Solving problems set in a practical context
This is A Level (IA2) content, and everything from the AS units is assumed: the mole calculations of Unit 1, the tests and titrations of Unit 2, and the practical skills of the Unit 3 paper (notes unit U3).
This unit prepares you for the A Level practical paper, WCH16 (Unit 6: Practical Skills in Chemistry II): 1 hour 20 minutes, 50 marks, every question compulsory, no multiple choice, no asterisked extended-writing item and no Data Booklet.
Every number you need is printed in the question; the tests, colours and indicators are not, and must be recalled.
It is 20% of the A Level (IA2) and 10% of the full International A Level.
The core practicals behind the paper are taught in full in the theory notes units: CP9a, CP9b and CP10 in notes unit 11, CP11 in unit 14b, CP12, CP13a and CP13b in unit 16, CP14 in unit 17b, CP15 in unit 19 and CP16 in unit 20.
These U6 sections do not teach each one again.
They teach the skills that every WCH16 question tests, drawing the core practicals and the further suggested practicals together, and they practise those skills in the paper's own shapes.
You need to solve problems set in a practical context.
This first section shows what that means on WCH16 and how to set out a practical calculation so that every scoring point is visible.
- In the 2024 and 2025 WCH16 mark schemes each scoring point earned one mark, printed "(1)". Longer answers were point-marked ("any three of five points"); there were no letter codes, no levels and no asterisked question.
- In calculations an earlier wrong value used correctly later still earned the later marks (TE), unless it gave an impossible result such as a yield above 100% or a negative activation energy. Significant figures were ignored except an answer to 1 s.f., unless the stem asked for "an appropriate number of significant figures".
- State symbols were marked only when the stem asked for them; every equation item on these papers printed "State symbols are not required".
- A second, wrong answer beside a right one cancelled it, and when a name and a formula were both given, both had to be right.
- Observations earned marks only in exact words, with the colour and the state ("green precipitate", "deep blue solution"), and reasons for steps only when they named the chemicals in that step.
The idea: every WCH16 question is one experiment, seen from the inside
A WCH16 question describes one experiment: a sentence of context, an equation, often a table of data (molar masses, densities, hazards, melting temperatures), and then numbered steps (Step 1, Step 2 ...).
The parts then ask you to think as the chemist who planned it.
Which reagent is in excess, and why was that choice made?
How much product should form?
What does this titre tell you about the sample?
Would the method still work if one step changed?
In 2024 and 2025 the three papers were built from the same few kinds of experiment: an investigation of a reaction rate, an organic preparation and purification, a qualitative analysis of an inorganic or organic unknown, and often a titration.
Every one of them carried a three- or four-mark mole calculation.
About a quarter of the marks came from Units 1 and 2 (moles, yields, back titrations, ion tests, hazards), so the AS skills are not finished with.
The problem types that recur
Problem
What you do
Typical tariff
Show that a reagent is in excess
find the amount of both reagents (use the density for a liquid), apply the ratio, and finish with a sentence that compares them
3 or 4
Theoretical yield and percentage yield
amount of the limiting reagent, ratio, mass of product; then actual ÷ theoretical × 100
2 or 3
Back titration
amount of reagent added, amount left over (from the titre), amount used, then the amount and mass of the sample
4
Identify from a molar mass
amount from a titre or a gas volume, then M = m ÷ n, then match it to a formula that fits the information
2 or 3
Expected titre
from a concentration (or a solubility in g dm⁻³), amount in the portion, ratio, then volume of titrant
4
Formula from data
masses or percentages, moles, simplest ratio (combustion analysis, water of crystallisation)
2 to 4
Why the order of working matters
Every practical calculation is a chain: a measured quantity becomes an amount in moles, the equation turns one amount into another, and the amount becomes a mass, a volume or a molar mass at the end.
Two habits make the chain safe.
Start from the quantity the question gives you. If the stem gives a mass of solid and a volume of solution, start from those, not from a value you will only reach later.
In 2025 some candidates began in the middle of the chain and could not finish it.
Keep track of which amount belongs to which portion. In a back titration the solution is often made up to 250.0 cm³ and only 25.0 cm³ portions are titrated.
The titre then gives the amount in one portion, which must be multiplied by 10 before it is compared with the amount added to the whole sample.
Dividing a mass by the wrong amount (the amount added, not the amount that reacted with the sample) was the commonest error in the 2024 back-titration item.
Checking that an answer can be right
A practical answer can be tested against what you know before you move on.
A percentage purity above 100% means the impurity was counted with the product, or a ratio was missed.
A molar mass of 37 for a "carboxylic acid" cannot be right, because the smallest, methanoic acid, has M = 46.0.
A titre of 3 cm³ or of 70 cm³ from a 50 cm³ burette tells you that the planned concentrations do not suit the apparatus.
Where a number cannot be right, say so in your answer: in a "suggest" part that is often the point.
Where this leads
Every later section uses these chains.
The data-processing sections work through each calculation type in full; the evaluation and uncertainty sections ask how far a result can be trusted; and the measurement and preparation sections build whole WCH16 questions around them.
- Limiting reagent: the reactant that is completely used up; it fixes the amount of product formed.
- Back titration: a method in which a known excess of a reagent is added to a substance, and the amount left over is found by titration, so that the amount that reacted can be found by subtraction.
- Aliquot: a measured portion (sample) of a solution, taken with a pipette.
A student prepares methyl benzoate by heating 6.10 g of benzoic acid (Mr 122.0) under reflux with 20.0 cm³ of methanol (Mr 32.0, density 0.790 g cm⁻³) and a little concentrated sulfuric acid.
C₆H₅COOH + CH₃OH ⇌ C₆H₅COOCH₃ + H₂O
Show, by calculation, that the methanol is in excess.
(3)
1. n(benzoic acid) = 6.10 ÷ 122.0 = 0.0500 mol (1)
2. mass of methanol = 20.0 × 0.790 = 15.8 g, so n(methanol) = 15.8 ÷ 32.0 = 0.494 mol (1)
3. the ratio is 1 : 1 and 0.494 mol is more than 0.0500 mol, so the methanol is in excess (by almost ten times) (1)
Answer: the methanol is in excess.
The third mark is for the comparison and the conclusion: two correct amounts with no sentence comparing them leave the question unanswered.
A 1.25 g sample of impure zinc oxide was added to 100.0 cm³ of 0.500 mol dm⁻³ hydrochloric acid, an excess.
When all the solid had reacted, the solution was made up to 250.0 cm³ with distilled water.
25.0 cm³ portions of this solution needed a mean titre of 21.30 cm³ of 0.100 mol dm⁻³ sodium hydroxide.
ZnO + 2HCl → ZnCl₂ + H₂O, then HCl + NaOH → NaCl + H₂O
Calculate the percentage by mass of zinc oxide in the sample.
[Ar: O 16.0, Zn 65.4] (4)
1. n(HCl) added = 0.500 × 100.0 ÷ 1000 = 0.0500 mol (1)
2. n(HCl) left in 25.0 cm³ = n(NaOH) = 0.100 × 21.30 ÷ 1000 = 2.13 × 10⁻³ mol, so in 250.0 cm³ = 0.0213 mol (1)
3. n(HCl) that reacted = 0.0500 − 0.0213 = 0.0287 mol, so n(ZnO) = 0.0287 ÷ 2 = 0.01435 mol (1)
4. mass of ZnO = 0.01435 × 81.4 = 1.168 g; percentage = 1.168 ÷ 1.25 × 100 = 93.4% (1)
Answer: 93.4%.
Forgetting the factor of 10 for the aliquot gives n(HCl) left = 0.00213 mol and a percentage above 100%, which is impossible and loses the TE.
0.185 g of a pure monoprotic carboxylic acid, RCOOH, was dissolved in water and titrated with 0.100 mol dm⁻³ sodium hydroxide.
The titre was 25.00 cm³.
(a) Calculate the molar mass of the acid. (2)
(b) Identify the acid. (1)
1. (a) n(NaOH) = 0.100 × 25.00 ÷ 1000 = 2.50 × 10⁻³ mol = n(RCOOH), since the ratio is 1 : 1 (1)
2. M = 0.185 ÷ 2.50 × 10⁻³ = 74.0 g mol⁻¹ (1)
3. (b) COOH = 45.0, so R = 29.0 = C₂H₅: the acid is propanoic acid, C₂H₅COOH (1)
Answer: (a) 74.0 g mol⁻¹; (b) propanoic acid.
A name with a wrong formula, or a formula with a wrong name, would lose the mark: give the one you are sure of, or make both right.
A student plans to titrate 25.0 cm³ portions of a saturated solution of benzoic acid at 25 °C with 0.0200 mol dm⁻³ sodium hydroxide, using a 50 cm³ burette.
The question gives the solubility of benzoic acid at 25 °C as 3.4 g dm⁻³ (the CRC Handbook value) and Mr = 122.0.
(a) Calculate the expected titre. (3)
(b) State whether the plan is suitable, giving a reason. (1)
1. (a) concentration = 3.4 ÷ 122.0 = 0.0279 mol dm⁻³ (1)
2. n(acid) in 25.0 cm³ = 0.0279 × 25.0 ÷ 1000 = 6.97 × 10⁻⁴ mol = n(NaOH) (1 : 1) (1)
3. volume of NaOH = 6.97 × 10⁻⁴ ÷ 0.0200 × 1000 = 34.8 cm³ (1)
4. (b) suitable: 34.8 cm³ fits in one filling of a 50 cm³ burette and is large enough for a small percentage uncertainty (1)
Answer: (a) 34.8 cm³; (b) yes.
Keeping the unrounded concentration in the calculator gives 34.84 cm³; rounding it to 0.028 first gives 35.0 cm³.
Carry the calculator value and round only at the end.
When a part says "Show, by calculation" or "You must show your working", a correct conclusion with no numbers earns nothing.
Write each amount with its unit on its own line, show the ratio from the equation in words, and end with a sentence that states the comparison or the answer.
- Dividing by the amount added. In a back titration the sample reacts with the amount used up (added minus left over), not with the amount added. Find the difference first.
- Forgetting the aliquot factor. If 25.0 cm³ of a 250.0 cm³ solution was titrated, the titre gives one tenth of the amount in the whole solution.
- Using a volume of liquid as if it were a mass. For a pure liquid reagent, mass = volume × density; only then divide by M.
- Stopping at the numbers. An excess-reagent answer needs the comparison and the conclusion; in 2025 many answers stopped at two correct amounts.
Check your understanding
1. Ethyl ethanoate is made from 12.0 cm³ of ethanoic acid (Mr 60.0, density 1.05 g cm⁻³) and 15.0 cm³ of ethanol (Mr 46.0, density 0.789 g cm⁻³), with a few drops of concentrated sulfuric acid. CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O Show, by calculation, that the ethanol is in excess. (3) Total [3]
2. A 0.600 g sample of impure magnesium hydroxide, Mg(OH)₂, was added to 50.0 cm³ of 0.500 mol dm⁻³ hydrochloric acid (an excess). The excess acid needed 15.20 cm³ of 0.500 mol dm⁻³ sodium hydroxide.
Mg(OH)₂ + 2HCl → MgCl₂ + 2H₂O
(a) Calculate the amount of hydrochloric acid that reacted with the sample. (2)
(b) Calculate the percentage by mass of magnesium hydroxide in the sample. [Ar: H 1.0, O 16.0, Mg 24.3] (2)
Total [4]
3. 0.225 g of a pure dicarboxylic acid, H₂X, needed 25.00 cm³ of 0.200 mol dm⁻³ sodium hydroxide for complete neutralisation.
(a) Calculate the molar mass of H₂X. (2)
(b) Deduce the structural formula of H₂X. (1)
Total [3]
4. A student's back titration gives the purity of a sample as 104%. Explain why this result cannot be correct, and suggest one error in the working that could have caused it. (2) Total [2]
5. A student plans to find the concentration of a saturated solution of iodine in water by titrating 25.0 cm³ portions with 0.00500 mol dm⁻³ sodium thiosulfate. The question gives the solubility of iodine as 0.300 g dm⁻³ at the temperature used (a value for this question). [Ar: I 126.9]
I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻
(a) Calculate the expected titre. (3)
(b) The student says a 0.100 mol dm⁻³ thiosulfate solution would be "more accurate because it is more concentrated". Explain why the student is wrong. (2)
Total [5]
6. Complete the table for portions of a 0.0400 mol dm⁻³ solution of a monoprotic acid titrated with 0.100 mol dm⁻³ sodium hydroxide. (2)
Volume of acid portion / cm³
Amount of acid / mol
Expected titre / cm³
10.0
25.0
Total [2]
Answers
1. n(ethanoic acid) = 12.0 × 1.05 ÷ 60.0 = 0.210 mol (1); n(ethanol) = 15.0 × 0.789 ÷ 46.0 = 0.257 mol (1); the ratio is 1 : 1 and 0.257 mol > 0.210 mol, so the ethanol is in excess (1).
2. (a) n(HCl) added = 0.0250 mol; n(HCl) left = n(NaOH) = 0.500 × 15.20 ÷ 1000 = 7.60 × 10⁻³ mol (1); reacted = 0.0250 − 0.00760 = 0.0174 mol (1). (b) n(Mg(OH)₂) = 0.0174 ÷ 2 = 8.70 × 10⁻³ mol; M = 58.3 g mol⁻¹, so mass = 0.507 g (1); percentage = 0.507 ÷ 0.600 × 100 = 84.5% (1).
3. (a) n(NaOH) = 0.200 × 25.00 ÷ 1000 = 5.00 × 10⁻³ mol, so n(H₂X) = 2.50 × 10⁻³ mol (ratio 1 : 2) (1); M = 0.225 ÷ 2.50 × 10⁻³ = 90.0 g mol⁻¹ (1). (b) two COOH groups have M = 90.0, so nothing else is present: HOOC–COOH, ethanedioic acid (1).
4. The sample cannot contain more of the substance than its own mass, so a purity above 100% is impossible (1); any one of: the amount added was used in place of the amount that reacted; the aliquot factor of 10 was missed; the ratio in the equation was not applied; the sample was weighed wet (1).
5. (a) concentration = 0.300 ÷ 253.8 = 1.18 × 10⁻³ mol dm⁻³ (1); n(I₂) in 25.0 cm³ = 2.96 × 10⁻⁵ mol, so n(S₂O₃²⁻) = 5.91 × 10⁻⁵ mol (1); volume = 5.91 × 10⁻⁵ ÷ 0.00500 × 1000 = 11.8 cm³ (1). (b) the titre would be only about 0.59 cm³ (1), so the uncertainty of ±0.10 cm³ in the titre would be about 17% of it: a larger titre, from the dilute solution, has a far smaller percentage uncertainty (1).
6. 10.0 cm³: 4.00 × 10⁻⁴ mol and 4.00 cm³; 25.0 cm³: 1.00 × 10⁻³ mol and 10.0 cm³. Amounts (1); titres (1).