Quantitative chemistry
50 exam-style questions with answers for 4.3 Quantitative chemistry in AQA GCSE Chemistry (spec 8462). Practise with flashcards, a topic quiz, a random quiz across the course, 5 crosswords, or read every question and answer below. Not official AQA material.
All 50 questions and answers
Question 1 · Spec 4.3.1.1 · Conservation of mass
State the law of conservation of mass. [1 mark]
Answer:No atoms are lost or made during a chemical reaction, so the mass of the products equals the mass of the reactants.Question 2 · Spec 4.3.1.1 · Balancing
Balance the equation: N₂ + H₂ → NH₃. [1 mark]
Answer:N₂ + 3H₂ → 2NH₃.Question 3 · Spec 4.3.1.1 · Balancing
Balance the equation: CH₄ + O₂ → CO₂ + H₂O. [1 mark]
Answer:CH₄ + 2O₂ → CO₂ + 2H₂O.Question 4 · Spec 4.3.1.2 · Relative formula mass
Calculate the Mr of CaCO₃. (Ca = 40, C = 12, O = 16) [1 mark]
Answer:40 + 12 + (3 × 16) = 100.Question 5 · Spec 4.3.1.2 · Relative formula mass
Calculate the Mr of Mg(OH)₂. (Mg = 24, O = 16, H = 1) [1 mark]
Answer:24 + 2 × (16 + 1) = 58.Question 6 · Spec 4.3.1.2 · Percentage mass
Calculate the percentage by mass of carbon in CO₂. (C = 12, O = 16) [2 marks]
Answer:Mr = 44 (1); 12 ÷ 44 × 100 = 27.3% (1).Question 7 · Spec 4.3.1.3 · Mass changes
Why does the mass decrease when magnesium carbonate is heated in an open tube? [1 mark]
Answer:Carbon dioxide gas is produced and escapes into the air.Question 8 · Spec 4.3.1.3 · Mass changes
Why does the mass increase when magnesium burns in air? [1 mark]
Answer:Magnesium combines with oxygen from the air, which was not weighed at the start.Question 9 · Spec 4.3.1.4 · Uncertainty
Readings of 24.1, 24.5 and 24.3 s were taken. Calculate the mean and its uncertainty. [2 marks]
Answer:Mean = 24.3 s (1); uncertainty = ± half the range = ± 0.2 s (1).Question 10 · Spec 4.3.2.1 · Moles · Higher only
What is the Avogadro constant? [1 mark]
Answer:6.02 × 10²³: the number of particles in one mole of a substance.Question 11 · Spec 4.3.2.1 · Moles · Higher only
Calculate the number of moles in 22 g of CO₂. (Mr = 44) [2 marks]
Answer:moles = mass ÷ Mr = 22 ÷ 44 (1); = 0.5 mol (1).Question 12 · Spec 4.3.2.1 · Moles · Higher only
Calculate the mass of 3 moles of water. (Mr = 18) [2 marks]
Answer:mass = moles × Mr = 3 × 18 (1); = 54 g (1).Question 13 · Spec 4.3.2.1 · Moles · Higher only
How many atoms are in 2 moles of helium? [1 mark]
Answer:2 × 6.02 × 10²³ = 1.204 × 10²⁴.Question 14 · Spec 4.3.2.2 · Reacting masses · Higher only
Calculate the mass of CaO made from 50 g of CaCO₃. CaCO₃ → CaO + CO₂ (Mr: CaCO₃ = 100, CaO = 56) [3 marks]
Answer:Moles CaCO₃ = 50 ÷ 100 = 0.5 mol (1); ratio 1:1, so 0.5 mol CaO (1); mass = 0.5 × 56 = 28 g (1).Question 15 · Spec 4.3.2.3 · Balancing with moles · Higher only
8 g of hydrogen (H₂) reacts with 64 g of oxygen (O₂) to make water. Use moles to write the balanced equation. (H₂ = 2, O₂ = 32, H₂O = 18) [3 marks]
Answer:Moles H₂ = 4, moles O₂ = 2, moles H₂O = 72 ÷ 18 = 4 (1); ratio 4:2:4 = 2:1:2 (1); 2H₂ + O₂ → 2H₂O (1).Question 16 · Spec 4.3.2.4 · Limiting reactants · Higher only
What is a limiting reactant? [1 mark]
Answer:The reactant that is completely used up, so it limits the amount of product formed.Question 17 · Spec 4.3.2.4 · Limiting reactants · Higher only
Why is one reactant often used in excess? [1 mark]
Answer:To make sure all of the other (more expensive) reactant reacts.Question 18 · Spec 4.3.2.5 · Concentration
Calculate the concentration of 10 g of salt dissolved in 0.25 dm³ of water in g/dm³. [2 marks]
Answer:Concentration = mass ÷ volume = 10 ÷ 0.25 (1); = 40 g/dm³ (1).Question 19 · Spec 4.3.2.5 · Units
Convert 500 cm³ to dm³. [1 mark]
Answer:0.5 dm³.Question 20 · Spec 4.3.2.5 · Concentration · Higher only
How would increasing the volume of water affect the concentration of a solution with the same mass of solute? [1 mark]
Answer:The concentration would decrease, as the same mass is spread through a larger volume.Question 21 · Spec 4.3.3.1 · Percentage yield
Why is the percentage yield often less than 100%? [3 marks]
Answer:Any three from: the reaction may not go to completion because it is reversible (1); some product may be lost when separated (1); some reactants may react in unexpected ways (1).Question 22 · Spec 4.3.3.1 · Percentage yield
A reaction makes 6 g of product. The maximum possible is 8 g. Calculate the percentage yield. [2 marks]
Answer:6 ÷ 8 × 100 (1); = 75% (1).Question 23 · Spec 4.3.3.1 · Theoretical yield · Higher only
Calculate the theoretical mass of MgO from 12 g of Mg. 2Mg + O₂ → 2MgO (Mg = 24, MgO = 40) [2 marks]
Answer:Moles Mg = 12 ÷ 24 = 0.5, so 0.5 mol MgO (1); mass = 0.5 × 40 = 20 g (1).Question 24 · Spec 4.3.3.2 · Atom economy
Define atom economy. [1 mark]
Answer:The percentage of the mass of reactants that ends up as the useful (desired) product.Question 25 · Spec 4.3.3.2 · Atom economy
Calculate the atom economy for making CaO: CaCO₃ → CaO + CO₂ (Mr: CaCO₃ = 100, CaO = 56) [2 marks]
Answer:56 ÷ 100 × 100 (1); = 56% (1).Question 26 · Spec 4.3.3.2 · Atom economy
Why is a high atom economy important? [2 marks]
Answer:It is better for sustainable development, as fewer resources are wasted (1); and it is more economical (1).Question 27 · Spec 4.3.3.2 · Reaction pathways · Higher only
Give three factors when choosing a reaction pathway. [3 marks]
Answer:Atom economy (1); yield (1); rate, equilibrium position and usefulness of by-products (1).Question 28 · Spec 4.3.4 · Concentration in mol/dm³ · Higher only
Calculate the concentration of 0.2 mol of NaOH in 500 cm³ of solution. [2 marks]
Answer:500 cm³ = 0.5 dm³ (1); 0.2 ÷ 0.5 = 0.4 mol/dm³ (1).Question 29 · Spec 4.3.4 · Titration calculation · Higher only
25.0 cm³ of NaOH is neutralised by 20.0 cm³ of 0.10 mol/dm³ HCl. Calculate the concentration of NaOH. [3 marks]
Answer:Moles HCl = 0.10 × 0.0200 = 0.0020 mol (1); NaOH + HCl reacts 1:1, so 0.0020 mol NaOH (1); 0.0020 ÷ 0.0250 = 0.080 mol/dm³ (1).Question 30 · Spec 4.3.4 · Concentration · Higher only
Convert 0.5 mol/dm³ of NaOH to g/dm³. (Mr = 40) [1 mark]
Answer:0.5 × 40 = 20 g/dm³.Question 31 · Spec 4.3.5 · Gas volumes · Higher only
What volume does one mole of any gas occupy at room temperature and pressure? [1 mark]
Answer:24 dm³.Question 32 · Spec 4.3.5 · Gas volumes · Higher only
Calculate the volume of 0.25 mol of CO₂ at room temperature and pressure. [1 mark]
Answer:0.25 × 24 = 6 dm³.Question 33 · Spec 4.3.5 · Gas volumes · Higher only
Calculate the volume of 8 g of oxygen (O₂ = 32) at room temperature and pressure. [2 marks]
Answer:Moles = 8 ÷ 32 = 0.25 mol (1); volume = 0.25 × 24 = 6 dm³ (1).Question 34 · Spec 4.3.1.2 · Relative formula mass
Calculate the Mr of H₂SO₄. (H = 1, S = 32, O = 16) [1 mark]
Answer:2 + 32 + 64 = 98.Question 35 · Spec 4.3.1.1 · Conservation of mass
12 g of magnesium reacts with 8 g of oxygen. What mass of magnesium oxide forms? [1 mark]
Answer:20 g.Question 36 · Spec 4.3.1.4 · Uncertainty
Why is there always some uncertainty in measurements? [1 mark]
Answer:The instruments and methods have limits, so readings vary.Question 37 · Spec 4.3.2.1 · Moles · Higher only
State the equation linking moles, mass and Mr. [1 mark]
Answer:Number of moles = mass (g) ÷ Mr.Question 38 · Spec 4.3.2.2 · Reacting masses · Higher only
What mass of hydrogen is needed to make 34 g of NH₃? N₂ + 3H₂ → 2NH₃ (NH₃ = 17, H₂ = 2) [3 marks]
Answer:Moles NH₃ = 34 ÷ 17 = 2 mol (1); moles H₂ = 2 × 3 ÷ 2 = 3 mol (1); mass = 3 × 2 = 6 g (1).Question 39 · Spec 4.3.2.4 · Limiting reactants · Higher only
0.5 mol of Mg reacts with 0.8 mol of HCl. Mg + 2HCl → MgCl₂ + H₂. Which is limiting? [2 marks]
Answer:0.5 mol Mg needs 1.0 mol HCl (1); only 0.8 mol HCl is present, so HCl is limiting (1).Question 40 · Spec 4.3.2.5 · Concentration
Calculate the mass of solute in 0.2 dm³ of a 50 g/dm³ solution. [1 mark]
Answer:50 × 0.2 = 10 g.Question 41 · Spec 4.3.3.1 · Percentage yield
State the equation for percentage yield. [1 mark]
Answer:Percentage yield = mass of product actually made ÷ maximum theoretical mass of product × 100.Question 42 · Spec 4.3.3.2 · Atom economy
What is the atom economy of a reaction with only one product? [1 mark]
Answer:100%.Question 43 · Spec 4.3.5 · Gas volumes · Higher only
Calculate the number of moles in 48 dm³ of hydrogen at room temperature and pressure. [1 mark]
Answer:48 ÷ 24 = 2 mol.Question 44 · Spec 4.3.1.2 · Percentage mass
Calculate the percentage by mass of oxygen in water (H₂O = 18, O = 16). [2 marks]
Answer:16 ÷ 18 (1); × 100 = 88.9% (1).Question 45 · Spec 4.3.1.3 · Mass changes
Explain why the mass stays the same when a reaction happens in a sealed flask. [1 mark]
Answer:No gases can escape or enter, so all atoms are kept in the flask.Question 46 · Spec 4.3.2.1 · Moles · Higher only
How many moles are in 9.2 g of sodium (Ar = 23)? [1 mark]
Answer:9.2 ÷ 23 = 0.4 mol.Question 47 · Spec 4.3.2.5 · Concentration
State the equation for concentration in g/dm³. [1 mark]
Answer:Concentration = mass of solute (g) ÷ volume of solution (dm³).Question 48 · Spec 4.3.3.1 · Percentage yield
What is the theoretical yield? [1 mark]
Answer:The maximum mass of product that could be made from the reactants.Question 49 · Spec 4.3.3.2 · Atom economy
State the equation for atom economy. [1 mark]
Answer:Atom economy = Mr of desired product ÷ sum of Mr of all reactants × 100.Question 50 · Spec 4.3.4 · Concentration · Higher only
Calculate the moles of solute in 25 cm³ of a 2.0 mol/dm³ solution. [1 mark]
Answer:2.0 × 0.025 = 0.05 mol.