OCR GCSE Chemistry A: Introducing chemical reactions (J248) – Revision Notes

These notes condense sub-topic C3.1 Introducing chemical reactions of the OCR GCSE (9-1) Chemistry A (Gateway Science) J248 specification, Version 4.0 (August 2026). They cover statements C3.1a to C3.1l and the maths outcomes CM3.1i to CM3.1iv. Statements C3.1e, C3.1g, C3.1h, C3.1k and C3.1l are Higher Tier only; the rest apply to both tiers. The C3.1 study guide has the full explanations.

Other links: practice questions | J248 course page | J248 checklist | free diagnostics

C3.1a and C3.1c: formulae of elements and compounds

• Metals and noble gases: single symbol (Mg, Ca, Ne).
• Diatomic elements: H₂, N₂, O₂, F₂, Cl₂, Br₂, I₂. A memory hook: the Group 7 elements plus hydrogen, nitrogen and oxygen.
• Simple covalent compounds: learn the common ones (H₂O, CO₂, NH₃, CH₄, HCl).
• Ionic compounds: simplest ratio of ions, no charges shown in the final formula.
• Read symbols from the supplied Periodic Table. Know the first 20 elements and Groups 1, 7 and 0 without looking.

C3.1d: formulae from ions

Charge	Ions to know
1+	H⁺, Li⁺, Na⁺, K⁺, NH₄⁺
2+	Mg²⁺, Ca²⁺, Cu²⁺, Zn²⁺, Fe²⁺
3+	Al³⁺, Fe³⁺
1−	F⁻, Cl⁻, Br⁻, I⁻, OH⁻, NO₃⁻
2−	O²⁻, S²⁻, SO₄²⁻, CO₃²⁻

Method in steps

1. Write both ions with their charges.
2. Find the lowest common multiple of the two charge sizes.
3. Take enough of each ion to reach it.
4. Use brackets round a group ion if you need two or more.

Reminder: iron(III) oxide is Fe³⁺ and O²⁻; LCM 6; 2 Fe and 3 O, so Fe₂O₃. Magnesium nitrate is Mg²⁺ and 2 NO₃⁻, so Mg(NO₃)₂.

C3.1b: balanced equations and half equations

Balancing in steps

1. Correct formulae first; never alter them.
2. Balance elements found in one compound on each side.
3. Balance free elements (O₂, H₂, metals) last.
4. If you need a half, double every coefficient.
5. Recount.

Reminder: Al + O₂ → Al₂O₃ becomes 4Al + 3O₂ → 2Al₂O₃ (Al 4 = 4, O 6 = 6).

Half equations show electrons (e⁻). Atoms and charges must both balance.

Change	Half equation
Zinc ion to zinc	Zn²⁺ + 2e⁻ → Zn
Chloride ion to chlorine	2Cl⁻ → Cl₂ + 2e⁻
Iron(III) ion to iron(II) ion	Fe³⁺ + e⁻ → Fe²⁺

C3.1f: state symbols

(s) solid, (l) liquid, (g) gas, (aq) dissolved in water. Pure water is (l). A precipitate is (s). A solution is (aq).

C3.1e: ionic equations – Higher Tier only

Method in steps

1. Write the balanced full equation with state symbols.
2. Split only (aq) ionic substances into separate ions.
3. Keep solids, liquids, gases and covalent molecules as they are.
4. Cancel ions that appear identically on both sides (spectator ions).
5. Check atoms and total charge.

Reminder: barium chloride solution + sodium sulfate solution gives Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s). Spectators: Na⁺ and Cl⁻.

C3.1g and C3.1h: the mole – Higher Tier only

• The mole (revised, IUPAC): exactly 6.02214076 × 10²³ elementary entities.
• The mole (classical): as many entities as there are atoms in 12 g of carbon-12.
• The specification accepts either definition.
• Avogadro constant: about 6.02 × 10²³ particles per mole.

Quantity	Relationship
Amount from mass	amount (mol) = mass (g) ÷ Mr
Mass from amount	mass (g) = amount (mol) × Mr
Number of particles	particles = amount (mol) × 6.02 × 10²³
Mass of one particle	mass (g) = Mr ÷ (6.02 × 10²³)

Reminder: 0.500 mol of Ca(NO₃)₂ (Mr = 164) has a mass of 82.0 g.

Must-know distinction: amount (in moles) is not the same as mass (in grams). Two substances with the same mass usually contain different numbers of moles.

C3.1i and C3.1j: conservation of mass

• Total mass of products = total mass of reactants. Atoms are rearranged, not made or destroyed.
• Closed system: the balance reading stays the same.
• Open system, reading falls: a gas product escapes (for example, carbon dioxide from an acid and a carbonate).
• Open system, reading rises: a gas from the air joins on (for example, a metal reacting with oxygen to make an oxide).
• Particle-model explanation: the same atoms exist before and after; the reading changes only because particles enter or leave the container.

C3.1k and C3.1l: stoichiometry, limiting reactants, reacting masses – Higher Tier only

Deducing an equation from masses

1. Convert each mass to moles.
2. Divide by the smallest amount.
3. Scale to whole numbers. These are the coefficients.

Finding the limiting reactant

1. Moles of each reactant.
2. Use the equation ratio to find how much of the second reactant the first would need.
3. If less than that is present, the second is limiting.
4. Base product amounts on the limiting reactant only.

Reminder: 0.78 g of potassium reacts with 0.16 g of oxygen to make 0.94 g of potassium oxide (Ar: K = 39, O = 16). Moles: K 0.020, O₂ 0.0050, K₂O 0.010. Dividing by 0.0050 gives 4 : 1 : 2, so 4K + O₂ → 2K₂O. The masses also add up: 0.78 + 0.16 = 0.94 g.

Reacting masses: mass → moles → ratio → moles → mass.

Reminder: to make 6.8 g of ammonia by N₂ + 3H₂ → 2NH₃, you need 6.8 ÷ 17 = 0.40 mol NH₃, so 0.40 × 3/2 = 0.60 mol H₂, which is 0.60 × 2 = 1.2 g of hydrogen.

What a limiting reactant does: once it is used up the reaction stops, so the amount of product is fixed by it. Adding more of the reactant in excess makes no extra product; adding more of the limiting one does, until the other runs out.

Must-know distinction: the limiting reactant is not always the one with the smaller mass. Compare moles against the ratio in the equation.

Maths skills for C3.1 (CM3.1i to CM3.1iv)

Outcome	What it looks like in a question
CM3.1i	Ratios when balancing equations or deducing a formula from ion charges
CM3.1ii	Standard form, for example 6.02 × 10²³ or 2.66 × 10⁻²³ g
CM3.1iii	An answer to sensible significant figures, usually matching the data
CM3.1iv	Converting a mass in grams to an amount in moles

Must-know distinctions

• Atom vs molecule: O is one oxygen atom; O₂ is one oxygen molecule made of two atoms.
• Full vs ionic equation: a full equation shows every substance; an ionic equation (Higher Tier only) shows only the particles that change.
• Closed vs open system: mass is conserved in both, but only a closed system shows a constant balance reading.
• Coefficient vs subscript: a coefficient (the big number in front) changes how many units react; a subscript changes what the substance is.

Quick self-test

1. Write the formula of calcium nitrate.
2. Write the formula of potassium oxide.
3. Balance: Fe + Cl₂ → FeCl₃
4. Write the half equation for bromide ions forming bromine.
5. What state symbol does sodium chloride have when dissolved in water?
6. A student heats a metal in an open crucible and the mass increases. Explain why.
7. (Higher Tier only) Write the ionic equation for copper(II) sulfate solution reacting with sodium hydroxide solution to form a blue precipitate of copper(II) hydroxide.
8. (Higher Tier only) How many moles are in 4.25 g of ammonia (Mr = 17)?
9. (Higher Tier only) What is the mass of 0.0400 mol of potassium carbonate, K₂CO₃? (Ar: K = 39, C = 12, O = 16)
10. (Higher Tier only) Calculate the mass of one methane molecule (Mr = 16).
11. (Higher Tier only) How many oxygen atoms are in 0.0500 mol of O₂?
12. (Higher Tier only) 0.30 mol of hydrogen reacts with 0.12 mol of oxygen: 2H₂ + O₂ → 2H₂O. Which is limiting, and how many moles of water form?

Answers

1. Ca²⁺ and two NO₃⁻: Ca(NO₃)₂
2. Two K⁺ and one O²⁻: K₂O
3. 2Fe + 3Cl₂ → 2FeCl₃ (Cl 6 = 6, Fe 2 = 2)
4. 2Br⁻ → Br₂ + 2e⁻
5. (aq)
6. Oxygen from the air combines with the metal to form an oxide. The oxygen atoms are now part of the solid, so they are weighed.
7. Cu²⁺(aq) + 2OH⁻(aq) → Cu(OH)₂(s)
8. 4.25 ÷ 17 = 0.250 mol
9. Mr = (2 × 39) + 12 + (3 × 16) = 138; 0.0400 × 138 = 5.52 g
10. 16 ÷ (6.02 × 10²³) = 2.66 × 10⁻²³ g
11. Each molecule has 2 atoms: 0.0500 × 2 × 6.02 × 10²³ = 6.02 × 10²² atoms
12. 0.30 mol H₂ needs 0.15 mol O₂, but only 0.12 mol is present, so oxygen is limiting. Water = 2 × 0.12 = 0.24 mol (0.06 mol of hydrogen is left over).

Where marks are usually lost

• Changing a formula (for example writing AlO for aluminium oxide) to make an equation balance.
• Missing brackets: writing MgNO₃₂ instead of Mg(NO₃)₂.
• Leaving charges unbalanced in half equations, or adding electrons where they should be removed.
• Splitting a precipitate or water into ions in an ionic equation.
• Writing (aq) for water, or (l) for a solution.
• Saying mass is “lost” or “destroyed” instead of explaining that a gas escaped or was gained from the air.
• Using the equation’s numbers as mass ratios instead of mole ratios.
• Choosing a limiting reactant by comparing masses rather than moles.
• Giving a long string of digits instead of an answer to sensible significant figures, or dropping standard form for very small masses.
• Forgetting that Cl₂ has Mr 71, not 35.5, when converting chlorine gas to moles.

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.