Pearson Edexcel International GCSE Chemistry 4CH1: Chemical formulae, equations and calculations – 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 from your board or school.

These questions cover sub-topic 1(e), Chemical formulae, equations and calculations (points 1.25 to 1.36), of the Pearson Edexcel International GCSE Chemistry (4CH1) specification, Issue 3 (September 2024). The qualification is untiered. Points 1.34C and 1.35C carry a C reference, so they are Paper 2 only; questions 9 and 10 are labelled. A calculator may be used on both papers. Give answers to 3 significant figures unless told otherwise.

This set adds to the Principles of chemistry practice set. Learn the content in the study guide and revision notes. Course hub: Edexcel IGCSE Chemistry; printable checklist.

Questions

1. Aluminium powder burns in oxygen to form solid aluminium oxide, Al₂O₃.

(a) Write the word equation for this reaction. [1] (b) Write the balanced equation, including state symbols. [2] (c) Solid lithium nitride, Li₃N, reacts with water to form lithium hydroxide solution and ammonia gas, NH₃. Write the balanced equation, including state symbols. [2]

2. (A_r: H = 1, N = 14, O = 16, S = 32, Ca = 40)

(a) Calculate the relative formula mass of ammonium sulfate, (NH₄)₂SO₄. [1] (b) Calculate the relative formula mass of calcium nitrate, Ca(NO₃)₂. [1] (c) State the unit used for the amount of a substance. [1]

3. (A_r: H = 1, C = 12, N = 14, O = 16, Mg = 24, Cl = 35.5)

(a) Calculate the amount, in moles, in 8.5 g of ammonia, NH₃. [1] (b) Calculate the mass of 0.25 mol of magnesium carbonate, MgCO₃. [1] (c) 0.200 mol of a Group 1 metal chloride has a mass of 14.9 g. Calculate its relative formula mass and identify the metal. (A_r: Li = 7, Na = 23, K = 39) [2]

4. Sodium hydrogencarbonate decomposes on heating:

2NaHCO₃(s) → Na₂CO₃(s) + H₂O(g) + CO₂(g)

(A_r: H = 1, C = 12, O = 16, Na = 23)

(a) Calculate the maximum mass of sodium carbonate that can form from 16.8 g of sodium hydrogencarbonate. [3] (b) A student obtains 9.54 g of sodium carbonate. Calculate the percentage yield. [2]

5. A hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass. Its relative molecular mass is 56. (A_r: H = 1, C = 12)

(a) State what is meant by the terms empirical formula and molecular formula. [2] (b) Calculate the empirical formula and the molecular formula of the hydrocarbon. [4]

6. A student finds the formula of magnesium oxide by heating magnesium ribbon in a crucible with a lid.

Measurement	Mass / g
empty crucible and lid	25.20
crucible, lid and magnesium	25.92
crucible, lid and magnesium oxide after heating to constant mass	26.40

(A_r: O = 16, Mg = 24)

(a) Explain why the lid is lifted from time to time but not removed. [2] (b) Use the results to show that the empirical formula is MgO. [3] (c) Some white smoke escapes when the lid is lifted. Explain the effect on the calculated ratio of magnesium to oxygen. [2]

7. A sample of hydrated magnesium sulfate, MgSO₄·xH₂O, has a mass of 4.92 g. After heating to constant mass, 2.40 g of anhydrous magnesium sulfate remains. (A_r: H = 1, O = 16, Mg = 24, S = 32)

(a) Calculate the value of x. [3] (b) Give the name for the water that is removed on heating. [1]

8. 2.86 g of an oxide of copper is heated in hydrogen until no further change occurs, leaving 2.54 g of copper. (A_r: O = 16, Cu = 63.5)

(a) Calculate the empirical formula of the oxide. [3] (b) Explain why hydrogen is kept flowing over the copper while it cools. [1] (c) Write the balanced equation for the reaction of this oxide with hydrogen. [1]

9. (Paper 2 only) 2.65 g of anhydrous sodium carbonate, Na₂CO₃, is dissolved in water and made up to 250 cm³. (A_r: C = 12, O = 16, Na = 23)

(a) Calculate the concentration of the solution in mol/dm³. [2] (b) 25.0 cm³ of this solution reacts exactly with 20.0 cm³ of hydrochloric acid: Na₂CO₃(aq) + 2HCl(aq) → 2NaCl(aq) + H₂O(l) + CO₂(g) Calculate the concentration of the hydrochloric acid in mol/dm³. [3]

10. (Paper 2 only) One mole of any gas occupies 24 dm³ (24 000 cm³) at rtp. (A_r: Mg = 24)

(a) Calculate the volume, in dm³, of 0.150 mol of carbon dioxide at rtp. [1] (b) 0.36 g of magnesium reacts with excess hydrochloric acid: Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g) Calculate the volume of hydrogen formed, in cm³, at rtp. [2] (c) Excess calcium carbonate is added to 50.0 cm³ of 2.00 mol/dm³ hydrochloric acid: CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g) Calculate the volume of carbon dioxide formed, in dm³, at rtp. [3]

11. Titanium is made by heating titanium(IV) chloride with sodium:

TiCl₄ + 4Na → Ti + 4NaCl

(A_r: Na = 23, Cl = 35.5, Ti = 48)

(a) Calculate the mass of sodium, in kg, needed to react with 380 kg of titanium(IV) chloride. [3] (b) The process gives 84.0 kg of titanium. Calculate the percentage yield. [3] (c) Suggest one reason why the yield is less than 100%. [1]

Answers

1. (a) aluminium + oxygen → aluminium oxide [1]. (b) 4Al(s) + 3O₂(g) → 2Al₂O₃(s): correct balancing [1]; correct state symbols [1]. (c) Li₃N(s) + 3H₂O(l) → 3LiOH(aq) + NH₃(g): correct formulae and balancing [1]; correct state symbols [1]. [5] Examiner insight: State symbols earn a separate mark: a balanced equation showing water as (aq) instead of (l) loses it.

2. (a) (14 + 4 × 1) × 2 + 32 + 4 × 16 = 132 [1]. (b) 40 + 2 × (14 + 3 × 16) = 164 [1]. (c) The mole (mol) [1]. [3] Examiner insight: Relative formula mass has no units; the bracket multiplier must apply to every atom inside it.

3. (a) n = 8.5 ÷ 17 = 0.500 mol [1]. (b) M_r(MgCO₃) = 84, so mass = 0.25 × 84 = 21.0 g [1]. (c) M_r = 14.9 ÷ 0.200 = 74.5; 74.5 − 35.5 = 39 [1], so the metal is potassium (KCl) [1]. [4] Examiner insight: In (c) show the subtraction of 35.5; a bare “potassium” risks losing the method mark.

4. (a) M_r(NaHCO₃) = 84; n = 16.8 ÷ 84 = 0.200 mol [1]. Ratio 2 : 1, so n(Na₂CO₃) = 0.100 mol [1]. M_r(Na₂CO₃) = 106; mass = 0.100 × 106 = 10.6 g [1]. (b) Percentage yield = (9.54 ÷ 10.6) × 100 [1] = 90.0% [1]. [5] Examiner insight: Missing the 2 : 1 ratio gives 21.2 g and loses one mark, but (b) still scores in full as error carried forward.

5. (a) Empirical formula: the simplest whole-number ratio of atoms of each element in a compound [1]. Molecular formula: the actual number of atoms of each element in one molecule [1]. (b) C: 85.7 ÷ 12 = 7.14; H: 14.3 ÷ 1 = 14.3 [1]. Ratio 7.14 : 14.3 = 1 : 2, so empirical formula CH₂ [1]. M_r(CH₂) = 14; 56 ÷ 14 = 4 [1]. Molecular formula C₄H₈ [1]. [6] Examiner insight: Each definition scores separately; in (b) the molecular formula mark needs the 56 ÷ 14 step shown.

6. (a) Lifting it lets oxygen in so all the magnesium reacts [1]; keeping it on stops magnesium oxide escaping [1]. (b) Mass of Mg = 25.92 − 25.20 = 0.72 g; mass of oxide = 1.20 g, so mass of O = 0.48 g [1]. Moles: Mg = 0.72 ÷ 24 = 0.030; O = 0.48 ÷ 16 = 0.030 [1]. Ratio 1 : 1, so the formula is MgO [1]. (c) The final mass is too low, so the calculated mass of oxygen is too small [1]; the ratio Mg : O comes out greater than 1 : 1 (too much magnesium) [1]. [7] Examiner insight: On a “show that” no mark goes to restating MgO; the moles of both elements and the 1 : 1 ratio must be shown.

7. (a) Mass of water = 4.92 − 2.40 = 2.52 g [1]. n(MgSO₄) = 2.40 ÷ 120 = 0.020 mol; n(H₂O) = 2.52 ÷ 18 = 0.14 mol [1]. x = 0.14 ÷ 0.020 = 7 [1]. (b) Water of crystallisation [1]. [4] Examiner insight: Dividing the masses (2.52 ÷ 2.40) without converting to moles scores only the first mark.

8. (a) Mass of O = 2.86 − 2.54 = 0.32 g [1]. Moles: Cu = 2.54 ÷ 63.5 = 0.040; O = 0.32 ÷ 16 = 0.020 [1]. Ratio 2 : 1, so Cu₂O [1]. (b) It stops hot copper reacting with oxygen in the air [1]. (c) Cu₂O + H₂ → 2Cu + H₂O [1]. [5] Examiner insight: Keep at least 2 significant figures in the moles; rounding too early can turn a 2 : 1 ratio into a wrong one.

9. (a) n(Na₂CO₃) = 2.65 ÷ 106 = 0.0250 mol [1]. Concentration = 0.0250 ÷ 0.250 = 0.100 mol/dm³ [1]. (b) n(Na₂CO₃) in 25.0 cm³ = 0.100 × 25.0 ÷ 1000 = 0.00250 mol [1]. n(HCl) = 2 × 0.00250 = 0.00500 mol [1]. Concentration = 0.00500 ÷ 0.0200 = 0.250 mol/dm³ [1]. [5] Examiner insight: Dividing by 250 instead of 0.250 gives an answer 1000 times too small and loses the accuracy mark; ECF applies from (a) to (b).

10. (a) 0.150 × 24 = 3.60 dm³ [1]. (b) n(Mg) = 0.36 ÷ 24 = 0.015 mol = n(H₂) [1]; volume = 0.015 × 24 000 = 360 cm³ [1]. (c) n(HCl) = 2.00 × 50.0 ÷ 1000 = 0.100 mol [1]. n(CO₂) = 0.100 ÷ 2 = 0.0500 mol [1]. Volume = 0.0500 × 24 = 1.20 dm³ [1]. [6] Examiner insight: Using 24 when cm³ is asked (or 24 000 for dm³) loses the final mark even when the moles are right.

11. (a) M_r(TiCl₄) = 48 + 4 × 35.5 = 190; n = 380 000 ÷ 190 = 2000 mol [1]. n(Na) = 4 × 2000 = 8000 mol [1]. Mass = 8000 × 23 = 184 000 g = 184 kg [1]. (b) n(Ti) = 2000 mol; theoretical mass = 2000 × 48 = 96 000 g = 96.0 kg [1]. Percentage yield = (84.0 ÷ 96.0) × 100 [1] = 87.5% [1]. (c) Any one: titanium lost when separating it from the sodium chloride; incomplete reaction; side reactions with air or water [1]. [7] Examiner insight: Working in kg throughout is fine if consistent; mixing g and kg gives an answer 1000 times out and loses the accuracy mark.

Where marks are usually lost

• Using (aq) for water or an insoluble solid.
• Ignoring the mole ratio, so a reacting mass is out by a factor of 2 or 4.
• Calculating percentage yield upside down (over 100%).
• Dividing percentages by atomic number instead of A_r.
• Rounding moles too early, so a 1 : 2 ratio looks like 1 : 1.8.
• Not converting cm³ to dm³ (÷ 1000) before using mol/dm³.
• Mixing up 24 dm³ and 24 000 cm³.

Next steps

• Principles of chemistry revision notes
• Principles of chemistry study guide
• Principles of chemistry practice questions
• Free 4CH1 all-topics diagnostic
• Edexcel IGCSE Chemistry course hub
• Printable Edexcel IGCSE Chemistry checklist
• Book a free trial class

Official syllabus

Pearson Edexcel International GCSE in Chemistry (4CH1) specification, Issue 3, September 2024, published by Pearson Education Limited – Topic 1 Principles of chemistry, sub-topic (e) Chemical formulae, equations and calculations, points 1.25 to 1.36.