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

For full explanations and longer worked examples, read the Monitoring chemical reactions study guide first.

These notes condense sub-topic C5.1 Monitoring chemical reactions, taken from OCR’s J248 specification for Chemistry A (Gateway Science) at GCSE (9–1), Version 4.0, dated August 2026. They cover outcomes C5.1a–C5.1k, mathematical outcomes CM5.1i–CM5.1vi and the PAG C6 (titration) and PAG C8 (gas volume) practical suggestions. C5.1a, C5.1c, C5.1d, C5.1e, C5.1f, C5.1g and C5.1k are Higher Tier only; C5.1b, C5.1h, C5.1i and C5.1j are tested on both tiers. Foundation candidates meet Topic C5 in Paper 2; Higher candidates meet it in Paper 4.

When you are ready, test yourself with the Monitoring chemical reactions practice questions. The course hub for OCR J248 chemistry links every topic, the printable J248 statement checklist lists each outcome, and our free 10-minute diagnostic shows which chemistry topics need work.

Key definitions

• Amount of substance: measured in moles (mol). A mole is not a mass; amount = mass ÷ Mr links the two.
• Concentration: how much solute is in 1 dm³ of solution, in g/dm³ or mol/dm³.
• Standard solution: a solution whose concentration is known accurately.
• Titre: the volume added from the burette to reach the end point.
• Concordant titres: accurate titres that agree closely, usually within 0.10 cm³.
• Molar gas volume: the volume of one mole of any gas; assume 24 dm³ at room temperature and pressure (RTP).
• Theoretical yield (theoretical mass): the most product the reactants could form.
• Actual yield: the mass of product really collected.
• Atom economy: the percentage of the total mass of reactants that ends up in the desired product, from the balanced equation.

Formula table

Quantity	Formula	Units	Tier
Volume conversion	dm³ = cm³ ÷ 1000	dm³	Both tiers
Mass concentration (C5.1f)	mass of solute ÷ volume of solution	g/dm³	Higher Tier only
Molar concentration (C5.1a)	amount of solute ÷ volume of solution	mol/dm³	Higher Tier only
Converting the two	mol/dm³ = (g/dm³) ÷ Mr	mol/dm³	Higher Tier only
Gas amount (C5.1d)	amount = volume (dm³) ÷ 24	mol	Higher Tier only
Gas volume (C5.1e)	volume = amount × 24 000	cm³	Higher Tier only
Percentage yield (C5.1h)	(actual yield ÷ theoretical yield) × 100	%	Both tiers
Atom economy (C5.1j)	(total Mr of desired product ÷ total Mr of reactants) × 100	%	Both tiers
Particles (CM5.1i)	number = amount × 6.02 × 10²³	none	Both tiers

C5.1a and C5.1f: concentration (Higher Tier only)

Method in steps

1. Convert the volume to dm³.
2. For g/dm³, divide the mass by the volume.
3. For mol/dm³, first find the amount (mass ÷ Mr), then divide by the volume.
4. To find a mass needed, rearrange: amount = concentration × volume, then mass = amount × Mr.

Small worked reminder. 2.92 g of hydrogen chloride (Mr 36.5) in 400 cm³ of solution: amount = 0.0800 mol; concentration = 0.0800 ÷ 0.400 = 0.200 mol/dm³, which is 2.92 ÷ 0.400 = 7.30 g/dm³.

Standard solution checklist: weigh accurately; dissolve in a little distilled water; transfer with all rinsings to a volumetric flask; make up to the line, bottom of meniscus at eye level; stopper and invert.

C5.1b: titration technique (both tiers, PAG C6)

Method in steps

1. Pipette a measured volume of one solution (for example the alkali) into a conical flask.
2. Add two or three drops of one indicator, either phenolphthalein (pink in alkali, colourless in acid) or methyl orange (yellow in alkali, red in acid).
3. Fill a rinsed burette with the other solution; no air bubble below the tap; read the bottom of the meniscus at eye level.
4. Run in the solution, swirling, over a white tile; slow to single drops as the colour starts to linger.
5. Stop when the colour change no longer fades. Titre = final reading − initial reading.
6. Do one rough run, then accurate runs until titres are concordant. Average the concordant ones only.

Record burette readings to the nearest 0.05 cm³ (two decimal places).

C5.1c: titration calculations (Higher Tier only)

Method in steps

1. Balance the equation first, then read off the mole ratio.
2. Amount of the known solution = concentration × titre (dm³).
3. Scale by the ratio to get the amount of the unknown.
4. Divide by the unknown’s volume in dm³.

Small worked reminder. 25.0 cm³ of potassium hydroxide needs 16.00 cm³ of 0.125 mol/dm³ nitric acid (1 : 1). Amount of HNO₃ = 0.125 × 0.01600 = 0.00200 mol, so amount of KOH = 0.00200 mol; concentration = 0.00200 ÷ 0.0250 = 0.0800 mol/dm³.

C5.1d and C5.1e: gas volumes (Higher Tier only)

• At RTP, each mole of gas fills 24 dm³, or 24 000 cm³.
• Gas volumes react in the same ratio as the moles in the equation. Count only gases.
• In PAG C8, gas goes into a gas syringe or an inverted measuring cylinder full of water; wait for it to cool to room temperature before reading.

Small worked reminder. 2H₂(g) + O₂(g) → 2H₂O(l). 80 cm³ of hydrogen needs 40 cm³ of oxygen. 0.0250 mol of oxygen at RTP occupies 0.0250 × 24 = 0.600 dm³.

C5.1g: theoretical mass (Higher Tier only)

Method in steps: mass of reactant → moles (÷ Mr) → moles of product (ratio) → mass of product (× Mr).

Small worked reminder. CuO + H₂ → Cu + H₂O. From 3.18 g of copper(II) oxide (Mr 79.5): 0.0400 mol CuO gives 0.0400 mol Cu, so theoretical mass = 0.0400 × 63.5 = 2.54 g.

C5.1h: percentage yield (both tiers)

Reasons the actual yield is lower than theoretical:

• the reaction is incomplete or reversible;
• product is lost during filtering, transferring or crystallising;
• side reactions form unwanted products.

A value above 100% means the product was weighed wet or impure.

C5.1i and C5.1j: atom economy (both tiers)

• Use the balanced equation and multiply each Mr by its coefficient.
• One product only means 100% atom economy (addition reactions, for example).
• High atom economy means less waste, so it is better for sustainability and cost.

Small worked reminder. Making ammonia in the laboratory: 2NH₄Cl + Ca(OH)₂ → CaCl₂ + 2H₂O + 2NH₃. Total Mr of reactants = (2 × 53.5) + 74.1 = 181.1; desired product 2NH₃ = 2 × 17.0 = 34.0. Atom economy = 34.0 ÷ 181.1 × 100 = 18.8%.

C5.1k: choosing a pathway (Higher Tier only)

Weigh these factors for each route, using the data given:

• atom economy (waste built into the equation);
• percentage yield (what is really obtained);
• rate (how fast product forms);
• equilibrium position (for reversible reactions);
• usefulness of by-products (a saleable by-product reduces waste).

Compare the routes factor by factor, then state and justify one choice. Rate is taught in C5.2 Controlling reactions and equilibrium in C5.3 Equilibria.

Must-know distinctions

• Atom economy vs percentage yield: atom economy is fixed by the equation; yield is measured in the experiment.
• g/dm³ vs mol/dm³: convert using Mr.
• cm³ vs dm³: divide by 1000 going from cm³ to dm³.
• Rough vs concordant titres: only concordant titres are averaged.
• Gas volume ratio vs mass ratio: only gases follow the volume ratio, and only at the same temperature and pressure.

Quick self-test

1. 5.85 g of sodium chloride (Mr 58.5) is dissolved to make 500 cm³ of solution. Find its concentration in mol/dm³.
2. A sodium hydroxide solution contains 4.0 g/dm³ (Mr 40.0). Give its concentration in mol/dm³.
3. How many moles of solute does 18.0 cm³ of 0.150 mol/dm³ solution contain?
4. 25.0 cm³ of sodium hydroxide is neutralised by 12.50 cm³ of 0.100 mol/dm³ hydrochloric acid. Find the concentration of the alkali.
5. What volume does 0.250 mol of oxygen occupy at RTP?
6. How many moles of gas are in 96 cm³ at RTP?
7. Oxygen is made by 2H₂O₂ → 2H₂O + O₂. Find the atom economy for oxygen (Mr: H₂O₂ = 34.0, O₂ = 32.0).
8. The theoretical yield is 5.6 g and the actual yield is 4.2 g. Find the percentage yield.
9. 2Na + Cl₂ → 2NaCl. Find the theoretical mass of sodium chloride from 1.38 g of sodium (Ar Na = 23.0).
10. Accurate titres are 24.15, 24.40 and 24.25 cm³. Which are concordant, and what is the mean?

Answers

1. 0.100 mol ÷ 0.500 dm³ = 0.200 mol/dm³
2. 4.0 ÷ 40.0 = 0.10 mol/dm³
3. 0.150 × 0.0180 = 0.00270 mol
4. 0.00125 mol HCl, so 0.00125 mol NaOH; 0.00125 ÷ 0.0250 = 0.0500 mol/dm³
5. 0.250 × 24 = 6.00 dm³
6. 96 ÷ 24 000 = 0.00400 mol
7. 32.0 ÷ (2 × 34.0) × 100 = 47.1%
8. 4.2 ÷ 5.6 × 100 = 75%
9. 0.0600 mol Na gives 0.0600 mol NaCl; 0.0600 × 58.5 = 3.51 g
10. 24.15 and 24.25 cm³ (within 0.10 cm³); mean = 24.20 cm³

Where marks are usually lost

• Dividing by a volume still in cm³, giving a concentration 1000 times too small.
• Using a 1 : 1 ratio when the equation shows 2 : 1, as with sulfuric acid and a hydroxide.
• Averaging all titres, including the rough one or an outlier.
• Recording burette readings as 24 or 24.2 instead of 24.20 cm³.
• Using 24 when the volume is in cm³ (it should be 24 000), or the reverse.
• Counting liquid water in a gas-volume ratio.
• Putting the theoretical yield on top in a percentage yield calculation.
• Leaving out a coefficient, such as the 2 in 2NH₃, when totalling Mr values for atom economy.
• In pathway questions, naming the route with the best atom economy without using the other data given.

Official syllabus

Aligned to OCR’s J248 specification for Chemistry A (Gateway Science) at GCSE (9–1): Version 4.0, dated August 2026, for first assessment in 2018. Sub-topic: C5.1 Monitoring chemical reactions.