Solving problems set in a practical context
This unit is the practical skills unit for AS Chemistry, and it is examined on its own written paper, WCH13 (Unit 3: Practical Skills in Chemistry I): 1 hour 20 minutes, 50 marks, every question compulsory, no laboratory work.
The paper draws on the practical work of Units 1 and 2: the eight core practicals (CP1 to CP8), the further suggested practicals, and settings you have never met.
It is 20% of the International AS and 10% of the full International A Level.
The core practicals themselves are taught in full in the theory notes units (CP1 in notes unit 1, CP2 in unit 6, CP3 and CP4 in unit 8b, CP5 and CP6 in unit 10a, CP7 and CP8 in unit 10b); these sections teach the skills that every WCH13 question tests, and practise them in the paper's own shapes.
You need to solve problems set in a practical context.
This first section shows what that means on the paper and how to read a WCH13 stem.
- In the 2024 and 2025 WCH13 mark schemes each scoring point earned one mark, printed as "(1)". Longer answers were point-marked (M1, M2 ... as labels only); 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). Significant figures were ignored except an answer to 1 s.f., unless the stem asked for "an appropriate number of significant figures", when 2 or 3 s.f. were needed.
- State symbols were marked only when the stem asked for them; "State symbols are not required" meant wrong ones were ignored.
- A second, wrong answer beside a right one cancelled it (the list principle), and when a name and a formula were both given, both had to be right.
- Observations earned marks only in precise words: "effervescence", "white precipitate", "from orange to green".
The idea: every practical question is a small puzzle
A WCH13 question describes one experiment, usually in a few lines of past-tense text and a set of numbered Steps, with the equation printed.
Then it asks you to think like the person who planned it: why is this reagent in excess, will the gas fit in the syringe, which reactant decides how much product forms, what would happen if a step were left out?
None of this needs new chemistry.
It needs you to use the information you are given.
The stem plants the facts that later parts need: the word "excess", a ΔH value that tells you whether the temperature rises, a density that tells you which layer is on top, the capacity of a syringe.
In the 2024 and 2025 papers many marks were lost by candidates who had the knowledge but ignored a fact printed in the question.
Read the stem twice, and underline every number and every word such as excess, dry, cooled, gas, solution and identify.
Four problem types that recur
Problem
What you do
Typical tariff
Which reagent is in excess?
Find the moles of both reagents, use the ratio in the equation, and state the comparison
2 or 3
Will it fit? (capacity)
Work out the largest volume or mass the apparatus can take, or the volume the plan would make
1 to 3
What is the unknown?
Find the moles from the measured quantity, then a molar mass, then match it to an element in the stated group
3
Why does it behave like that?
Use one piece of theory (solubility, rate, equilibrium, redox) to explain an observation in this apparatus
1 or 2
Why the limiting reagent decides everything
When one reagent is in excess, some of it is left over at the end, so the amount of product is fixed by the other one, the limiting reagent.
This is why the method for measuring the molar volume puts the acid in excess: then the metal's mass alone decides the moles of gas, and the acid's volume need not be measured accurately.
If you used the moles of the reagent in excess, you would be calculating an amount of product that never formed.
This one error cost marks on every WCH13 paper in 2024 and 2025, in gas-volume questions and in calorimetry.
Checking that an answer is sensible
A practical answer can be tested against what you know.
One mole of gas at room conditions occupies about 24 dm³, so a molar volume of 2.4 or 240 dm³ mol⁻¹ means a slip in converting cm³ to dm³.
An Ar of 40 for a "Group 2 metal" means calcium, not potassium, even though potassium's Ar (39.1) is close: the stem told you the group.
A percentage yield above 100% is impossible for the pure product, so it tells you something extra was weighed (the evaluation section shows what).
Where this leads
The same reasoning runs through every later section: evaluation asks which way an error moves a result, apparatus questions ask whether a piece of equipment can do the job, and the measurement experiments chain these problems together into whole questions.
At A Level, the Unit 6 paper (WCH16) asks the same kind of questions about the Units 4 and 5 practicals.
- Limiting reagent: the reactant that is completely used up; it fixes the amount of product formed.
- Excess: more of a reactant than the equation needs, so some is left over at the end of the reaction.
- Molar volume: the volume occupied by one mole of any gas at a stated temperature and pressure; about 24 dm³ mol⁻¹ at room temperature and pressure.
A student plans to react 0.500 g of zinc granules with 30.0 cm³ of 0.500 mol dm⁻³ hydrochloric acid and says the acid is in excess.
Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)
Explain, by calculation, whether the student is correct.
(3)
1. n(Zn) = 0.500 ÷ 65.4 = 7.65 × 10⁻³ mol (1)
2. HCl needed = 2 × 7.65 × 10⁻³ = 0.0153 mol; HCl present = 0.500 × 30.0 ÷ 1000 = 0.0150 mol (1)
3. 0.0150 mol is less than the 0.0153 mol needed, so the acid is not in excess: the zinc is (slightly) in excess, and the student is wrong (1)
Answer: the student is wrong.
The ratio of 2 : 1 is the trap: comparing 0.0150 mol of acid with 7.65 × 10⁻³ mol of zinc without the ratio makes the acid look twice as plentiful as it is.
Calcium reacts with water: Ca(s) + 2H₂O(l) → Ca(OH)₂(s) + H₂(g).
A student plans to add 0.25 g of calcium to excess water and collect the hydrogen in a 100 cm³ gas syringe.
(Molar volume of gas = 24.0 dm³ mol⁻¹)
(a) Show, by calculation, that the plan would not work. (2)
(b) Calculate the largest mass of calcium that could be used. (2)
1. (a) n(Ca) = 0.25 ÷ 40.1 = 6.23 × 10⁻³ mol = n(H₂) (1 : 1) (1)
2. volume = 6.23 × 10⁻³ × 24 000 = 150 cm³, which is more than the 100 cm³ the syringe holds (1)
3. (b) largest n(H₂) = 100 ÷ 24 000 = 4.17 × 10⁻³ mol = n(Ca) (1)
4. mass = 4.17 × 10⁻³ × 40.1 = 0.167 g (1)
Answer: (a) about 150 cm³ would form, more than the syringe can hold; (b) 0.167 g, so in practice about 0.15 g, leaving a margin.
0.150 g of a Group 2 metal, M, reacted with excess dilute hydrochloric acid, M(s) + 2HCl(aq) → MCl₂(aq) + H₂(g).
89.5 cm³ of hydrogen was collected at room temperature and pressure.
Identify M. You must show your working.
(Molar volume of gas = 24.0 dm³ mol⁻¹) (3)
1. n(H₂) = 89.5 ÷ 24 000 = 3.73 × 10⁻³ mol (1)
2. n(M) = 3.73 × 10⁻³ mol (1 : 1), so Ar = 0.150 ÷ 3.73 × 10⁻³ = 40.2 (1)
3. the Group 2 metal with Ar closest to 40.2 is calcium (40.1) (1)
Answer: calcium.
Potassium (39.1) is nearer than you might expect, but it is in Group 1 and its chloride is KCl, which does not fit the equation.
A bare "calcium" with no working is worth much less than the three steps.
Sodium hydrogencarbonate decomposes on heating: 2NaHCO₃(s) → Na₂CO₃(s) + H₂O(g) + CO₂(g).
A student heats 1.00 g in a boiling tube connected to a 100 cm³ gas syringe and plans to use the volume of gas to confirm the equation.
(Molar volume of gas = 24.0 dm³ mol⁻¹)
(a) Calculate the volume of carbon dioxide the equation predicts. (2)
(b) Explain two problems with the plan, and how to solve each. (4)
1. (a) n(NaHCO₃) = 1.00 ÷ 84.0 = 0.0119 mol, so n(CO₂) = 0.0119 ÷ 2 = 5.95 × 10⁻³ mol (1)
2. volume = 5.95 × 10⁻³ × 24 000 = 143 cm³ (1)
3. (b) 143 cm³ is more than the syringe holds (1); use 0.65 g or less (0.700 g would give exactly 100 cm³) (1)
4. while the tube is hot, the gas in the apparatus is warm and contains steam, so a reading taken then is too large (1); let the apparatus cool to room temperature, when the steam has condensed and only the CO₂ is left as gas, before reading the syringe (1)
Answer: (a) 143 cm³; (b) the gas would overflow the syringe, so use a smaller mass, and the volume must be read only after cooling to room temperature, when the steam has condensed.
When a part says "Show, by calculation" or "Your answer must include a calculation", a correct statement with no numbers earns nothing.
Give the moles of both reagents, apply the ratio from the equation in writing, and finish with a sentence that states the comparison ("0.0150 mol is less than 0.0153 mol, so ...").
- Using the moles of the reagent in excess. It is tempting because its volume and concentration are printed, but only the limiting reagent decides the amount of product. Find the limiting reagent first, every time.
- Comparing moles without the ratio. Two moles of HCl react with one mole of zinc, so the numbers can only be compared after the ratio is applied.
- Choosing the nearest Ar from any group. The stem names the group; pick the element from that group, and say which group you checked.
- Accepting an answer that cannot be right. A molar volume ten times too small, or a yield above 100%, should send you back to your working.
- Forgetting why a metal reacts slowly at first in an inverted burette. When a metal is placed at the top of a burette full of water and acid is added, the metal sits in water: nothing happens until the acid has diffused down to it.
Check your understanding
1. 0.0600 g of magnesium ribbon is added to 20.0 cm³ of 0.200 mol dm⁻³ sulfuric acid. Mg(s) + H₂SO₄(aq) → MgSO₄(aq) + H₂(g). (Molar volume of gas = 24.0 dm³ mol⁻¹)
(a) Show, by calculation, that the acid is in excess. (2)
(b) Calculate the volume of hydrogen formed, in cm³. (2)
(c) Explain why this volume could not be collected in an inverted 50 cm³ burette. (1)
Total [5]
2. 0.0832 g of a Group 2 metal reacts with excess dilute hydrochloric acid and gives 82.0 cm³ of hydrogen at room temperature and pressure. M(s) + 2HCl(aq) → MCl₂(aq) + H₂(g). Identify the metal. You must show your working. (Molar volume of gas = 24.0 dm³ mol⁻¹) (3) Total [3]
3. In a molar volume experiment, a student measures the acid with a beaker, not a measuring cylinder. Explain why this does not affect the result. (2) Total [2]
4. Calcium carbonate reacts with excess dilute hydrochloric acid: CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g). Complete the table, using a molar volume of 24.0 dm³ mol⁻¹. (3)
Mass of CaCO₃ / g
Amount of CaCO₃ / mol
Volume of CO₂ / cm³
0.100
0.250
0.400
Total [3]
5. Small lumps of calcium are added to water in a flask. After a minute the fizzing slows down long before the calcium has been used up, even though plenty of water is left. Suggest a reason. (1) Total [1]
6. A student wants the largest volume of carbon dioxide that a 100 cm³ gas syringe can measure, from calcium carbonate and excess acid.
(a) Calculate the largest mass of calcium carbonate that can be used. (Molar volume of gas = 24.0 dm³ mol⁻¹) (2)
(b) A second student says it is better to use a much smaller mass, for example 0.050 g, "to be safe". Explain why this is a poor choice. (2)
Total [4]
Answers
1. (a) n(Mg) = 0.0600 ÷ 24.3 = 2.47 × 10⁻³ mol; n(H₂SO₄) = 0.200 × 20.0 ÷ 1000 = 4.00 × 10⁻³ mol (1); the ratio is 1 : 1 and 4.00 × 10⁻³ > 2.47 × 10⁻³, so the acid is in excess (1). (b) n(H₂) = n(Mg) = 2.47 × 10⁻³ mol (1); volume = 2.469 × 10⁻³ × 24 000 = 59.3 cm³ (1). (c) 59.3 cm³ is more than the 50 cm³ the burette can hold, so gas would escape from the bottom (1).
2. n(H₂) = 82.0 ÷ 24 000 = 3.42 × 10⁻³ mol (1); Ar = 0.0832 ÷ 3.417 × 10⁻³ = 24.3 (1); magnesium (1). Sodium (23.0) is in Group 1, so it is not the answer.
3. The acid is in excess, so the amount of gas is fixed by the mass of the metal (the limiting reagent) (1); the exact volume of acid does not change the moles of gas formed, as long as there is enough acid for all the metal to react (1).
4. Amounts, using M(CaCO₃) = 100.1 g mol⁻¹: 9.99 × 10⁻⁴, 2.50 × 10⁻³ and 4.00 × 10⁻³ mol (1); volumes: 24.0 cm³ (1); 59.9 and 95.9 cm³ (1). Each amount × 24 000 gives the volume, because the ratio CaCO₃ : CO₂ is 1 : 1.
5. The calcium hydroxide formed is only sparingly soluble, so it coats the calcium and slows contact with the water (1).
6. (a) n(CO₂) = 100 ÷ 24 000 = 4.17 × 10⁻³ mol = n(CaCO₃) (1); mass = 4.17 × 10⁻³ × 100.1 = 0.417 g (1). (b) only about 12 cm³ of gas would form (1), so the uncertainty of reading the syringe, and of weighing 0.050 g, would be a large percentage of each measurement; a mass a little below 0.417 g keeps the percentage uncertainties small and still fits the syringe (1).