OCR GCSE Chemistry A: Introducing chemical reactions (J248)

This guide teaches sub-topic C3.1 Introducing chemical reactions from Topic C3 of the OCR GCSE (9-1) Chemistry A (Gateway Science) J248 specification, Version 4.0 (August 2026). It covers every learning outcome from C3.1a to C3.1l, plus the mathematical learning outcomes CM3.1i to CM3.1iv. Five statements (C3.1e, C3.1g, C3.1h, C3.1k and C3.1l) are tested only in the Higher Tier papers and are labelled Higher Tier only below; everything else is for both tiers. The specification says Topic C3 is assessed in Paper 1 (Foundation) and Paper 3 (Higher), with Papers 2 and 4 assuming knowledge of it.

Use it with the C3.1 revision notes and practice questions. See also the J248 course page, the statement checklist and the free diagnostics.

C3.1 at a glance

Ref	Skill	Tier
C3.1a	Write formulae of elements, simple covalent and ionic compounds	Both
C3.1b	Write formulae, balanced equations and half equations	Both
C3.1c	Use a supplied Periodic Table to write formulae and equations	Both
C3.1d	Deduce the formula of a compound from the formulae of its ions	Both
C3.1e	Construct balanced ionic equations	Higher Tier only
C3.1f	Use state symbols (s), (l), (g) and (aq)	Both
C3.1g	Use the definitions of the Avogadro constant and the mole	Higher Tier only
C3.1h	Explain how mass and amount in moles are related, both ways	Higher Tier only
C3.1i	Recall and use the law of conservation of mass	Both
C3.1j	Explain mass changes in open systems with the particle model	Both
C3.1k	Deduce stoichiometry from masses; explain limiting reactants	Higher Tier only
C3.1l	Use a balanced equation to calculate masses of reactants or products	Higher Tier only

Maths outcomes: ratio when balancing (CM3.1i), standard form with the Avogadro constant (CM3.1ii), significant figures (CM3.1iii) and mass-to-moles conversion (CM3.1iv).

Formulae of elements and compounds (C3.1a, C3.1c)

• Metals and noble gases are written as a single symbol: Na, Fe, Cu, He, Ar.
• Most non-metal elements that exist as molecules are diatomic: H₂, N₂, O₂, F₂, Cl₂, Br₂, I₂.
• Simple covalent compounds have a fixed molecular formula you learn or are told: H₂O, CO₂, NH₃, CH₄, HCl.
• Ionic compounds are written as the simplest whole-number ratio of ions, for example NaCl or MgO.

For C3.1c, know the first 20 elements and Groups 1, 7 and 0 well enough not to confuse K (potassium) with P (phosphorus).

Formulae from ions (C3.1d)

An ionic compound has no overall charge, so the ion charges must cancel.

Positive ions	Negative ions
H⁺, Li⁺, Na⁺, K⁺, NH₄⁺	F⁻, Cl⁻, Br⁻, I⁻, OH⁻, NO₃⁻
Mg²⁺, Ca²⁺, Cu²⁺, Zn²⁺, Fe²⁺	O²⁻, S²⁻, SO₄²⁻, CO₃²⁻
Al³⁺, Fe³⁺

Worked example 1. Write the formulae of aluminium sulfate and ammonium carbonate.

• Aluminium sulfate: Al³⁺ and SO₄²⁻. Charges of 3 and 2 meet at 6, so take 2 × Al³⁺ (+6) and 3 × SO₄²⁻ (−6). Formula: Al₂(SO₄)₃.
• Ammonium carbonate: NH₄⁺ and CO₃²⁻. You need 2 × NH₄⁺ to cancel one CO₃²⁻. Formula: (NH₄)₂CO₃.

Brackets go round a group ion only when you need more than one: Ca(OH)₂ but NaOH.

Balanced equations (C3.1b, CM3.1i)

Atoms are rearranged, not made or destroyed, so each type of atom must appear in equal numbers on both sides.

Method:

1. Write the correct formula for every reactant and product. Never change a formula to make it balance.
2. Count atoms of each element on each side.
3. Put whole-number coefficients in front of formulae. Balance elements that appear in only one substance on each side first; leave elements that appear on their own (like O₂) until last.
4. Recount every element.

Worked example 2. Balance the combustion of ethane: C₂H₆ + O₂ → CO₂ + H₂O.

• Carbon: 2 on the left, so put 2 in front of CO₂.
• Hydrogen: 6 on the left, so put 3 in front of H₂O.
• Oxygen on the right is now (2 × 2) + 3 = 7 atoms. That would need 3.5 O₂, which is not a whole number, so double everything.
• Result: 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O.
• Check: C 4 and 4; H 12 and 12; O 14 and 14.

Half equations (C3.1b)

In a half equation, one species gains or loses electrons (e⁻). Atoms and charges must both balance.

• Copper ions gaining electrons: Cu²⁺ + 2e⁻ → Cu
• Bromide ions losing electrons: 2Br⁻ → Br₂ + 2e⁻
• Hydrogen ions gaining electrons: 2H⁺ + 2e⁻ → H₂

Check by adding charges: for bromide, −2 on the left and 0 + (−2) on the right. You meet these again in electrolysis (C3.4).

State symbols (C3.1f)

Symbol	Meaning	Example
(s)	solid	a precipitate, a metal
(l)	liquid	water, molten salt
(g)	gas	hydrogen, carbon dioxide
(aq)	aqueous: dissolved in water	sodium chloride solution

Water itself is (l), never (aq).

Example: Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)

Ionic equations (C3.1e) – Higher Tier only

Ions that take no part in a reaction are spectator ions. An ionic equation leaves them out.

Worked example 3. Silver nitrate solution is added to potassium iodide solution, and a yellow precipitate of silver iodide forms.

1. Full equation: AgNO₃(aq) + KI(aq) → AgI(s) + KNO₃(aq)
2. Split the dissolved ionic substances into ions: Ag⁺(aq) + NO₃⁻(aq) + K⁺(aq) + I⁻(aq) → AgI(s) + K⁺(aq) + NO₃⁻(aq)
3. Delete the spectators, K⁺ and NO₃⁻, which are identical on each side.
4. Ionic equation: Ag⁺(aq) + I⁻(aq) → AgI(s)

Others to build: neutralisation, H⁺(aq) + OH⁻(aq) → H₂O(l), and halogen displacement, Cl₂(aq) + 2I⁻(aq) → 2Cl⁻(aq) + I₂(aq).

The mole and the Avogadro constant (C3.1g) – Higher Tier only

The mole is the unit for amount of substance. The specification accepts two definitions:

• Revised (Avogadro constant based): one mole contains exactly 6.02214076 × 10²³ elementary entities (atoms, molecules, ions or electrons). This is the IUPAC definition.
• Classical (carbon-12 based): one mole is the amount of substance that contains as many entities as there are atoms in exactly 12 g of carbon-12.

The Avogadro constant is the number of particles per mole, about 6.02 × 10²³ per mole. One mole of a substance has a mass in grams equal to its relative formula mass, Mr.

So the mass of one particle = Mr ÷ (6.02 × 10²³) grams.

Worked example 4. Use Ar of Au = 197.

• Mass of one gold atom = 197 ÷ (6.02 × 10²³) = 3.27 × 10⁻²² g (3 significant figures).
• Mass of one water molecule (Mr = 18) = 18 ÷ (6.02 × 10²³) = 2.99 × 10⁻²³ g.

To count particles, multiply moles by the Avogadro constant: 5.10 g of ammonia (Mr = 17) is 0.300 mol, so 0.300 × 6.02 × 10²³ = 1.81 × 10²³ molecules.

Mass and moles (C3.1h, CM3.1iv) – Higher Tier only

The link between mass and amount is:

amount (mol) = mass (g) ÷ Mr

Rearranged: mass (g) = amount (mol) × Mr.

Worked example 5.

• How many moles are in 8.19 g of sodium chloride? Mr = 23 + 35.5 = 58.5. Amount = 8.19 ÷ 58.5 = 0.140 mol.
• What is the mass of 0.150 mol of carbon dioxide? Mr = 12 + (2 × 16) = 44. Mass = 0.150 × 44 = 6.60 g.

Match significant figures to the least precise data (CM3.1iii).

Conservation of mass (C3.1i, C3.1j)

The law of conservation of mass: total mass of products equals total mass of reactants, because atoms are only rearranged.

In a closed system (a sealed flask), the balance reading does not change during a reaction, even if a gas forms.

In a non-enclosed system (an open container), the reading can change:

• Mass goes down when a product is a gas that escapes. Zinc reacting with hydrochloric acid in an open flask gives hydrogen gas. The hydrogen molecules move into the air, so they are no longer weighed.
• Mass goes up when a reactant is a gas from the air. Iron wool heated in air joins with oxygen molecules to form iron oxide. The oxygen atoms are now part of the solid on the balance.

Worked example 6. An open flask of zinc and acid reads 148.36 g at the start and 148.24 g at the end. The 0.12 g decrease is hydrogen that left the flask. The atoms are in the air, not destroyed; a balance only shows a change when particles cross the container boundary.

Practical work

The specification gives no practical suggestion against C3.1. A simple check is to weigh a gas-producing reaction in a sealed bottle, then in an open one, and explain the difference.

Stoichiometry from masses and limiting reactants (C3.1k) – Higher Tier only

Stoichiometry is the mole ratio in which substances react.

Worked example 7. 1.08 g of aluminium reacts with 4.26 g of chlorine to give 5.34 g of aluminium chloride. Use Ar: Al = 27, Cl = 35.5. Deduce the balanced equation.

1. Amount of Al = 1.08 ÷ 27 = 0.0400 mol
2. Amount of Cl₂ = 4.26 ÷ 71 = 0.0600 mol
3. Amount of AlCl₃ = 5.34 ÷ 133.5 = 0.0400 mol
4. Divide by the smallest (0.0400): ratio 1 : 1.5 : 1, which doubles to 2 : 3 : 2.
5. Equation: 2Al + 3Cl₂ → 2AlCl₃. Check: 1.08 + 4.26 = 5.34 g, so mass is conserved.

A limiting reactant is the one that is used up first. It sets the maximum amount of product. The other reactant is in excess, and some of it is left over.

Worked example 8. 4.6 g of sodium reacts with 5.68 g of chlorine: 2Na + Cl₂ → 2NaCl. Which reactant is limiting, and what mass of sodium chloride forms?

1. Amount of Na = 4.6 ÷ 23 = 0.20 mol. Amount of Cl₂ = 5.68 ÷ 71 = 0.080 mol.
2. 0.20 mol Na would need 0.10 mol Cl₂. Only 0.080 mol is present, so chlorine is limiting.
3. Amount of NaCl = 2 × 0.080 = 0.16 mol. Mass = 0.16 × 58.5 = 9.4 g (2 significant figures, matching 4.6 g).
4. Sodium used = 0.16 × 23 = 3.68 g, so about 0.92 g of sodium is left over.

Reacting masses (C3.1l) – Higher Tier only

Steps: mass → moles of the known substance → mole ratio from the equation → moles of the unknown → mass.

Worked example 9. What mass of carbon dioxide forms when 6.0 g of ethane burns completely? Use the equation from worked example 2.

1. Mr of C₂H₆ = 30. Amount = 6.0 ÷ 30 = 0.20 mol.
2. Ratio C₂H₆ : CO₂ = 2 : 4 = 1 : 2, so CO₂ = 0.40 mol.
3. Mr of CO₂ = 44. Mass = 0.40 × 44 = 18 g (2 significant figures; 17.6 g before rounding).

Common errors

• Changing a subscript to balance an equation (writing H₂O₂ instead of putting 2 in front of H₂O).
• Forgetting brackets: writing CaOH₂ for calcium hydroxide.
• Leaving charges unbalanced in a half equation or ionic equation.
• Using the mass ratio from the equation instead of the mole ratio.
• Saying mass is “lost” or “destroyed” in an open container, instead of saying a gas escaped.
• Picking the reactant with the smaller mass as limiting, instead of comparing moles against the equation ratio.

Next steps

Move on to the revision notes and practice questions. Revisit relative formula mass in C2.1 if Mr slows you down.

Official syllabus

Source document: the J248 specification for GCSE (9-1) Chemistry A (Gateway Science) from OCR, Version 4.0, dated August 2026 and in use since first assessment in 2018. This page follows its sub-topic C3.1 Introducing chemical reactions.