Overview
What you’ll learn
- State the symbols of the elements and the formulae of the compounds in the syllabus, deduce formulae from atom counts and ion charges, and construct chemical equations with state symbols, including ionic equations.
- Define relative atomic mass and relative molecular mass, and calculate them from atomic masses.
- Define the mole in terms of the Avogadro constant.
- Calculate stoichiometric reacting masses and volumes of gases (one mole occupies 24 dm³ at room temperature and pressure), including limiting-reactant ideas.
- Apply solution concentrations in mol/dm³ and g/dm³ to process the results of volumetric experiments and solve simple problems.
Key points
- 1Balance equations by counting atoms of each element — never change a formula to make numbers fit; add coefficients only.
- 2State symbols: (s) solid, (l) liquid, (g) gas, (aq) dissolved in water — every fully written equation carries them.
- 3An ionic equation keeps only the species that change; unchanged spectator ions are removed, and the charges must balance too.
- 4Ar is the average mass of an element's atoms relative to carbon-12; Mr is the sum of the Ar values in a formula (Mr of H₂SO₄ = 2(1) + 32 + 4(16) = 98).
- 5One mole contains the Avogadro constant (6.02 × 10²³) of particles; moles = mass ÷ Mr.
- 6One mole of any gas occupies 24 dm³ at room temperature and pressure; gas volumes follow the mole ratio of the equation.
- 7moles of solution = concentration (mol/dm³) × volume (dm³); convert g/dm³ to mol/dm³ by dividing by Mr.
- 8The limiting reactant is the one that runs out first — it fixes the moles of every product; work in moles, never in grams.
Notes · 1 of 5
Formulae and equations
Formulae from charges, then balanced equations with state symbols.
Ionic formulae come from balancing charges: Na⁺ with Cl⁻ → NaCl; Ca²⁺ with Cl⁻ → CaCl₂; Al³⁺ with O²⁻ → Al₂O₃ (two +3 charges balance three −2 charges). Molecular formulae come from the atom counts: water is H₂O because two hydrogens share electrons with one oxygen.
- 1Write the formulae
Magnesium burning: Mg + O₂ → MgO (unbalanced).
- 2Balance by counting atoms
Two oxygens on the left means two MgO: Mg + O₂ → 2MgO, then double the magnesium: 2Mg + O₂ → 2MgO.
- 3Add state symbols
2Mg(s) + O₂(g) → 2MgO(s).
An ionic equation strips the equation down to what actually reacts. In Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g), the chloride ions sit unchanged on both sides — they are spectator ions. Removing them: Mg(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂(g). Check atoms and charges balance: left +2, right +2. ✓
Notes · 2 of 5
Relative atomic and molecular mass
Ar from the data sheet; Mr is their sum.
Relative atomic mass (Ar) compares an element's atoms with carbon-12 — the Periodic Table data sheet lists every value (H = 1, C = 12, O = 16, Na = 23, Cl = 35.5). Relative molecular mass (Mr) — relative formula mass for ionic compounds — is the sum of the Ar values in the formula.
| Compound | Sum | Mr |
|---|---|---|
| H₂O | 2(1) + 16 | 18 |
| CO₂ | 12 + 2(16) | 44 |
| CaCO₃ | 40 + 12 + 3(16) | 100 |
| H₂SO₄ | 2(1) + 32 + 4(16) | 98 |
| Mg(NO₃)₂ | 24 + 2(14 + 3(16)) | 148 |
Notes · 3 of 5
The mole
A chemist's counting unit: 6.02 × 10²³ particles per mole.
A mole is the amount of substance containing the Avogadro constant (6.02 × 10²³) of particles — atoms, molecules or ions. One mole of water molecules (18 g) contains 6.02 × 10²³ water molecules and three times as many atoms. The bridge between grams and moles is: moles = mass ÷ Mr, so mass = moles × Mr.
- 9 g of water: 9 ÷ 18 = 0.5 mol (3.01 × 10²³ molecules).
- 4.4 g of CO₂: 4.4 ÷ 44 = 0.1 mol.
- 10 g of CaCO₃: 10 ÷ 100 = 0.1 mol.
For gases at room temperature and pressure, one mole occupies 24 dm³ (24 000 cm³) whatever the gas is. Volume = moles × 24 dm³. Equal volumes of gases therefore contain equal moles — which is why gas volumes in an equation follow the big coefficients: burning 500 cm³ of methane (CH₄ + 2O₂ → CO₂ + 2H₂O) needs 1000 cm³ of oxygen and makes 500 cm³ of carbon dioxide.
Notes · 4 of 5
Reacting masses and limiting reactants
Equations are recipes in moles; convert grams to moles first.
- 1Balance the equation
Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g).
- 2Convert given masses to moles
2.40 g Mg = 2.40 ÷ 24 = 0.100 mol.
- 3Use the mole ratio
1 Mg : 1 H₂, so 0.100 mol H₂ forms; MgCl₂ also 0.100 mol = 0.100 × 95 = 9.50 g.
- 4Convert back
H₂ volume = 0.100 × 24 = 2.40 dm³ at r.t.p.
Notes · 5 of 5
Solution concentrations
mol/dm³ for the chemistry; g/dm³ for the label; titrations tie them together.
Concentration has two everyday forms: moles per dm³ and grams per dm³. They connect through Mr: g/dm³ = mol/dm³ × Mr. So 0.5 mol/dm³ of NaCl (Mr 58.5) is 29.25 g/dm³, and 8 g/dm³ of NaOH (Mr 40) is 8 ÷ 40 = 0.2 mol/dm³. The volume bridge: moles = concentration × volume (in dm³), and 1000 cm³ = 1 dm³.
Revise
Key terms
The words examiners expect you to use precisely.
- state symbol
- (s), (l), (g) or (aq) written after each formula to show the physical state of every species in an equation.
- spectator ion
- An ion present but unchanged on both sides of an equation; removed when writing the ionic equation.
- ionic equation
- An equation showing only the reacting species and charges, with spectators removed.
- relative atomic mass (Ar)
- The average mass of an element's atoms compared with carbon-12; printed on the Periodic Table data sheet.
- relative molecular mass (Mr)
- The sum of the Ar values in a formula (relative formula mass for ionic compounds).
- mole
- The amount of substance containing the Avogadro constant (6.02 × 10²³) of particles.
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- Avogadro constant
- 6.02 × 10²³ — the number of particles in one mole.
- limiting reactant
- The reactant that is completely used up first; it fixes the maximum moles of the products.
- molar gas volume
- One mole of any gas occupies 24 dm³ at room temperature and pressure.
- concentration
- Amount of solute per dm³ of solution, quoted in mol/dm³ or g/dm³ (g/dm³ = mol/dm³ × Mr).
- titration
- A volumetric method where one solution of known concentration is run into another until neutralisation, to find an unknown concentration.
Revise
Common mistakes
Where students most often lose marks in this chapter — and what to do instead.
Mistake: Balancing an equation by changing subscripts.
Editing H₂O to H₂O₂ changes the substance; only coefficients may change.
Instead: Adjust the big numbers in front of formulae; the formulae themselves are fixed by bonding.
Mistake: Leaving state symbols out of "write the equation" answers.
The question asked for them; the information (like "dissolved in water") is in the stem.
Instead: Carry (s), (l), (g), (aq) on every species — (aq) for anything dissolved.
Show 10 more mistakesShow fewer mistakes
Mistake: Keeping spectator ions in an ionic equation.
Ions that appear unchanged on both sides did nothing.
Instead: Cancel spectators, then balance atoms and charges in what remains.
Mistake: Reading the nucleon-style error of adding Ar values without multiplying by subscripts.
Each element's Ar must be multiplied by its subscript count.
Instead: Mr = Σ (subscript × Ar) — including bracket multipliers like (NO₃)₂.
Mistake: Dividing Mr by mass to find moles.
The division is upside down, giving mol⁻¹-style nonsense.
Instead: moles = mass ÷ Mr (smaller over bigger for grams-to-moles).
Mistake: Using 24 dm³ as if it applied to liquids or solids.
The molar volume is a gas-only shortcut.
Instead: 24 dm³ per mole applies only to gases at r.t.p.; liquids and solids use density or mass.
Mistake: Applying the mole ratio to volumes of solutions of different concentrations.
Coefficients connect moles, not cm³.
Instead: Convert each solution to moles first (conc × volume), then apply the ratio.
Mistake: Forgetting to halve (or double) when a ratio of 2:1 applies.
Assuming every reaction is 1:1.
Instead: Write the ratio from the balanced equation before converting back to mass or volume.
Mistake: Choosing the limiting reactant by comparing masses.
Grams of different substances are different-sized counting units.
Instead: Convert both reactants to moles, then compare against the equation ratio.
Mistake: Mixing up cm³ and dm³ in concentration calculations.
A factor of 1000 error either way.
Instead: Convert cm³ ÷ 1000 = dm³ before multiplying by mol/dm³.
Mistake: Quoting a gas volume in dm³ when the question gave cm³.
The answer looks ten-to-a-hundred times off against the mark scheme.
Instead: Match the question's unit: 0.0750 mol × 24 000 cm³ = 1800 cm³ = 1.80 dm³.
Mistake: Rounding too early in a multi-step titration calculation.
Rounded intermediates drift the final answer outside the allowed range.
Instead: Carry at least 3–4 significant figures through; round only the final answer.
Check yourself
Quick quiz
20 questions with instant explanations. About 5 minutes.
Question 1 of 3
- Question 1: current question
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- Question 3: not answered
State the symbols of the elements and the formulae of the compounds in the syllabus, deduce formulae from atom counts and ion charges, and construct chemical equations with state symbols, including ionic equations.
The formula of aluminium oxide, built from Al³⁺ and O²⁻ ions, is
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Practise
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Chapter 4 Practice Paper
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5 multiple-choice questions and structured questions with answer lines. Chapter 4 (formulae and equations with state symbols, ionic equations, Ar and Mr, the mole, reacting masses, gas volumes at 24 dm³/mol, limiting reactants, concentrations in mol/dm³ and g/dm³).
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Questions
Frequently asked questions
What order should I follow for any mole calculation?
Write the balanced equation, convert everything given to moles (mass ÷ Mr; volume ÷ 24 for gases; concentration × volume for solutions), apply the coefficient ratio, then convert the answer back to whatever the question asks for.
How do I spot the limiting reactant?
Convert both reactants to moles. Divide each by its coefficient in the equation. The smaller result runs out first — that reactant limits every product.
When can I use gas volumes directly in a ratio?
Only when both gases are at the same temperature and pressure: volumes then follow the coefficients (500 cm³ of methane needs 1000 cm³ of oxygen in a 1:2 reaction). Solids and liquids never work this way.
How do the two concentration units connect?
Multiply mol/dm³ by Mr to get g/dm³; divide g/dm³ by Mr to get mol/dm³. Check by units: (g/dm³) ÷ (g/mol) = mol/dm³.
Why did my titration answer come out exactly half (or double) the mark scheme?
Almost always the coefficient ratio. In H₂SO₄ + 2NaOH, alkali moles are twice acid moles. Write the equation above your working and arrow the ratio before dividing by the volume.