IGCSE Chemistry: Formulae, Equations and the Mole — Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Formulae, Equations and the Mole revision notes

Tier note: questions and parts marked (0620 Extended, 5070 required) go beyond 0620 Core. All other questions are answerable by a 0620 Core candidate.

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Questions

1. Balance these equations:

(a) ___ Al + ___ O₂ → ___ Al₂O₃ [1] (b) ___ C₃H₈ + ___ O₂ → ___ CO₂ + ___ H₂O [1] (c) ___ Fe + ___ HCl → ___ FeCl₂ + ___ H₂ [1]

2. Explain, in terms of atoms, why mass is conserved in a closed chemical reaction. [2]

3. A metal carbonate is heated in an open crucible and the mass decreases.

(a) Explain why. [2] (b) Explain what would happen to the mass if a metal were heated in air instead, and why. [2]

4. (0620 Extended, 5070 required) 8.00 g of magnesium oxide is formed when magnesium burns. (A_r: Mg = 24.0, O = 16.0)

(a) Calculate the M_r of MgO. [1] (b) Calculate the moles of MgO. [2] (c) Calculate the mass of magnesium that reacted. [3]

5. (0620 Extended, 5070 required) 5.00 g of calcium carbonate is strongly heated (thermally decomposed). 1.80 g of calcium oxide is obtained, against a theoretical maximum of 2.80 g.

(a) Calculate the percentage yield. [2] (b) Give three reasons why the yield is below 100%. [3]

6. (0620 Extended, 5070 required) Calculate the concentration in mol dm⁻³ of a solution containing 4.00 g of NaOH in 250 cm³. (M_r = 40.0) [3]

7. (0620 Extended, 5070 required) Using the ions Al³⁺ and SO₄²⁻, deduce the formula of aluminium sulfate, explaining your method step by step. [3]

8. (0620 Extended, 5070 required) A sample of impure zinc carbonate has a total mass of 12.0 g. Analysis shows it contains 10.2 g of pure zinc carbonate, the rest being insoluble rock impurities.

(a) Calculate the percentage purity of the sample. [2]

(b) Explain why percentage purity and percentage yield are different calculations, even though both are expressed as a percentage and both compare a real quantity against a maximum. [2]

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9. Calculate the relative formula mass, Mᵣ, of calcium nitrate, Ca(NO₃)₂. [Aᵣ: Ca = 40, N = 14, O = 16] [2]

10. (0620 Extended, 5070 required) 0.050 mol of calcium carbonate reacts completely with excess dilute hydrochloric acid.

CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

Calculate the volume of carbon dioxide produced at room temperature and pressure, in cm³. (One mole of any gas occupies 24 dm³ at r.t.p.) [2]

11. (0620 Extended, 5070 required) 1.35 g of calcium reacts completely with excess dilute hydrochloric acid: Ca + 2HCl → CaCl₂ + H₂. Calculate the mass of calcium chloride formed. [Aᵣ: Ca = 40, Cl = 35.5] [3]

Answers

1. (a) 4Al + 3O₂ → 2Al₂O₃ [1]. (b) C₃H₈ + 5O₂ → 3CO₂ + 4H₂O [1]. (c) Fe + 2HCl → FeCl₂ + H₂ [1].

2. Atoms are rearranged, not created or destroyed [1]; the same number and type of each atom is present before and after, so the total mass is unchanged [1].

3. (a) Carbon dioxide gas is produced [1] and escapes from the open crucible, so the mass remaining falls [1]. (b) The mass would increase [1], because the metal combines with oxygen from the air, which adds to the mass of the solid [1].

4. (a) 24.0 + 16.0 = 40.0 [1]. (b) n = 8.00 ÷ 40.0 [1] = 0.200 mol [1]. (c) From 2Mg + O₂ → 2MgO, n(Mg) = n(MgO) = 0.200 mol [1] m = 0.200 × 24.0 [1] = 4.80 g [1].

5. (a) (1.80 ÷ 2.80) × 100 [1] = 64.3% [1]. (b) Any three: the decomposition may be incomplete (not heated hot enough or long enough) [1]; some product is lost during transfer between containers [1]; side reactions produce other products [1]; the calcium carbonate may be impure; the reaction is reversible, so some CaO can recombine with CO₂ on cooling.

6. n = 4.00 ÷ 40.0 = 0.100 mol [1] V = 250 ÷ 1000 = 0.250 dm³ [1] c = 0.100 ÷ 0.250 = 0.400 mol dm⁻³ [1].

7. Swap the charges of the two ions to become the subscripts of the other: Al³⁺ (charge 3) becomes the subscript on SO₄²⁻, and SO₄²⁻ (charge 2) becomes the subscript on Al³⁺, giving subscripts 2 and 3 respectively [1]. Since more than one SO₄²⁻ polyatomic ion is needed, it is enclosed in brackets [1]: Al₂(SO₄)₃ [1].

8. (a) (10.2 ÷ 12.0) × 100 [1] = 85.0% [1]. The remaining 1.8 g is insoluble rock impurity that never reacted. (b) Percentage purity compares the mass of the pure substance actually present against the total mass of the sample, including impurities that were never meant to be there [1]; percentage yield compares the actual mass of product obtained in a reaction against the maximum mass the balanced equation predicts, assuming the reaction went to completion with no losses — a different comparison, since yield concerns how much of a reaction happened, not how clean the product is [1].

9. Count the atoms: 1 × Ca, 2 × N and 6 × O (the 2 outside the bracket multiplies everything inside it) [1]. Mᵣ = 40 + (2 × 14) + (6 × 16) = 40 + 28 + 96 = 164 [1].

Examiner insight (June 2025): a common error was to use proton numbers (for example 20 for calcium) instead of the relative atomic masses and the number of atoms in the formula, which gives very different answers. Try the real question next: Cambridge IGCSE Chemistry 0620, June 2025, Paper 32, Question 3.

10. (0620 Extended, 5070 required) The mole ratio CaCO₃ : CO₂ is 1 : 1, so 0.050 mol of CO₂ is produced [1]. Volume = 0.050 × 24 = 1.2 dm³ = 1200 cm³ [1].

Examiner insight (June 2025): a common error was to mix up 24 (for dm³) and 24 000 (for cm³), or to divide by the molar volume instead of multiplying. Check which unit the question asks for. Try the real question next: Cambridge IGCSE Chemistry 0620, June 2025, Paper 42, Question 4.

11. (0620 Extended, 5070 required) Moles of Ca = 1.35 ÷ 40 = 0.03375 mol [1]. Moles of CaCl₂ = 0.03375 mol (1 : 1) and Mᵣ of CaCl₂ = 40 + 71 = 111 [1]. Mass = 0.03375 × 111 = 3.75 g [1].

Try the real question next: Cambridge O Level Chemistry 5070, June 2025, Paper 22, Question 4.

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Where marks are usually lost

• Changing formulae instead of coefficients when balancing.
• Not identifying the gas when explaining a mass change.
• Working from mass ratios rather than converting to moles.
• Forgetting to convert cm³ to dm³ for concentration.
• Adding proton (atomic) numbers instead of mass numbers when working out an Ar or Mr — mass number is the one relevant to relative mass; proton number only identifies the element.
• Forgetting brackets around a polyatomic ion when more than one is needed in a formula — Al₂SO₄₃ is meaningless and unreadable; it must be written Al₂(SO₄)₃.
• Swapping the charges onto the wrong ion, or forgetting to simplify the resulting subscripts to their lowest whole-number ratio where they share a common factor.
• Treating percentage purity and percentage yield as interchangeable — one is about how clean a sample is, the other about how much of a reaction actually happened.

Questions 7 and 8 draw on the common-ions table and percentage purity formula in the Formulae, Equations and the Mole revision notes, material the earlier balancing and mole-calculation questions on this page don’t reach.