OCR GCSE Chemistry: Chemical Reactions Guide (C3)

Topic C3 — Chemical reactions is the engine room of the OCR Gateway Science A GCSE Chemistry specification (J248). It takes you from the bedrock principle that mass is conserved in every reaction, through the mathematics of the mole, into the energy changes that drive reactions, and on to the great families of reaction — oxidation and reduction, acids and bases, and electrolysis. Students often find C3 daunting because it is where the maths of chemistry really arrives, but it is in fact one of the most rewarding topics to revise. The methods are fixed, the equations follow rules, and once you can balance an equation and use the mole relationship confidently you can answer a huge slice of the paper.

This guide walks through every major idea in C3 at GCSE depth: conservation of mass and balanced equations, the mole and reacting masses, exothermic and endothermic reactions and reaction profiles, oxidation and reduction, acids and bases and the pH scale, neutralisation and making soluble salts, and electrolysis. Higher-tier-only material is flagged with [H] throughout. For structured practice alongside this guide, work through the LearningBro OCR GCSE Chemistry: Chemical Reactions course, which covers every section below with exam-style questions in the OCR format.

How C3 Fits the J248 Specification

OCR Gateway Science A GCSE Chemistry (J248) is assessed by two papers. Paper 1 covers topics C1–C3, and Paper 2 covers topics C4–C6, so C3 sits firmly on Paper 1. Each paper is worth 90 marks, lasts 1 hour 45 minutes, and counts for 50% of the qualification. The same content is examined on Foundation and Higher tiers, but Higher papers reach further into the more demanding ideas — the mole calculations, bond-energy sums and the half equations of electrolysis are largely Higher territory, which is why the [H] flags below matter for targeting your revision. At least 20% of the marks across the qualification reward maths, and a great many of those marks live in C3.

Chemistry questions use the OCR command words precisely, so read them with care. "Balance the equation" asks for the correct big numbers in front of the formulae; "Calculate" asks for a number with working and units; "Describe" asks you to say what happens; and "Explain" asks for the reason why. Knowing exactly what each word wants will stop you writing a paragraph of theory when a single calculated value was all that was needed.

Conservation of Mass and Balanced Equations

The foundation of all reaction chemistry is the law of conservation of mass: in a chemical reaction, no atoms are created or destroyed, so the total mass of the products equals the total mass of the reactants. The atoms are simply rearranged into new substances.

This is why chemical equations must balance — there must be the same number of each type of atom on both sides. You balance an equation by placing big numbers (multipliers) in front of the formulae; you must never change the small subscript numbers inside a formula, because that would change the substance itself.

Take the combustion of methane. The unbalanced equation is $\text{CH}_4 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}$CH4​+O2​→CO2​+H2​O. Counting atoms shows the hydrogens and oxygens do not match. Balancing gives:

$\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$CH4​+2O2​→CO2​+2H2​O

Now each side has 1 carbon, 4 hydrogens and 4 oxygens — the equation balances. Another worked example is the formation of magnesium oxide, where two magnesium atoms react with one oxygen molecule:

$2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$2Mg+O2​→2MgO

State symbols add useful detail, shown in subscript after each formula: $(s)$(s) solid, $(l)$(l) liquid, $(g)$(g) gas and $(aq)$(aq) aqueous (dissolved in water). For example, neutralising hydrochloric acid with sodium hydroxide is written $\text{HCl}_{(aq)} + \text{NaOH}_{(aq)} \rightarrow \text{NaCl}_{(aq)} + \text{H}_2\text{O}_{(l)}$HCl(aq)​+NaOH(aq)​→NaCl(aq)​+H2​O(l)​.

Common misconception: when a metal burns and gains mass, mass has not been created — the metal has combined with oxygen from the air, and if you weighed the oxygen too, the total would be unchanged. Likewise, when a reaction in an open flask seems to lose mass, a gas has escaped. Conservation of mass always holds in a closed system.

The Mole and Reacting Masses [H]

The relative formula mass ($M_r$Mr​) of a compound is the sum of the relative atomic masses ($A_r$Ar​) of all its atoms, read from the periodic table. For water, $M_r(\text{H}_2\text{O}) = (2 \times 1) + 16 = 18$Mr​(H2​O)=(2×1)+16=18. For carbon dioxide, $M_r(\text{CO}_2) = 12 + (2 \times 16) = 44$Mr​(CO2​)=12+(2×16)=44.

The mole is the unit chemists use to count particles. One mole of any substance contains the same number of particles — the Avogadro constant, $6.02 \times 10^{23}$6.02×1023 — and has a mass in grams equal to its $A_r$Ar​ or $M_r$Mr​. The central relationship to learn is:

$\text{moles} = \dfrac{\text{mass}}{M_r}$moles=Mr​mass​

which rearranges to $\text{mass} = \text{moles} \times M_r$mass=moles×Mr​ and $M_r = \dfrac{\text{mass}}{\text{moles}}$Mr​=molesmass​.

Worked example: moles from mass [H]

Calculate the number of moles in 88 g of carbon dioxide ($\text{CO}_2$CO2​).

First find $M_r(\text{CO}_2) = 12 + (2 \times 16) = 44$Mr​(CO2​)=12+(2×16)=44. Then:

$\text{moles} = \dfrac{\text{mass}}{M_r} = \dfrac{88}{44} = 2 \text{ mol}$moles=Mr​mass​=4488​=2 mol

So 88 g of carbon dioxide is 2 moles.

Worked example: reacting masses [H]

What mass of magnesium oxide is formed when 48 g of magnesium burns completely in oxygen? The equation is $2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$2Mg+O2​→2MgO.

Find the moles of magnesium: $A_r(\text{Mg}) = 24$Ar​(Mg)=24, so $\text{moles of Mg} = \dfrac{48}{24} = 2 \text{ mol}$moles of Mg=2448​=2 mol. The equation shows that 2 mol of Mg produces 2 mol of MgO, so 2 mol of MgO forms. Now convert back to mass: $M_r(\text{MgO}) = 24 + 16 = 40$Mr​(MgO)=24+16=40, so $\text{mass} = \text{moles} \times M_r = 2 \times 40 = 80 \text{ g}$mass=moles×Mr​=2×40=80 g.

So 80 g of magnesium oxide forms. The method is always the same: mass → moles → use the balanced equation ratio → moles → mass. The most common error is to skip the equation ratio and assume the moles stay the same on both sides — always check the big numbers in the balanced equation.

Energetics: Exothermic and Endothermic Reactions

Every reaction involves an energy change, and reactions fall into two kinds:

• An exothermic reaction transfers energy to the surroundings, usually as heat, so the temperature rises. Combustion, neutralisation and most oxidation reactions are exothermic. Hand warmers and self-heating cans use exothermic reactions.
• An endothermic reaction takes in energy from the surroundings, so the temperature falls. Thermal decomposition (such as breaking down a metal carbonate) and the reaction in some sports cold-packs are endothermic.

A reaction profile is a diagram showing the energy of the reactants and products. In an exothermic reaction the products sit at a lower energy than the reactants (energy was released); in an endothermic reaction the products sit at a higher energy. In both cases there is an initial "hump" — the activation energy, the minimum energy the reacting particles must have for a successful collision that starts the reaction.

Bond Energies [H]

Energy changes can be calculated from bond energies because breaking bonds takes in energy (endothermic) while making bonds releases energy (exothermic). The overall change is:

$\Delta H = (\text{energy to break bonds}) - (\text{energy released making bonds})$ΔH=(energy to break bonds)−(energy released making bonds)

If more energy is released making the new bonds than was needed to break the old ones, the reaction is exothermic (a negative $\Delta H$ΔH); if breaking bonds costs more than making bonds releases, it is endothermic (a positive $\Delta H$ΔH). The common slip is to get the subtraction the wrong way round — it is always bonds broken minus bonds made.

Common misconception: "exothermic" does not mean "explosive" or "fast", and "endothermic" does not mean "slow". The terms describe the direction of energy transfer (out or in), not the speed of the reaction.

Oxidation and Reduction

Oxidation and reduction always happen together — a redox reaction. There are two ways to define them, and you need both:

• In terms of oxygen: oxidation is the gain of oxygen; reduction is the loss of oxygen. When magnesium burns it gains oxygen, so it is oxidised; when copper oxide is heated with carbon, the copper oxide loses oxygen, so it is reduced.
• In terms of electrons [H]: oxidation is loss of electrons; reduction is gain of electrons. The memory aid is OIL RIG — Oxidation Is Loss, Reduction Is Gain (of electrons).

When magnesium reacts with oxygen, each magnesium atom loses two electrons to form $\text{Mg}^{2+}$Mg2+ (oxidation) while each oxygen atom gains two electrons to form $\text{O}^{2-}$O2− (reduction). Because one species loses the electrons that the other gains, oxidation and reduction must occur together.

Common misconception: students often think a substance can be oxidised on its own. It cannot — if something is oxidised, something else must be reduced at the same time, because electrons (or oxygen) have to go somewhere.

Acids, Bases and the pH Scale

An acid is a substance that releases hydrogen ions ($\text{H}^{+}$H+) when dissolved in water. A base is a substance that neutralises an acid; a base that is soluble in water is called an alkali, and an alkali releases hydroxide ions ($\text{OH}^{-}$OH−) in solution.

The pH scale runs from 0 to 14 and measures how acidic or alkaline a solution is:

pH	Nature	Examples
0–6	Acidic	hydrochloric acid, vinegar, lemon juice
7	Neutral	pure water
8–14	Alkaline	sodium hydroxide, ammonia, soap

The lower the pH, the more acidic (the higher the concentration of $\text{H}^{+}$H+ ions); the higher the pH, the more alkaline. pH can be measured with universal indicator (which changes through a range of colours and gives an approximate value) or, more accurately, with a pH probe and meter.

A useful distinction at GCSE is between a strong and a weak acid. A strong acid (such as hydrochloric, nitric or sulfuric acid) ionises completely in water, releasing all its hydrogen ions. A weak acid (such as ethanoic or citric acid) only partially ionises, so at the same concentration it gives a higher (less acidic) pH. Do not confuse this with concentration — strength is about the degree of ionisation, concentration is about how much acid is dissolved per unit volume.

Neutralisation and Making Soluble Salts

When an acid reacts with a base or alkali, they neutralise each other to produce a salt and water. The essential ionic process is hydrogen ions reacting with hydroxide ions to form water:

$\text{H}^{+} + \text{OH}^{-} \rightarrow \text{H}_2\text{O}$H++OH−→H2​O

The name of the salt comes from the acid used: hydrochloric acid gives chlorides, sulfuric acid gives sulfates, and nitric acid gives nitrates. There are three general routes to a salt, which you should know as word equations:

• Acid + metal → salt + hydrogen. For example, $\text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2$Mg+2HCl→MgCl2​+H2​.
• Acid + base (metal oxide or hydroxide) → salt + water. For example, $\text{CuO} + \text{H}_2\text{SO}_4 \rightarrow \text{CuSO}_4 + \text{H}_2\text{O}$CuO+H2​SO4​→CuSO4​+H2​O.
• Acid + metal carbonate → salt + water + carbon dioxide. For example, $\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2$CaCO3​+2HCl→CaCl2​+H2​O+CO2​.

Required practical: preparing a pure, dry salt

A standard required practical is preparing copper sulfate crystals from copper oxide (an insoluble base) and sulfuric acid. The method illustrates several techniques examiners test:

1. Warm the dilute sulfuric acid, then add copper oxide a little at a time, stirring, until no more reacts — the excess solid sitting at the bottom shows all the acid has been used up.
2. Filter to remove the unreacted excess copper oxide, leaving a blue copper sulfate solution.
3. Heat the solution gently to evaporate some of the water and form a saturated solution, then leave it to crystallise slowly as it cools.
4. Pat the crystals dry between filter paper.

Adding the base in excess and then filtering it off guarantees that all the acid has reacted, so no acid contaminates the salt. The reason for slow crystallisation rather than fast evaporation is to grow larger, purer crystals. This practical is a reliable source of "explain why this step is done" marks.

Electrolysis

Electrolysis is the breaking down of an ionic compound using electricity. It only works when the ions are free to move, so the compound must be molten or dissolved in water. The liquid or solution is called the electrolyte, and two electrodes are dipped into it: the cathode (negative) and the anode (positive).

The rule for what is produced is governed by charge: positive ions move to the cathode (the negative electrode) and negative ions move to the anode (the positive electrode), where they gain or lose electrons.

Electrolysis of Molten Compounds

For a molten ionic compound, the products are simply the two elements. Electrolysing molten lead bromide, for example, gives lead at the cathode and bromine at the anode. This is how reactive metals like aluminium are extracted from their molten ores — a process explored further in C4.

Electrolysis of Aqueous Solutions

When the electrolyte is a solution, water also provides $\text{H}^{+}$H+ and $\text{OH}^{-}$OH− ions, so the rules are slightly more involved:

• At the cathode: hydrogen is produced unless the metal is less reactive than hydrogen, in which case the metal is deposited. So copper sulfate solution gives copper at the cathode (copper is below hydrogen in reactivity), but sodium chloride solution gives hydrogen.
• At the anode: oxygen is produced unless a halide (chloride, bromide or iodide) is present, in which case the halogen is produced. So copper sulfate solution gives oxygen, but sodium chloride solution gives chlorine.

Half Equations [H]

The reaction at each electrode can be written as a half equation showing the electrons gained or lost. Reduction (gain of electrons) happens at the cathode; oxidation (loss of electrons) happens at the anode. For the electrolysis of molten sodium chloride:

• At the cathode (reduction): $\text{Na}^{+} + e^{-} \rightarrow \text{Na}$Na++e−→Na
• At the anode (oxidation): $2\text{Cl}^{-} \rightarrow \text{Cl}_2 + 2e^{-}$2Cl−→Cl2​+2e−

For hydrogen forming at the cathode in an aqueous solution, the half equation is $2\text{H}^{+} + 2e^{-} \rightarrow \text{H}_2$2H++2e−→H2​. The standard exam check is that the charges balance on both sides of a half equation: include enough electrons so the total charge is equal left and right.

Common misconception: the products of electrolysis are not always the elements written in the formula. In solution, water competes — so electrolysing sodium chloride solution gives hydrogen and chlorine (and sodium hydroxide is left behind), not sodium and chlorine.

Common Mistakes in C3

The same slips recur every year. Knowing them is half the battle.

• Changing subscripts to balance an equation. Only ever change the big numbers in front; altering a subscript changes the substance.
• Skipping the equation ratio in mole calculations. Always go mass → moles → apply the balanced ratio → moles → mass.
• Getting the bond-energy subtraction backwards. It is bonds broken minus bonds made; a negative answer means exothermic.
• Confusing strength with concentration in acids. Strength is the degree of ionisation; concentration is the amount dissolved.
• Forgetting that something is reduced when something is oxidised. Redox reactions always pair the two together.
• Assuming electrolysis of a solution gives the elements in the formula. Water competes, so check the reactivity and halide rules for aqueous electrolytes.
• Thinking exothermic means fast or explosive. It only describes energy released to the surroundings.

Exam Technique for C3 on J248

• Show the full calculation method. Mole and reacting-mass questions carry method marks at each stage — write the $M_r$Mr​, the moles, the ratio and the final mass with units, so a slip in arithmetic still scores most of the marks.
• Balance before you calculate. A reacting-mass answer is only as good as the equation it rests on, so check every equation balances first.
• Match the answer form to the command word. "Balance" wants numbers in the equation; "calculate" wants a value with units; "explain" wants reasons.
• Learn the salt-preparation steps and their reasons. The required practical is examined as method and evaluation — know why the base is added in excess and why crystallisation is slow.
• State electrode products with a reason. For aqueous electrolysis, justify the product using the reactivity-than-hydrogen and halide rules, and balance any half equation for charge.

Prepare with LearningBro

The LearningBro OCR GCSE Chemistry: Chemical Reactions course covers every part of C3 — conservation of mass and balancing, the mole and reacting masses, energetics and reaction profiles, oxidation and reduction, acids and the pH scale, neutralisation and making salts, and electrolysis — with worked examples and practice questions that match the OCR J248 format, plus immediate feedback on your answers.

For broader preparation across the whole specification and both papers, the OCR GCSE Chemistry Exam Prep course walks you through the paper structure, command words and answering technique. And for the wider picture of the entire subject, start with our OCR GCSE Chemistry complete revision guide.

Chemical reactions reward practice above almost any other topic. The more equations you balance and the more mole calculations you work in full, the more automatic the methods become — and C3 turns from a worry into one of the most reliable places on the paper to score.

Good luck with your revision.

Related Reading

• OCR GCSE Chemistry (J248): Complete Revision Guide
• OCR GCSE Chemistry: Exam Technique
• OCR GCSE Chemistry: Atomic Structure and Bonding (C1–C2)
• OCR GCSE Chemistry: Predicting and Identifying Reactions (C4)
• AQA vs Edexcel vs OCR GCSE Chemistry: Which Is Right for You?