Moles, reacting masses, limiting reactants & concentration in g/dm³ – GCSE Chemistry exam questions with mark scheme

1Grade 4–54 marks

A student dissolved 12.5 g of sodium hydroxide in water and made the solution up to 250 cm³.

(a) Calculate the concentration of the solution in g/dm³. [3]

(b) Calculate the mass of sodium hydroxide in 40.0 cm³ of this solution. [1]

(a) volume $= 250 \div 1000 = 0.250\ \text{dm}^3$(1)
(a) concentration $= 12.5 \div 0.250$(1)
(a) $= 50\ \text{g/dm}^3$ allow 50 with no working for 3 marks(1)
(b) $50 \times 0.0400 = 2.0\ \text{g}$ allow ecf from (a)(1)
Answer: (a) 50 g/dm³ (b) 2.0 g
Examiner tip: Convert cm³ to dm³ (÷ 1000) before dividing — 12.5 ÷ 250 = 0.05 is the classic wrong answer.

2Grade 6–7Higher3 marks

$A_r$ of Na = 23. The Avogadro constant is $6.02 \times 10^{23}$ per mole.

Calculate the number of atoms in 4.6 g of sodium. Give your answer in standard form to 2 significant figures. [3]

moles $= 4.6 \div 23 = 0.20$ mol(1)
atoms $= 0.20 \times 6.02 \times 10^{23}$ ($= 1.204 \times 10^{23}$)(1)
$1.2 \times 10^{23}$(1)
Answer: $1.2 \times 10^{23}$ atoms
Examiner tip: Final mark needs standard form AND 2 s.f. — 1.204 × 10²³ is not to 2 s.f.

3Grade 6–7Higher4 marks

Iron is extracted from iron(III) oxide in a blast furnace.

$$\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2$$

Relative formula mass ($M_r$): $\text{Fe}_2\text{O}_3$ = 160. $A_r$: Fe = 56

Calculate the maximum mass of iron, in kg, that can be produced from 800 kg of iron(III) oxide. [4]

moles $\text{Fe}_2\text{O}_3 = 800\,000 \div 160 = 5000$ mol allow 800 ÷ 160 = 5 (kmol); allow the mass-ratio method: 160 g Fe₂O₃ gives 2 × 56 = 112 g Fe, so 800 kg gives 800 × 112 ÷ 160 (scores marks 1–3)(1)
moles Fe $= 2 \times 5000 = 10\,000$ mol(1)
mass Fe $= 10\,000 \times 56 = 560\,000$ g(1)
= 560 kg allow 560 kg with no working for 4 marks; ecf throughout(1)
Answer: 560 kg
Examiner tip: Use the 1 : 2 mole ratio from the equation — forgetting to double gives 280 kg and loses the ratio mark (the later marks can still be given by ecf, but only if your working is shown).

4Grade 8–9Higher4 marks

A student burned 2.4 g of magnesium. It reacted with exactly 1.6 g of oxygen, $\text{O}_2$, to produce 4.0 g of magnesium oxide, MgO.

$A_r$: Mg = 24, O = 16

Use these masses to work out the balanced equation for the reaction. Show your working. [4]

moles Mg $= 2.4 \div 24 = 0.10$(1)
moles $\text{O}_2 = 1.6 \div 32 = 0.050$ and moles MgO $= 4.0 \div 40 = 0.10$(1)
ratio Mg : O₂ : MgO $= 0.10 : 0.050 : 0.10 = 2 : 1 : 2$(1)
$2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$(1)
Answer: 2Mg + O₂ → 2MgO
Examiner tip: Use $M_r$ = 32 for O₂, not 16 — the gas is diatomic.

5Grade 1–34 marks

A sports drink contains glucose dissolved in water. The concentration of glucose in drink A is 30 g/dm³.

(a) A bottle holds 500 cm³ of drink A.

Calculate the mass of glucose in the bottle. [2]

(b) A different drink, B, contains 18 g of glucose in 400 cm³.

Which drink, A or B, has the higher concentration of glucose?

Use a calculation to support your answer. [2]

(a) $500 \div 1000 = 0.5$ (dm³) or $30 \div 2$ allow $30 \times 0.5$(1)
(a) $15$ (g) allow 15 with no working for 2 marks(1)
(b) concentration of B $= 18 \div 0.4 = 45$ (g/dm³) allow mass of glucose in 400 cm³ of A $= 12$ g or mass in 500 cm³ of B $= 22.5$ g or any other valid like-for-like comparison(1)
(b) B (has the higher concentration) must be supported by a correct calculation; ecf from an incorrect calculated value only if the comparison is consistent with it(1)
Answer: (a) 15 g (b) B — 45 g/dm³, which is higher than 30 g/dm³
Examiner tip: Change cm³ to dm³ first: 30 × 500 = 15 000 g is the most common wrong answer, and in (b) a bare "B" with no calculation scores 0.

6Grade 4–5Higher4 marks

Calcium reacts with hydrochloric acid.

$$\text{Ca} + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2$$

A student reacts 0.25 mol of calcium with excess hydrochloric acid.

(a) Give the number of moles of hydrochloric acid that react with 0.25 mol of calcium. [1]

(b) Calculate the mass of 0.25 mol of calcium. $A_r$ of Ca = 40 [1]

Calculate the number of chloride ions in the calcium chloride produced.

The Avogadro constant is $6.02 \times 10^{23}$ per mole. Give your answer in standard form. [2]

(a) 0.50 (mol) allow 0.5(1)
(b) $0.25 \times 40 = 10$ (g)(1)
(c) moles of $\text{Cl}^-$ $= 2 \times 0.25 = 0.50$ (mol) may be implied by the final answer(1)
(c) $0.50 \times 6.02 \times 10^{23} = 3.01 \times 10^{23}$ allow $3.0 \times 10^{23}$ ; allow $1.505 \times 10^{23}$ / $1.5 \times 10^{23}$ for 1 mark only (moles of CaCl₂ used instead of moles of Cl⁻)(1)
Answer: (a) 0.50 mol (b) 10 g (c) $3.01 \times 10^{23}$ chloride ions
Examiner tip: Each formula unit of CaCl₂ contains two chloride ions, so double the moles before multiplying by the Avogadro constant.

7Grade 6–7Higher5 marks

A student heats a mixture of 5.6 g of iron powder and 4.0 g of sulfur powder. They react to form iron sulfide.

$$\text{Fe} + \text{S} \rightarrow \text{FeS}$$

$A_r$: Fe = 56, S = 32

(a) Show that iron is the limiting reactant and calculate the maximum mass of iron sulfide that can be made. [4]

(b) Calculate the mass of the reactant that is left over at the end of the reaction. [1]

(a) moles of Fe $= 5.6 \div 56 = 0.10$ (mol)(1)
(a) moles of S $= 4.0 \div 32 = 0.125$ (mol)(1)
(a) iron is limiting because the reactants react in a 1 : 1 ratio and there are fewer moles of iron (than sulfur) allow 0.10 mol Fe needs only 0.10 mol S or 4.0 g S would need 7.0 g Fe ; ignore "less mass of iron" without moles(1)
(a) $M_r$ of FeS $= 88$ and mass of FeS $= 0.10 \times 88 = 8.8$ (g) allow ecf from moles of Fe; do not accept 9.6 g (from adding the masses)(1)
(b) $(0.125 - 0.10) \times 32 = 0.80$ g of sulfur allow $4.0 - 3.2 = 0.8$ g ; must identify sulfur (or S) for the mark; allow ecf from (a)(1)
Answer: (a) Fe 0.10 mol, S 0.125 mol; 1 : 1 ratio so Fe is used up first; 8.8 g FeS (b) 0.80 g of sulfur
Examiner tip: Comparing masses (5.6 g vs 4.0 g) gets no credit — you must compare moles against the 1 : 1 ratio in the equation.

8Grade 8–9Higher7 marks

A student added different masses of magnesium powder to 100 cm³ portions of the same copper sulfate solution. For each portion, the student collected, dried and weighed the copper produced.

$$\text{Mg} + \text{CuSO}_4 \rightarrow \text{MgSO}_4 + \text{Cu}$$

The graph shows the results.

(a) Explain why the mass of copper produced stops increasing when more than about 0.5 g of magnesium is added. [2]

(b) Use the graph to find the mass of magnesium that reacts exactly with all the copper sulfate in 100 cm³ of solution. [1]

$A_r$ of Mg = 24. $M_r$ of $\text{CuSO}_4$ = 159.5 [4]

(a) (all) the copper sulfate has (reacted / been used up) allow copper sulfate is the limiting reactant(1)
(a) so the (extra) magnesium is in excess / has nothing left to react with allow no more copper ions to be displaced ; ignore "the reaction has finished" alone(1)
(b) 0.48 g allow 0.46–0.50 g(1)
(c) moles of Mg $= 0.48 \div 24 = 0.020$ (mol) allow ecf from (b)(1)
(c) moles of $\text{CuSO}_4$ $= 0.020$ (mol) (1 : 1 ratio)(1)
(c) mass of $\text{CuSO}_4$ in 100 cm³ $= 0.020 \times 159.5 = 3.19$ (g) allow 3.2(1)
(c) concentration $= 3.19 \div 0.100 = 31.9$ (g/dm³) allow 32; allow ecf from (b) — e.g. 0.50 g gives 33.2 g/dm³; 3.19 (mass in 100 cm³ not divided by the volume) or 0.0319 (divided by 100 cm³ instead of 0.100 dm³) scores 3 marks(1)
Answer: (a) The copper sulfate is all used up (limiting); the extra magnesium is in excess. (b) 0.48 g (c) 31.9 g/dm³
Examiner tip: Read the mass of magnesium where the two lines meet, not where the last point is plotted — then remember 100 cm³ is 0.100 dm³.

9Grade 8–9Higher4 marks

A Group 2 metal, M, burns in oxygen to form the oxide MO.

$$2\text{M} + \text{O}_2 \rightarrow 2\text{MO}$$

A student burned 2.00 g of M. The mass of MO produced was 2.80 g.

Calculate the relative atomic mass ($A_r$) of M and identify metal M.

$A_r$ of O = 16. You may use the periodic table. [4]

mass of oxygen $= 2.80 - 2.00 = 0.80$ (g)(1)
moles of $\text{O}_2$ $= 0.80 \div 32 = 0.025$ (mol) allow moles of O atoms $= 0.80 \div 16 = 0.050$(1)
moles of M $= 2 \times 0.025 = 0.050$ (mol) allow ecf(1)
$A_r = 2.00 \div 0.050 = 40$ and M is calcium allow ecf for identity only if the calculated $A_r$ matches a Group 2 metal; an answer of 80 (2 : 1 ratio not used) scores 2 marks(1)
Answer: $A_r = 40$, so M is calcium
Examiner tip: Use conservation of mass to find the oxygen first, then use the 2 : 1 ratio — forgetting to double the moles gives A_r = 80.

10Grade 6–7Higher6 marks

A student needs to make 250 cm³ of copper sulfate solution with a concentration of 40 g/dm³, using solid copper sulfate and distilled water.

Describe how the student should make the solution. Include a calculation of the mass of copper sulfate needed.

Explain how the student could change the method to make 250 cm³ of solution with a concentration of 80 g/dm³, and how using a larger volume of water with the same mass of solid would affect the concentration. [6]

Calculation: volume $= 250 \div 1000 = 0.250$ dm³(1)
Calculation: mass $= 40 \times 0.250 = 10$ g (10.0 g) of copper sulfate(1)
Method: weigh 10.0 g of copper sulfate using a (2 d.p.) balance, e.g. in a weighing boat / by difference(1)
Method: dissolve the solid in a smaller volume of distilled water (less than 250 cm³) in a beaker, stirring with a glass rod(1)
Method: transfer the solution to a 250 cm³ volumetric flask, rinsing the beaker and rod with distilled water and adding the washings to the flask(1)
Method: add distilled water until the bottom of the meniscus is on the line (the last few drops with a dropping pipette) and then stopper and invert to mix(1)
Explanation: concentration = mass of solute ÷ volume of solution(1)
Explanation: doubling the mass of solute (20 g) in the same volume (250 cm³) doubles the concentration to 80 g/dm³(1)
Explanation: the same mass in a larger volume gives a lower concentration (the solute particles are more spread out / concentration is inversely proportional to volume), e.g. 10 g in 500 cm³ = 20 g/dm³(1)
Ignore references to moles; do not credit adding 250 cm³ of water to the solid (final volume would not be 250 cm³)(1)

Level	Marks	What it looks like
Level 3	5–6	A clear, logically sequenced method that would produce an accurate 40 g/dm³ solution, including the correct mass of 10 g and making up to a final volume of 250 cm³. The effect of changing the mass and the volume on concentration is correctly explained using the relationship between mass, volume and concentration.
Level 2	3–4	A method with most of the main steps in a sensible order, which may contain an error or omission (e.g. dissolving in 250 cm³ of water rather than making up to 250 cm³), with a correct or nearly correct mass. Some correct explanation of how mass or volume affects concentration.
Level 1	1–2	Some relevant points, such as weighing the solid and dissolving it in water, or a correct calculation of the mass, or a simple statement that more solid gives a more concentrated solution. Logic and sequence are unclear.
No relevant content	0	No relevant content.

Answer: 10 g of copper sulfate; weigh, dissolve in less than 250 cm³ of distilled water, transfer with washings to a 250 cm³ volumetric flask, make up to the mark and mix. Use 20 g for 80 g/dm³ (double the mass in the same volume doubles the concentration); the same mass in a larger volume gives a lower concentration.
Examiner tip: Dissolving the solid in 250 cm³ of water is wrong — the solution must be made up TO a total volume of 250 cm³, otherwise the concentration will be lower than 40 g/dm³.

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