Amount of Substance

物质的量

Converting between amount of substance, mass and molar mass, when molar gas volume applies, and using mole ratios read off a chemical equation.

11 real-exam questions · 4 sittings8 in mock papers19 practice

Formulas

Amount of substance from mass
Definition
$n=\frac{m}{M}$
Applies when：M is the molar mass in g/mol, numerically equal to the relative molecular (atomic) mass
Molar gas volume
Formula
$n=\frac{V}{V_m}$
Applies when：Gases only; the molar volume is 22.4 L/mol at STP (0 °C, 101 kPa) — that value does not carry over to other conditions
Avogadro constant and particle number
Definition
$N=n\,N_A$
Applies when：The Avogadro constant is about 6.02×10²³ mol⁻¹; be explicit about whether the particles counted are molecules, atoms or ions

How the real exam tests it

Questions from real CSCA papers that test this concept. Work each one out before you open the answer.

CSCA Chemistry December 2025 Past Paper
Using CO to reduce a mixture of CuO and Cu powder, 23.4 g of the mixture reacts completely to yield 20.2 g of pure copper powder. The mass of CuO in the original mixture is ( )
- A. 20 g
- B. 32 g
- C. 40 g
- D. 16 g
Show answer & solution
Answer: D
Answer: D CuO + CO → Cu + CO₂. The mass fraction of Cu in CuO is 64 / 80 = 0.8. Let the mass of CuO be x: the original Cu is 23.4 − x and the Cu produced is 0.8x. (23.4 − x) + 0.8x = 20.2 → x = 16 g.
See it in the full CSCA Chemistry December 2025 Past Paper
CSCA Chemistry April 2026 Past Paper
Excess CO reacts completely with 16 g of CuO powder at high temperature. The $\mathrm{CO_{2}}$ produced is passed into limewater. The mass of the precipitate formed is ( ).
- A. 10 g
- B. 15 g
- C. 20 g
- D. 25 g
Show answer & solution
Answer: C
Answer: C 16 g of CuO is 0.2 mol; CO + CuO →(high temperature) Cu + $\mathrm{CO_{2}}$ produces 0.2 mol $\mathrm{CO_{2}}$ ; the $\mathrm{CO_{2}}$ reacts with limewater to form 0.2 mol $\mathrm{CaCO_{3}}$ ; the molar mass of $\mathrm{CaCO_{3}}$ is 100 g/mol, so the precipitate mass = 0.2 × 100 = 20 g.
Take the full CSCA Chemistry April 2026 Past Paper — timed & graded
CSCA Chemistry March 2026 Past Paper
$2 \mathrm { H } _ { 2 } \mathrm {~S} + \mathrm { SO } _ { 2 } = 3 \mathrm {~S} + 2 \mathrm { H } _ { 2 } \mathrm { O }$ . If 96 g sulfur produced, $\mathrm { H } _ { 2 } \mathrm {~S}$ consumed is
- A．1 mol
- B．2 mol
- C．3 mol
- D．4 mol
Show answer & solution
Answer: B
$\mathrm { n } ( \mathrm { S } ) = 3 \mathrm {~mol}$ . $\mathrm { n } \left( \mathrm { H } _ { 2 } \mathrm {~S} \right) = 2 / 3 \times 3 = 2 \mathrm {~mol}$ . Answer: B．
Take the full CSCA Chemistry March 2026 Past Paper — timed & graded

Real-exam evidence

Recorded past-paper appearances are listed below; online verification is labelled separately.

11 questions across 4 sittings1 of 4 sittings available to verify free online

CSCA Chemistry April 2026 Past Paper3 questionsreading page not yet public
CSCA Chemistry March 2026 Past Paper2 questionsreading page not yet public
CSCA Chemistry January 2026 Past Paper4 questionsreading page not yet public
CSCA Chemistry December 2025 Past Paper2 questionsView question 40 and answer free (full paper)View question 45 and answer free (full paper)

8 questions across 5 mock papers — our own papers, so a coverage signal rather than real-exam evidence

19 related practice questions availablePractise free: Mole Calculations Practise free: Ideal Gas Law

Sitting evidence comes from the CSCA past papers on record; questions whose original paper is missing are excluded, and sprint papers are deliberately not counted.

Glossary

mole

Related concepts

Classification of Matter Solution Composition and Concentration

Formulas