What is the commonest lost mark in mole sums?

Forgetting to divide cm3 by 1 000, which makes concentration times volume a thousand times too large.

Why is a yield below 100 per cent?

The reaction is incomplete, product is lost in transfer, gas escapes or the weighing is inaccurate.

S4 Chemistry: The Mole (Paper 4)

2026-11-01

This paper mixes moles, concentration, gas volume, dilution and percentage yield, and repairs the common unit conversion error.

Direct answer　Write the moles first, then multiply by molar mass or divide by molar volume, and always change cm3 to dm3 before using concentration.

01　Paper details (check answers afterwards)

Item	Detail
Questions	10
Suggested time	25
Difficulty mix	Foundation 4 · Intermediate 3 · Challenge 2 · Error redo 1
Chapter	Chapter: S4 Chemistry: The Mole, Concentration and Percentage Yield
Level	Four basic, three advanced, two challenge and one error-repair item

02　Questions (10 items)

FoundationWhat is the mass of 0.25 mol of carbon dioxide with CO2 = 44?
FoundationHow many moles are in 5.6 g of iron with Fe = 56?
Foundation2.0 mol of sodium hydroxide is made up to 500 cm3. Find the concentration.
FoundationHow many moles of HCl are in 25.0 cm3 of 0.20 M hydrochloric acid?
IntermediateWhat volume do 3.0 mol of oxygen occupy at room conditions with a molar volume of 24.0 dm3?
Intermediate12.0 g of magnesium reacts with excess dilute acid. Find the moles and volume of hydrogen with Mg = 24 and a molar volume of 24.0 dm3.
Intermediate100 cm3 of 2.0 M solution is diluted to 500 cm3. Find the new concentration.
ChallengeHeating 5.0 g of calcium carbonate should give 2.2 g of carbon dioxide but 1.87 g is collected. Find the percentage yield to one decimal place.
Challenge0.050 mol of zinc reacts with 0.060 mol of copper(II) sulphate. Which is limiting and what mass of copper forms with Cu = 63.5?
Error redoError repair: a student writes that 250 cm3 of 0.40 M solution holds 0.40 times 250 or 100 mol. Give the mistake and the correct amount.

03　Answers and one-line notes

#	Answer	One-line note
1	11 g	Mass is moles times molar mass, so 0.25 times 44 gives 11 g.
2	0.10 mol	Divide the mass by the molar mass: 5.6 over 56 is 0.10 mol.
3	4.0 M	500 cm3 is 0.500 dm3, and 2.0 divided by 0.500 gives 4.0 mol dm-3.
4	0.0050 mol	Convert to 0.0250 dm3 first, then 0.20 times 0.0250 gives 0.0050 mol.
5	72 dm3	At room conditions each mole occupies 24.0 dm3, so 3.0 moles take 72 dm3.
6	0.50 mol; 12 dm3	12.0 g is 0.50 mol of magnesium, the ratio to hydrogen is one to one, so 0.50 mol or 12 dm3 forms.
7	0.40 M	The moles stay at 0.20, and dividing by the new volume of 0.500 dm3 gives 0.40 M.
8	85.0%	Divide the actual mass by the theoretical mass: 1.87 over 2.2 is 0.85, that is 85.0 per cent.
9	zinc limits; 3.18 g	The ratio is one to one, so 0.050 mol of zinc runs out first and 0.050 times 63.5 gives 3.175 g or 3.18 g of copper.
10	the volume must be changed from cm3 to dm3 first; the answer is 0.10 mol	The concentration is per dm3, so 250 cm3 is 0.250 dm3 and 0.40 times 0.250 gives 0.10 mol, not 100 mol.

04　What to do after marking

After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
Repeat the same item two days later; move to the next paper only when both attempts are correct.
Parents only need the answer and note columns; there is no need to explain every item.

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Questions and answers are written originally by the TutorZone team and are not taken from any school, publisher or examination authority paper. Every value is checked by computer.