Edexcel International A Level Chemistry: Formulae, equations and amount of substance (YCH11) – 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.

Here you practise Topic 1, Formulae, Equations and Amount of Substance, of the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017: outcomes 1.1 to 1.12, including Core Practical 1. The topic is part of Unit 1: Structure, Bonding and Introduction to Organic Chemistry (IAS). Learn the content first from the study guide and the revision notes. The course hub and the printable checklist list the rest of the course, and the free 10-minute diagnostics help you find weak areas.

Use these values: H 1.0, C 12.0, O 16.0, Na 23.0, Al 27.0, S 32.1, Cl 35.5, Ca 40.1, Fe 55.8, Zn 65.4, Ba 137.3. L = 6.02 × 10²³ mol⁻¹. R = 8.31 J mol⁻¹ K⁻¹. Molar volume at room temperature and pressure = 24.0 dm³ mol⁻¹.

Questions

1. State the empirical formula of hydrogen peroxide, H₂O₂, and explain why it is different from the molecular formula. [2]

2. Write ionic equations, with state symbols, and give observations for these test-tube reactions.

(a) Lead(II) nitrate solution is added to potassium iodide solution. Give one observation and the ionic equation. [2] (b) Iron filings are stirred into copper(II) sulfate solution. Give two observations and the ionic equation. [3] (c) Dilute nitric acid is added to sodium carbonate solution. Give the ionic equation only. [1]

3. Calculate the total number of ions in 5.684 g of sodium sulfate, Na₂SO₄. [3]

4. Parts per million.

(a) A 0.500 kg sample of tap water contains 0.60 mg of fluoride ions. Express this in ppm by mass. [1] (b) Air near a smelter contains 0.0150% sulfur dioxide by volume. Express this in ppm. [1] (c) Calculate the volume, in dm³, of sulfur dioxide in 2.00 m³ of this air. [2]

5. 4.282 g of barium hydroxide, Ba(OH)₂, is dissolved in water and made up to 500 cm³. Calculate the concentration in mol dm⁻³ and in g dm⁻³. [3]

6. Complete combustion of 0.580 g of a compound containing only carbon, hydrogen and oxygen gives 1.320 g of carbon dioxide and 0.540 g of water. The compound has Mr = 116. Determine its empirical and molecular formulae. [5]

7. In the thermite reaction, Fe₂O₃(s) + 2Al(s) → Al₂O₃(s) + 2Fe(l). Calculate the mass of aluminium needed to react with 8.00 g of iron(III) oxide, and the mass of iron formed. [4]

8. A 0.258 g sample of a volatile hydrocarbon is vaporised completely. The vapour occupies 94.9 cm³ at 90 °C and 100 kPa. The empirical formula is C₃H₅. Calculate the molar mass and deduce the molecular formula. [5]

9. Yield and atom economy.

(a) Zinc sulfate crystals are made by the reaction ZnO + H₂SO₄ + 6H₂O → ZnSO₄·7H₂O. A student starts with 3.00 g of zinc oxide and collects 7.85 g of dry crystals. Calculate the percentage yield. [4] (b) Electrolysis of brine is represented by 2NaCl + 2H₂O → 2NaOH + Cl₂ + H₂. Calculate the percentage atom economy for chlorine. [2] (c) The process in (a) has an atom economy of 100% but the yield is below 100%. Explain both facts. [2]

10. In a molar volume experiment, 0.150 g of calcium carbonate is reacted with excess hydrochloric acid in a flask connected to a gas syringe. The final volume of carbon dioxide is 34.0 cm³ at room temperature and pressure.

(a) Write the equation for the reaction, with state symbols. [2] (b) Calculate the molar volume of carbon dioxide from these results, in dm³ mol⁻¹. [3] (c) Calculate the percentage difference between your answer and 24.0 dm³ mol⁻¹. [1] (d) The mass was found by difference with a total uncertainty of ±0.001 g, and the syringe reading has an uncertainty of ±0.5 cm³. Calculate both percentage uncertainties. [2] (e) Use (c) and (d) to comment on the result, and suggest one improvement to the method. [2]

11. A student wants to confirm the equation for aluminium reacting with hydrochloric acid. 0.135 g of aluminium foil reacts completely with excess acid, giving 174 cm³ of hydrogen at room temperature and pressure.

(a) Calculate the amounts in moles of aluminium and of hydrogen. [2] (b) Use the ratio to deduce the balanced equation, with state symbols. [3] (c) Calculate the volume of hydrogen expected, and suggest why less was collected. [2]

Answers

1. Empirical formula HO [1]. The molecular formula gives the actual atoms in one molecule (two of each); the empirical formula is only the simplest whole-number ratio, 1 : 1 [1]. Examiner insight: “Simplest ratio” on its own may not score the second mark; contrast it with the actual number of atoms in a molecule.

2. (a) Yellow precipitate [1]. Pb²⁺(aq) + 2I⁻(aq) → PbI₂(s) [1] (b) Pink-brown solid forms [1]; solution changes from blue to pale green [1]. Fe(s) + Cu²⁺(aq) → Fe²⁺(aq) + Cu(s) [1] (c) CO₃²⁻(aq) + 2H⁺(aq) → CO₂(g) + H₂O(l) [1] Examiner insight: “Copper is formed” is a conclusion, not an observation; describe colour and state as seen, and keep spectator ions out of the equation.

3. M(Na₂SO₄) = 142.1 g mol⁻¹; n = 5.684 / 142.1 = 0.0400 mol [1]. Each formula unit contains 2 Na⁺ and 1 SO₄²⁻, so ions = 0.120 mol [1]. Number = 0.120 × 6.02 × 10²³ = 7.22 × 10²² ions [1] Examiner insight: Stopping at 2.41 × 10²² counts formula units, not ions, so the answer is wrong; read whether the question asks for ions, atoms or formula units.

4. (a) (0.60 × 10⁻³ g / 500 g) × 10⁶ = 1.2 ppm [1] (b) 0.0150 × 10⁴ = 150 ppm [1] (c) 2.00 m³ = 2000 dm³ [1]; 2000 × (150 / 10⁶) = 0.300 dm³ [1] Examiner insight: In (a), both masses must be in the same unit before dividing; mixing mg with kg gives an answer wrong by a factor of 10⁶.

5. M(Ba(OH)₂) = 171.3 g mol⁻¹; n = 4.282 / 171.3 = 0.0250 mol [1]. c = 0.0250 / 0.500 = 0.0500 mol dm⁻³ [1]. Mass concentration = 4.282 / 0.500 = 8.56 g dm⁻³ [1] Examiner insight: Dividing by 500 instead of 0.500 is the commonest slip; units are required on both answers.

6. Mass of C = 1.320 × (12.0 / 44.0) = 0.360 g [1]. Mass of H = 0.540 × (2.0 / 18.0) = 0.0600 g [1]. Mass of O = 0.580 − 0.360 − 0.0600 = 0.160 g [1]. Moles C : H : O = 0.0300 : 0.0600 : 0.0100 = 3 : 6 : 1, so empirical formula C₃H₆O [1]. Empirical mass 58.0; 116 / 58.0 = 2, so molecular formula C₆H₁₂O₂ [1] Examiner insight: Oxygen must be found by difference; using the oxygen in CO₂ and H₂O counts oxygen that came from the air.

7. M(Fe₂O₃) = 159.6; n = 8.00 / 159.6 = 0.0501 mol [1]. n(Al) = 2 × 0.0501 = 0.100 mol [1]. Mass of Al = 0.100 × 27.0 = 2.71 g [1]. n(Fe) = 0.100 mol; mass = 0.100 × 55.8 = 5.59 g [1] Examiner insight: Keep unrounded moles in the calculator; rounding 0.05013 to 0.05 gives 2.70 g and 5.58 g, which may not match the expected answer.

8. T = 363 K, p = 100 000 Pa, V = 94.9 × 10⁻⁶ m³ [1]. n = pV / RT = (100 000 × 94.9 × 10⁻⁶) / (8.31 × 363) = 3.146 × 10⁻³ mol [1]. M = 0.258 / 3.146 × 10⁻³ = 82.0 g mol⁻¹ [1]. Empirical formula mass of C₃H₅ = 41.0; 82.0 / 41.0 = 2 [1]. Molecular formula C₆H₁₀ [1] Examiner insight: Each unconverted quantity (°C, kPa or cm³) gives a wrong molar mass; show the converted values so an examiner can follow your method.

9. (a) n(ZnO) = 3.00 / 81.4 = 0.0369 mol [1]. M(ZnSO₄·7H₂O) = 287.5 g mol⁻¹ [1]. Theoretical mass = 0.0369 × 287.5 = 10.6 g [1]. Percentage yield = 7.85 / 10.6 × 100 = 74.1% [1] (b) Sum of product masses = (2 × 40.0) + 71.0 + 2.0 = 153.0 [1]. Atom economy = 71.0 / 153.0 × 100 = 46.4% [1] (c) There is only one product, so every reactant atom ends up in it [1]. Yield is lower because of practical losses, such as crystals left in the solution or lost during filtration [1]. Examiner insight: In (b), the 2 in front of NaOH must be used; including only one NaOH gives 62.8%, which is wrong.

10. (a) CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + CO₂(g) + H₂O(l): correct balanced species [1], all state symbols correct [1]. (b) n(CaCO₃) = 0.150 / 100.1 = 1.4985 × 10⁻³ mol [1]; n(CO₂) = n(CaCO₃) [1]. Molar volume = 34.0 / 1.4985 × 10⁻³ = 22 690 cm³ mol⁻¹ = 22.7 dm³ mol⁻¹ [1] (c) (24.0 − 22.69) / 24.0 × 100 = 5.5% [1] (5.4% if the rounded 22.7 is used) (d) Mass: 0.001 / 0.150 × 100 = 0.67% [1]. Volume: 0.5 / 34.0 × 100 = 1.5% [1] (e) The total uncertainty, about 2.1%, is smaller than the 5.5% difference, so there is a systematic error, such as gas escaping before the bung is fitted [1]. Improvement: put the solid in a small tube inside the sealed flask and tip it over, or use a larger mass with a larger syringe [1]. Examiner insight: An improvement must match the error you identified; “repeat the experiment” does not reduce a systematic error and is unlikely to gain credit.

11. (a) n(Al) = 0.135 / 27.0 = 0.00500 mol [1]; n(H₂) = 174 / 24 000 = 0.00725 mol [1] (b) Ratio H₂ : Al = 0.00725 / 0.00500 = 1.45, close to 1.5, so 2 : 3 [1]. 2Al(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂(g): balanced [1], state symbols [1] (c) Expected = 0.00750 × 24 000 = 180 cm³ [1]. Aluminium is coated with an oxide layer, so less metal is present than weighed and the start of the reaction is delayed [1]. Examiner insight: Round the ratio to the nearest simple ratio and say so; a ratio of 1.45 left as “about 1 : 1.45” does not confirm an equation.

Where marks are usually lost

• Counting formula units instead of ions or atoms in Avogadro constant questions.
• Not converting cm³ to dm³ (concentration) or to m³ (pV = nRT).
• Finding oxygen in combustion analysis from the products instead of by difference.
• Using the wrong coefficient when converting between moles of reactant and product.
• Leaving the multiplier off a product in the atom economy denominator.
• Observations written as product names rather than what is seen.
• Suggesting “repeat it” or “be more careful” as an improvement in practical questions.
• State symbols left out where they are requested.

Next steps

• Revision notes for this topic
• Study guide for this topic
• Atomic structure practice
• Edexcel IAL Chemistry course hub
• Printable checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017, Pearson Education Limited: Topic 1, Formulae, Equations and Amount of Substance, outcomes 1.1 to 1.12.