OCR GCSE Chemistry A: Monitoring chemical reactions (J248)

This guide covers sub-topic C5.1 Monitoring chemical reactions from OCR’s J248 specification for Chemistry A (Gateway Science) at GCSE (9–1), Version 4.0, dated August 2026. It teaches outcomes C5.1a to C5.1k, mathematical outcomes CM5.1i to CM5.1vi, and the practical work suggested for PAG C6 (titration) and PAG C8 (gas volumes). Higher Tier only statements are C5.1a, C5.1c, C5.1d, C5.1e, C5.1f, C5.1g and C5.1k. Topic C5 is examined in Paper 2 on the Foundation Tier and Paper 4 on the Higher Tier.

Afterwards, use the revision notes for a recap and the practice questions to test yourself. The course hub for OCR J248 chemistry gathers every topic, the printable J248 statement checklist helps you track progress, and a free 10-minute diagnostic shows your weakest areas. If moles feel unsure, revisit the C3.1 moles page first.

The C5.1 outcomes at a glance

Statement	You need to	Tier
C5.1a	Link mol/dm³ to mass of solute and volume	Higher Tier only
C5.1b	Describe the technique of titration (PAG C6)	Both tiers
C5.1c	Use titre data to find a volume or concentration	Higher Tier only
C5.1d	Convert between moles of gas and gas volume	Higher Tier only
C5.1e	Calculate reacting gas volumes using 24 dm³ per mole	Higher Tier only
C5.1f	Relate mass of solute, volume and concentration	Higher Tier only
C5.1g	Calculate theoretical mass of product	Higher Tier only
C5.1h	Calculate percentage yield from the actual yield	Both tiers
C5.1i	Define atom economy	Both tiers
C5.1j	Calculate atom economy from a balanced equation	Both tiers
C5.1k	Explain the choice of a reaction pathway from data	Higher Tier only
CM5.1i–vi	Standard form, significant figures, unit conversion, ratio, yield and atom economy arithmetic, rearranging	Both tiers

A warning about the mole

The specification warns that many learners think a mole is a mass. It is not. The mole is the unit of amount of substance; mass is in grams. They are linked by amount (mol) = mass (g) ÷ Mr. Most C5.1 calculations then follow one pattern: change what you know into moles, use the equation’s ratio, and change moles back into what is asked.

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

Concentration is the amount of solute in each cubic decimetre (dm³) of solution, given in two ways:

• concentration (g/dm³) = mass of solute (g) ÷ volume of solution (dm³)
• concentration (mol/dm³) = amount of solute (mol) ÷ volume of solution (dm³)

Since amount = mass ÷ Mr, the two are joined by: mol/dm³ = (g/dm³) ÷ Mr.

As 1 dm³ = 1000 cm³, divide cm³ by 1000 to get dm³ (CM5.1iii).

These are the relationships C5.1f asks about: dissolving more solute in an unchanged volume raises the concentration, while spreading a fixed mass through more solution lowers it. Rearranging (CM5.1vi) gives:

• mass of solute = concentration (g/dm³) × volume (dm³)
• amount of solute = concentration (mol/dm³) × volume (dm³)

Worked example 1. A technician dissolves 4.25 g of sodium nitrate, NaNO₃, to give exactly 250 cm³ of solution. Use Ar: Na = 23.0, N = 14.0, O = 16.0.

• Volume = 250 ÷ 1000 = 0.250 dm³
• Concentration = 4.25 ÷ 0.250 = 17.0 g/dm³
• Mr of NaNO₃ = 23.0 + 14.0 + (3 × 16.0) = 85.0
• Amount = 4.25 ÷ 85.0 = 0.0500 mol
• Concentration = 0.0500 ÷ 0.250 = 0.200 mol/dm³

Worked example 2. What mass of sodium nitrate makes 200 cm³ of 0.150 mol/dm³ solution?

• Amount = 0.150 × 0.200 = 0.0300 mol
• Mass = 0.0300 × 85.0 = 2.55 g

Making a standard solution

A standard solution has an accurately known concentration. To make one (the C5.1a practical suggestion):

1. Weigh the solid accurately and dissolve it in a little distilled water in a beaker.
2. Transfer it to a volumetric flask through a funnel, adding the rinsings from the beaker, rod and funnel so no solute is lost.
3. Top up with distilled water, finishing drop by drop, so the meniscus bottom rests on the graduation mark.
4. Stopper and invert the flask several times to mix.

C5.1b: the titration technique (both tiers)

A titration finds the exact volume of one solution that reacts completely with a measured volume of another. PAG C6 suggests titrating a strong acid against a strong alkali, with an indicator, to find the acid’s concentration.

1. Use a pipette and filler to measure a fixed volume of alkali, such as 25.0 cm³, and run it into a conical flask.
2. Add two or three drops of indicator. Phenolphthalein is pink in alkali and colourless in acid; methyl orange is yellow in alkali and red in acid. Universal indicator changes colour gradually, so it gives no sharp end point.
3. Rinse and fill a burette with the acid, with no air bubble in the jet. Read it at eye level, at the bottom of the meniscus.
4. Add acid while swirling, with the flask on a white tile. As the colour starts to linger, slow to single drops; stop when one drop gives a colour change that stays. Record the final reading.
5. Treat the first run as a rough titration. Repeat until you have concordant titres, ones that agree to within about 0.10 cm³, and average only those.

Worked example 3. A student’s titres are: rough 18.20 cm³, then 17.55, 17.45 and 17.50 cm³. All three accurate titres agree within 0.10 cm³, so the mean is (17.55 + 17.45 + 17.50) ÷ 3 = 17.50 cm³. The rough titre is left out.

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

The balanced equation fixes the ratio of the amounts that react. Knowing one concentration and both volumes, you can find the other concentration:

1. Amount of the known substance = concentration × volume (dm³).
2. Use the mole ratio from the equation to find the amount of the other substance.
3. Concentration of the other substance = amount ÷ its volume (dm³).

Worked example 4. 20.0 cm³ of lithium hydroxide solution is neutralised by 17.50 cm³ (the mean titre from worked example 3) of 0.0800 mol/dm³ sulfuric acid.

2LiOH + H₂SO₄ → Li₂SO₄ + 2H₂O

• Amount of H₂SO₄ = 0.0800 × (17.50 ÷ 1000) = 0.00140 mol
• The ratio LiOH : H₂SO₄ is 2 : 1, so amount of LiOH = 2 × 0.00140 = 0.00280 mol
• Concentration of LiOH = 0.00280 ÷ (20.0 ÷ 1000) = 0.140 mol/dm³
• In g/dm³, using Mr(LiOH) = 23.9: 0.140 × 23.9 = 3.35 g/dm³ (3 significant figures, CM5.1ii)

Ignoring the 2 : 1 ratio gives 0.0700 mol/dm³, half the right answer.

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

Equal amounts of different gases fill equal volumes when temperature and pressure match. The specification tells you to assume this molar gas volume is 24 dm³ at room temperature and pressure, which is 24 000 cm³.

• volume of gas (dm³) = amount (mol) × 24
• amount (mol) = volume of gas (dm³) ÷ 24

So gas volumes react in the same ratio as moles. For ethane:

2C₂H₆(g) + 7O₂(g) → 4CO₂(g) + 6H₂O(l)

40 cm³ of ethane needs (7 ÷ 2) × 40 = 140 cm³ of oxygen and forms 2 × 40 = 80 cm³ of carbon dioxide, at the same temperature and pressure. The water is liquid, so it is not counted.

Worked example 5. Calcium reacts with water: Ca + 2H₂O → Ca(OH)₂ + H₂. What volume of hydrogen, measured at room temperature and pressure, does 0.401 g of calcium give? (Ar Ca = 40.1)

• Amount of Ca = 0.401 ÷ 40.1 = 0.0100 mol
• Ratio Ca : H₂ is 1 : 1, so 0.0100 mol of H₂
• Volume = 0.0100 × 24 000 = 240 cm³

Measuring gas volumes (PAG C8)

Collect the gas in a gas syringe, or over water in an upturned measuring cylinder or burette. Read the volume once the reaction has stopped and the apparatus is back at room temperature, then divide by 24 000 to get moles. For example, 360 cm³ of hydrogen is 360 ÷ 24 000 = 0.0150 mol. Hydrogen is H₂, so that is 0.0150 × 6.02 × 10²³ = 9.03 × 10²¹ molecules (CM5.1i: standard form with the Avogadro constant).

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

The theoretical mass (theoretical yield) is the largest mass of product a given mass of reactant could form, with complete reaction and no losses.

Worked example 6. Zinc carbonate decomposes on heating: ZnCO₃ → ZnO + CO₂. Find the theoretical mass of zinc oxide from 6.27 g of zinc carbonate. Use Ar: Zn = 65.4, C = 12.0, O = 16.0.

• Mr(ZnCO₃) = 65.4 + 12.0 + 48.0 = 125.4; Mr(ZnO) = 65.4 + 16.0 = 81.4
• Amount of ZnCO₃ = 6.27 ÷ 125.4 = 0.0500 mol
• Ratio 1 : 1, so 0.0500 mol of ZnO can form
• Theoretical mass = 0.0500 × 81.4 = 4.07 g

The same 0.0500 mol of carbon dioxide would occupy 0.0500 × 24 = 1.20 dm³ at room temperature and pressure.

C5.1h: percentage yield (both tiers)

The actual yield is the mass you collect. It is usually below the theoretical yield because the reaction may be incomplete or reversible, product is lost in filtering and transferring, or side reactions make other products.

percentage yield = (actual yield ÷ theoretical yield) × 100

Worked example 7. The student in worked example 6 collects 3.62 g of zinc oxide. Percentage yield = (3.62 ÷ 4.07) × 100 = 88.9%.

A yield over 100% means a mistake, usually a damp or impure product. Working out the theoretical yield first (C5.1g) is Higher Tier only.

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

Atom economy is the percentage of the total mass of reactants that ends up in the desired product, using the balanced equation:

atom economy = (total Mr of desired product ÷ total Mr of all reactants) × 100

Multiply each Mr by its number in the balanced equation. A reaction with one product has 100% atom economy. High atom economy means less waste, fewer raw materials and lower disposal costs.

Worked example 8. Zinc chloride can be made from hydrochloric acid and three different zinc compounds. Use Mr: ZnCl₂ = 136.4, HCl = 36.5, Zn = 65.4, ZnO = 81.4, ZnCO₃ = 125.4.

Route	Equation	Atom economy
A	Zn + 2HCl → ZnCl₂ + H₂	136.4 ÷ (65.4 + 73.0) × 100 = 98.6%
B	ZnO + 2HCl → ZnCl₂ + H₂O	136.4 ÷ (81.4 + 73.0) × 100 = 88.3%
C	ZnCO₃ + 2HCl → ZnCl₂ + H₂O + CO₂	136.4 ÷ (125.4 + 73.0) × 100 = 68.8%

Atom economy comes from the equation; yield comes from the experiment. One can be high while the other is low.

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

A route is chosen by weighing atom economy and yield against rate, the position of any equilibrium, and whether the by-products have a use. No single figure decides it.

For the zinc chloride routes, route A has the highest atom economy. Now suppose a question adds that zinc metal costs more than zinc oxide, and that route A’s hydrogen is flammable. Route B might then be chosen: its only by-product is harmless water. Route C wastes the most mass as carbon dioxide, so it comes last unless zinc carbonate is much cheaper. Different data could point to a different route.

Compare the routes one factor at a time, then give a reasoned choice.

Common errors

• Using cm³ in a mol/dm³ calculation without dividing by 1000.
• Dropping a mole ratio that is not 1 : 1.
• Including the rough titre in the mean.
• Writing percentage yield upside down.
• Forgetting the equation’s numbers when adding Mr values for atom economy.

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.