A balanced equation tells you how many moles of each substance take part. In 2Mg + O₂ → 2MgO, the coefficients say: 2 mol of magnesium react with 1 mol of oxygen to give 2 mol of magnesium oxide. Double the magnesium and you must double the oxygen and the oxide. The ratio 2 : 1 : 2 is the recipe — and every reacting-mass question is just the recipe applied to your actual amounts.
| Equation | Mole ratio (reactants : products) |
|---|---|
| 2Mg + O₂ → 2MgO | 2 : 1 → 2 |
| CaCO₃ → CaO + CO₂ | 1 → 1 : 1 |
| CH₄ + 2O₂ → CO₂ + 2H₂O | 1 : 2 → 1 : 2 |
| 2H₂ + O₂ → 2H₂O | 2 : 1 → 2 |
Because atoms are merely rearranged — never created or destroyed — the mole picture is exactly why conservation of mass works. Count the atoms either side of the recipe and the totals match; weigh them either side of the experiment and the masses match too.
Notes · 3 of 5
Reacting-mass calculations (Higher)
Mass → moles → ratio → moles → mass: one road, walked the same way every time.
- 1Step 1 — known mass to moles
Divide by the Mr of the substance you were given.
- 2Step 2 — ride the ratio
Multiply (or divide) by the coefficient ratio from the balanced equation.
- 3Step 3 — moles back to mass
Multiply by the Mr of the substance asked for.
Notes · 4 of 5
Limiting reactants (Higher)
The recipe stops when the first ingredient runs out — everything else waits unused.
Real mixtures rarely match the recipe exactly. Whatever a reactant is in short supply relative to the equation ratio is the limiting reactant — it determines how much product can form. The reactant left over is in excess. To find which is which: convert each reactant’s mass to moles, then compare each to the equation ratio. For 2Mg + O₂ → 2MgO, 0.2 mol of Mg needs 0.1 mol of O₂; if only 0.05 mol of O₂ is present, oxygen is limiting — it can burn just 0.1 mol of Mg and the rest of the metal sits unreacted.
The limiting-reactant idea explains real industrial choices: factories feed the cheaper reactant in excess so the expensive one is fully used up — air is free, so processes burn fuel in excess air; in the Haber process hydrogen is the costly ingredient, so it is the one engineers try not to waste.
Notes · 5 of 5
Molar gas volume: 24 dm³ (Chemistry only, Higher)
At room temperature and pressure, one mole of any gas fills the same box: 24 litres.
Gases spread out to fill their container, and at room temperature and pressure (RTP) one mole of any gas occupies the same volume: 24 dm³, which is 24 000 cm³. The identity of the gas does not matter — a mole of hydrogen, a mole of chlorine and a mole of methane all fill 24 dm³ at RTP, even though the chlorine weighs 71 g and the hydrogen just 2 g. Volume of gas = moles × 24 dm³; moles of gas = volume ÷ 24.
Revise
Key terms
The words examiners expect you to use precisely.
- Mole
- The amount of substance containing 6.02 × 10²³ particles; the chemist’s counting unit.
- Avogadro constant
- The number of particles in one mole: 6.02 × 10²³ per mole.
- Molar mass
- The mass of one mole of a substance in grams, numerically equal to its relative formula mass.
- Mole ratio
- The ratio of moles of reactants and products given by the coefficients of a balanced equation.
- Limiting reactant
- The reactant that is completely used up first, fixing the maximum amount of product.
- Excess reactant
- A reactant left over at the end because the other reactant ran out first.
Show 6 more key termsShow fewer key terms
- Molar gas volume
- The volume occupied by one mole of any gas at room temperature and pressure: 24 dm³.
- Room temperature and pressure (RTP)
- About 20 °C and 1 atmosphere — the conditions for the 24 dm³ molar gas volume.
- Reacting-mass calculation
- Predicting the mass of product from a mass of reactant via moles and the equation ratio.
- Relative formula mass (Mr)
- The sum of the relative atomic masses in a formula; the mass in grams of one mole.
- Conservation of mass
- The total mass of reactants equals the total mass of products, because atoms are only rearranged.
- dm³
- A cubic decimetre — one litre, 1000 cm³.
Revise
Common mistakes
Where students most often lose marks in this chapter — and what to do instead.
Mistake: Dividing mass by 24 to find moles of a solid.
24 is the molar gas volume, nothing to do with solids.
Instead: moles = mass ÷ Mr for solids and liquids; volume ÷ 24 only for gases at RTP.
Mistake: Using atomic number instead of Ar as the Mr.
Atomic numbers count protons; Mr needs the relative atomic masses from the table.
Instead: Read Ar values off the periodic table and add them per the formula.
Show 10 more mistakesShow fewer mistakes
Mistake: Forgetting to halve or double when the ratio is not 1 : 1.
The coefficients are the recipe — ignoring them breaks the calculation.
Instead: Multiply the moles by (wanted coefficient ÷ given coefficient).
Mistake: Saying the limiting reactant is whichever has the smaller mass.
Masses are not comparable — different Mr values mean different particle counts.
Instead: Convert both to moles, then compare each against the equation ratio.
Mistake: Calculating product from the excess reactant.
Once the limiting reactant is gone, the reaction stops.
Instead: Product amount is always set by the limiting reactant.
Mistake: Quoting the Avogadro constant as 6.02 × 10²³ grams.
It is a count of particles, dimensionless — the mass depends on the substance.
Instead: 6.02 × 10²³ particles per mole; a mole of H₂ weighs 2 g, a mole of O₂ weighs 32 g.
Mistake: Using 24 dm³ for gases at any temperature and pressure.
24 dm³ holds only at room temperature and pressure.
Instead: At other conditions the molar volume changes; GCSE questions specify RTP for 24 dm³.
Mistake: Mixing up cm³ and dm³ in gas-volume work.
A factor of 1000 error follows.
Instead: 1 dm³ = 1000 cm³; 0.5 mol of gas = 12 dm³ = 12 000 cm³.
Mistake: Thinking equal masses of gases contain equal moles.
Heavier molecules pack fewer per gram.
Instead: Equal volumes of gas at RTP contain equal moles; equal masses do not (4 g H₂ is 2 mol, 4 g O₂ is 0.125 mol).
Mistake: Adding Mr values of reactants to predict product mass per mole incorrectly, e.g. expecting 2 mol Mg + 1 mol O₂ to give 3 mol of product.
Moles of products are fixed by the equation, not by adding reactant moles.
Instead: 2Mg + O₂ → 2MgO: 3 mol of reactant atoms end up in just 2 mol of MgO formula units — the total mass still matches.
Mistake: Rounding mid-calculation.
Rounded intermediates drift the final answer outside the mark range.
Instead: Carry extra digits through and round only the final answer to the asked precision.
Mistake: Assuming mass is lost when a solid dissolves or melts.
No particles leave — physical changes keep every atom in the beaker.
Instead: Only escaping gas changes an open system’s mass; dissolving and melting conserve it.
Check yourself
Quick quiz
30 questions with instant explanations. About 5 minutes.
Question 1 of 3
- Question 1: current question
- Question 2: not answered
- Question 3: not answered
Recall the Avogadro constant
A mole of oxygen molecules and a mole of hydrogen molecules both contain
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Practise
Topical questions
18 exam-style questions with model answers and mark schemes.
A set of structured and calculation questions matched to this chapter, in three tiers, each with a model answer, mark scheme and examiner tip. Higher-tier and Chemistry-only questions are tagged.
Practise topical questionsExam practice
Practice paper
A 30-mark paper in the style of a GCSE paper, with a separate mark scheme. Give yourself 40 min.
Question paper
Chapter 6 Practice Paper
30 marks · 40 min · 5 pages
10 multiple-choice questions and structured questions with answer lines. Chapter 6: the mole and Avogadro constant, mol = mass ÷ Mr, equations as mole recipes, reacting masses, limiting reactants, molar gas volume (Chemistry-only, Higher).
Mark scheme
PROChapter 6 Practice Paper — mark scheme
Answers · Marking points · 3 pages
Model answers with a mark for each creditworthy point — for marking at home or in class.
How is this chapter examined?
Paper 1
1 h 45 · 100 marks
Chapters 1–11: atoms, bonding, quantitative, chemical and energy changes
Paper 2
1 h 45 · 100 marks
Chapters 12–18: rates, organic, analysis, atmosphere, resources
Tiers
Foundation (grades 1–5) · Higher (grades 4–9)
Higher assumes all Foundation content; a periodic table is printed
Want a full paper? Try the Paper 1-style (Foundation), Paper 1-style (Higher), Paper 2-style (Foundation), Paper 2-style (Higher), Required practicals practice sets.
Questions
Frequently asked questions
Do I have to memorise 6.02 × 10²³?
Yes — it is one of the few numbers GCSE Chemistry expects from memory, along with 24 dm³ for the molar gas volume at RTP. Everything else (Ar values, formulae) is given.
Why does 1 mole of every gas fill the same volume?
Because gases are almost entirely empty space: the volume is set by how far apart the particles are, and at the same temperature and pressure equal numbers of particles spread out identically — whether they weigh 2 g per mole or 71 g.
What is the fastest way through a reacting-mass question?
The three-step road: mass → moles (÷ Mr), moles → moles (× ratio), moles → mass (× Mr). Write the equation, put the ratio over the substances, and quote each step — method marks are generous.
How do I spot the limiting reactant quickly?
Convert each reactant to moles, then divide each by its own coefficient. The smaller result is limiting. (For 2Mg + O₂: 0.2 mol Mg → 0.1; 0.15 mol O₂ → 0.15 — Mg’s 0.1 is smaller, so Mg limits.)
Does QuizQuip have any connection to the exam boards?
No. AQA, Pearson Edexcel and OCR are named only to identify the specifications this content supports. QuizQuip is not affiliated with, endorsed by, or connected to AQA, Pearson Edexcel or OCR, and no past-paper or specimen material is reproduced.
Is the mole a Higher-only idea?
Yes. The DfE subject content underlines the whole of the mole work — the Avogadro constant, mol = mass ÷ Mr, reacting-mass calculations, limiting reactants and molar gas volume — so it is assessed only on Higher papers (and molar gas volume is also Chemistry-only, separate science). Foundation students should master conservation of mass, relative formula mass and uncertainty instead.
Up next
Chapter 7: Acids, bases and salt preparations