AS Level Chemistry chapter 2: Atoms, molecules and stoichiometry MCQs with answers

By Team Megalecture · Updated 4 Nov 2026

88 exam-style questions on Atoms, molecules and stoichiometry, each with the answer and a short explanation. Ten of them are below to try. The chapter test draws ten fresh questions each time and is marked the moment you finish, with your score by topic.

88 questions · 4 topics · 10 per test

Test this chapter freeAll AS Level Chemistry chapters

What chapter 2 covers

Ten Atoms, molecules and stoichiometry questions with answers

Try each one, then open the answer. These ten stay the same; the test draws new ones every time.

1. What is the mass of one atom of carbon-12? (L = 6.02 × 1023 mol−1)

  1. A
    1.99 × 10−23 g
  2. B
    1.66 × 10−24 g
  3. C
    7.22 × 1024 g
  4. D
    12.0 g
Show answer

Answer: A. One mole of carbon-12 (12.0 g) contains 6.02 × 1023 atoms, so one atom has a mass of 12.0 ÷ (6.02 × 1023) = 1.99 × 10−23 g. 1.66 × 10−24 g is the mass of 1 u.

2. A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass, and has Mr = 60. What is its molecular formula?

  1. A
    CH2O
  2. B
    C3H8O
  3. C
    C2H4O2
  4. D
    C2H6O2
Show answer

Answer: C. Moles per 100 g: C 40.0/12.0 = 3.33, H 6.7/1.0 = 6.7, O 53.3/16.0 = 3.33, ratio 1:2:1, so the empirical formula is CH2O (mass 30). 60/30 = 2, so the molecular formula is C2H4O2. C3H8O also has Mr 60 but the wrong percentages.

3. The unified atomic mass unit is defined as

  1. A
    one twelfth of the mass of one atom of carbon-12
  2. B
    the mass of one proton
  3. C
    the mass of one atom of hydrogen-1
  4. D
    one twelfth of the mass of one mole of carbon-12
Show answer

Answer: A. By definition 1 u = 1/12 of the mass of a single 12C atom. It is close to, but not exactly, the mass of a proton or a hydrogen atom.

4. Equal masses of methane, CH4, and oxygen, O2, are taken. What is the ratio of the number of atoms in the methane sample to the number of atoms in the oxygen sample?

  1. A
    1 : 2
  2. B
    2 : 1
  3. C
    5 : 2
  4. D
    5 : 1
Show answer

Answer: D. For mass m: methane has m/16 mol × 5 atoms, oxygen has m/32 mol × 2 atoms. Ratio = (5/16) ÷ (2/32) = 5 : 1. 2 : 1 is the ratio of molecules only; 5 : 2 ignores the different molar masses.

5. What is the formula of iron(III) sulfate?

  1. A
    FeSO4
  2. B
    Fe(SO4)3
  3. C
    Fe3(SO4)2
  4. D
    Fe2(SO4)3
Show answer

Answer: D. Iron(III) is Fe3+ and sulfate is SO42−. Two Fe3+ (6+) balance three SO42− (6−). FeSO4 is iron(II) sulfate.

6. Aqueous lead(II) nitrate is added to aqueous potassium iodide and a yellow precipitate forms. What is the ionic equation, including state symbols?

  1. A
    Pb2+(aq) + I−(aq) → PbI(s)
  2. B
    Pb(NO3)2(aq) + 2KI(aq) → PbI2(s) + 2KNO3(aq)
  3. C
    Pb2+(aq) + 2I−(aq) → PbI2(aq)
  4. D
    Pb2+(aq) + 2I−(aq) → PbI2(s)
Show answer

Answer: D. K+ and NO3− are spectators. Pb2+ needs two I−, and the precipitate is a solid, so PbI2(s). The equation with KI is the full equation, and PbI2(aq) would mean no precipitate.

7. Chlorine consists of 35Cl and 37Cl atoms in the ratio 3 : 1. Chlorine molecules, Cl2, therefore have relative molecular masses of 70, 72 and 74. In what ratio are these three kinds of molecule present?

  1. A
    1 : 2 : 1
  2. B
    3 : 2 : 1
  3. C
    9 : 3 : 1
  4. D
    9 : 6 : 1
Show answer

Answer: D. The probability of each atom being 35Cl is 3/4. Mr 70: (3/4)2 = 9/16; Mr 72: 2 × (3/4)(1/4) = 6/16 (two ways to arrange the pair); Mr 74: (1/4)2 = 1/16. Forgetting the factor of 2 gives 9 : 3 : 1.

8. 50.0 cm3 of 0.200 mol dm−3 sodium chloride is mixed with 150 cm3 of 0.100 mol dm−3 calcium chloride. What is the concentration of chloride ions in the mixture?

  1. A
    0.125 mol dm−3
  2. B
    0.150 mol dm−3
  3. C
    0.200 mol dm−3
  4. D
    0.300 mol dm−3
Show answer

Answer: C. n(Cl−) = 0.0500 × 0.200 + 0.150 × 0.100 × 2 = 0.0400 mol in 0.200 dm3, so 0.200 mol dm−3. Forgetting the 2 Cl− in CaCl2 gives 0.125; averaging the two concentrations gives 0.150.

9. What volume does 0.880 g of carbon dioxide occupy at room conditions?

  1. A
    0.480 dm3
  2. B
    0.754 dm3
  3. C
    0.880 dm3
  4. D
    21.1 dm3
Show answer

Answer: A. n = 0.880 ÷ 44.0 = 0.0200 mol; V = 0.0200 × 24.0 = 0.480 dm3. Using Mr 28.0 (carbon monoxide) gives 0.754 dm3; multiplying the mass directly by 24.0 gives 21.1 dm3.

10. A hydrocarbon has the empirical formula CH2 and a relative molecular mass of 70.0. What is its molecular formula?

  1. A
    C4H8
  2. B
    C5H12
  3. C
    C5H10
  4. D
    C6H12
Show answer

Answer: C. The empirical formula mass is 12.0 + 2.0 = 14.0, and 70.0 ÷ 14.0 = 5, so the formula is 5 × CH2 = C5H10. C5H12 has Mr 72 and does not fit the CH2 ratio.

Start the chapter testAbout the AS Level Chemistry test

Read the notes for this chapter

Ways to practise this chapter

Atoms, molecules and stoichiometry: questions students ask

How many Atoms, molecules and stoichiometry MCQs are there?

88, each with the answer and a short explanation, across 4 topics: Relative masses of atoms and molecules, The mole and the Avogadro constant, Formulas, Reacting masses and volumes (of solutions and gases). Every chapter test draws ten of them, so you can take it many times.

Is the AS Level Chemistry chapter 2 test free?

Yes. The test, the answers and the explanations are free and need no sign-up. Sign in only if you want your scores kept across devices.

How is the chapter test marked?

The moment you press Finish: your score, the right answer and an explanation for every question, and the ones you got wrong saved in My mistakes to retry.

Keep going

← Chapter 1: Atomic structureChapter 3: Chemical bonding →All chaptersStuck? Ask a question