These are original Marlbridge practice questions, not past exam questions. All samples and data are invented for practice. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. Data for every question: C 12.01, O 16.00, Na 22.99, Mg 24.31, S 32.06, Cl 35.45, K 39.10, Ca 40.08, Fe 55.85, Cu 63.55 g mol⁻¹.
Question 1 (multiple choice · foundation)
Which of the following samples is a mixture?
- (A) Solid carbon dioxide, made only of CO₂ molecules
- (B) A gas made of N₂ molecules and O₂ molecules
- (C) Crystals of sodium chloride, made only of NaCl formula units
- (D) Liquid bromine, made only of Br₂ molecules
Answer and explanation
Answer: (B). A mixture contains two or more types of particle. Option (B) has two: N₂ molecules and O₂ molecules. Their proportions could change and it would still be “a mixture of nitrogen and oxygen”.
- (A) contains two elements, but every particle is an identical CO₂ molecule, so it is a pure compound.
- (C) is a pure ionic compound: one type of formula unit, even though it is made of two kinds of ion.
- (D) is a pure element. Bromine exists as Br₂ molecules, but they are all the same.
Question 2 (multiple choice · core)
A white powder is a mixture of magnesium oxide, MgO, and calcium oxide, CaO, only. Which of the following could be the mass percent of oxygen in the powder?
- (A) 25.0%
- (B) 34.0%
- (C) 42.0%
- (D) 68.2%
Answer and explanation
Answer: (B). Find the limits set by the two pure compounds.
- Pure MgO: 16.00 ÷ 40.31 × 100 = 39.69% O.
- Pure CaO: 16.00 ÷ 56.08 × 100 = 28.53% O.
Any mixture of only these two must lie between 28.53% and 39.69%. Only 34.0% is in that range.
- (A) is below the CaO value. A powder with this much less oxygen would need a third component with little or no oxygen.
- (C) is above the MgO value, which no mix of the two can reach.
- (D) adds the two percentages (39.69 + 28.53). Mass percents of a mixture are averaged (weighted by mass), never added.
Question 3 (multiple choice · core)
A 5.00 g sample of impure calcium carbonate, CaCO₃, is analysed and found to contain 1.80 g of calcium. The impurity contains no calcium. What is the percent by mass of CaCO₃ in the sample?
- (A) 36.0%
- (B) 40.0%
- (C) 89.9%
- (D) 111%
Answer and explanation
Answer: (C).
- Mass fraction of Ca in CaCO₃ = 40.08 ÷ 100.09 = 0.40044.
- Mass of CaCO₃ = 1.80 g ÷ 0.40044 = 4.495 g.
- Purity = 4.495 g ÷ 5.00 g × 100 = 89.9%.
- (A) is 1.80 ÷ 5.00 × 100: the mass percent of calcium in the sample, not of calcium carbonate.
- (B) is the mass percent of calcium in pure CaCO₃ (40.04%), which ignores the measured data.
- (D) multiplies by the mass fraction instead of dividing (5.00 × 0.40044 ÷ 1.80). A purity above 100% is impossible, which is a warning sign.
Question 4 (calculation · core)
A 3.000 g sample is a mixture of sodium sulfate, Na₂SO₄, and sodium chloride, NaCl. Elemental analysis shows that it contains 0.5000 g of sulfur.
(a) Calculate the percent by mass of Na₂SO₄ in the mixture. (b) Calculate the total mass of sodium in the sample.
Worked solution and suggested Marlbridge rubric
(a) Only Na₂SO₄ contains sulfur, so all 0.5000 g of S came from it.
- M(Na₂SO₄) = 2(22.99) + 32.06 + 4(16.00) = 142.04 g mol⁻¹.
- Mass fraction of S = 32.06 ÷ 142.04 = 0.22571.
- Mass of Na₂SO₄ = 0.5000 g ÷ 0.22571 = 2.2152 g.
- Percent by mass = 2.2152 ÷ 3.000 × 100 = 73.84%.
Mole route: n(S) = 0.5000 ÷ 32.06 = 0.015596 mol = n(Na₂SO₄); m = 0.015596 × 142.04 = 2.2152 g.
(b) Sodium comes from both salts.
- Mass of NaCl = 3.000 − 2.2152 = 0.7848 g.
- Na in Na₂SO₄ = 2.2152 g × (45.98 ÷ 142.04) = 0.7171 g.
- Na in NaCl = 0.7848 g × (22.99 ÷ 58.44) = 0.3087 g.
- Total Na = 0.7171 + 0.3087 = 1.026 g.
| Point | What earns it |
|---|---|
| 1 | Uses the sulfur mass to find the mass (or moles) of Na₂SO₄, with all S assigned to Na₂SO₄ |
| 1 | 73.84% Na₂SO₄ (accept 73.8%) |
| 1 | Finds the mass of NaCl by difference and recognises that both salts contain sodium |
| 1 | 1.026 g of sodium (accept 1.03 g), carrying forward the answer to (a) |
Common error: reporting 0.5000 ÷ 3.000 × 100 = 16.67% as the percent of Na₂SO₄. That is the percent of sulfur.
Question 5 (calculation · stretch)
A sample of black iron-oxide powder is a mixture of iron(II) oxide, FeO, and iron(III) oxide, Fe₂O₃, only. Analysis shows the powder is 74.00% iron by mass.
(a) Show that the powder cannot be pure FeO or pure Fe₂O₃. (b) Calculate the percent by mass of FeO in the powder.
Worked solution and suggested Marlbridge rubric
(a) M(FeO) = 55.85 + 16.00 = 71.85 g mol⁻¹, so FeO is 55.85 ÷ 71.85 × 100 = 77.73% Fe. M(Fe₂O₃) = 2(55.85) + 3(16.00) = 159.70 g mol⁻¹, so Fe₂O₃ is 111.70 ÷ 159.70 × 100 = 69.94% Fe. 74.00% matches neither fixed value, so the powder is not either pure compound. It lies between them, so a mixture of the two is possible.
(b) Let x be the mass fraction of FeO. Then (1 − x) is the mass fraction of Fe₂O₃.
0.77731x + 0.69944(1 − x) = 0.7400 0.07788x = 0.04056 x = 0.5209
The powder is 52.1% FeO by mass (and 47.9% Fe₂O₃).
Alternative: take a 100.0 g sample containing 74.00 g Fe and set up a two-equation system in moles, n(FeO) + 2n(Fe₂O₃) = 74.00 ÷ 55.85 and 71.85n(FeO) + 159.70n(Fe₂O₃) = 100.0. This gives the same 52.1 g of FeO.
| Point | What earns it |
|---|---|
| 1 | Both mass percents of iron calculated correctly |
| 1 | States that 74.00% matches neither pure value |
| 1 | Correct equation (mass or mole route) that accounts for iron from both oxides |
| 1 | 52.1% FeO (accept 52.0–52.2%) |
Check: 52.1% is just over half, and 74.00% is just over halfway from 69.94% to 77.73%, so the answer is sensible.
Question 6 (constructed response · core)
A technician has two bottles, P and Q, both labelled “copper(II) oxide, CuO”. She takes three samples from each bottle and measures the percent of copper by mass.
| Sample | Bottle P: % Cu | Bottle Q: % Cu |
|---|---|---|
| 1 | 79.9 | 75.2 |
| 2 | 79.8 | 77.0 |
| 3 | 79.9 | 73.9 |
(a) Calculate the percent by mass of copper in pure CuO. (b) Identify which bottle contains a mixture. Justify your answer with two pieces of evidence from the table. (c) Assume the impurity in bottle Q contains no copper. Calculate the percent of CuO in sample 1 from bottle Q.
Worked solution and suggested Marlbridge rubric
(a) M(CuO) = 63.55 + 16.00 = 79.55 g mol⁻¹. Percent Cu = 63.55 ÷ 79.55 × 100 = 79.89%.
(b) Bottle Q contains a mixture.
- Evidence 1: every Q value is below 79.89%, the fixed value for pure CuO. Something with less copper (or none) is present.
- Evidence 2: the Q values vary from sample to sample (73.9% to 77.0%). A pure compound has a fixed composition, so its samples agree, as bottle P’s do within 0.1%. Varying composition is a property of mixtures.
(c) Mass of CuO per 100 g of sample = 75.2 g ÷ 0.79887 = 94.13 g, so the sample is 94.1% CuO.
| Point | What earns it |
|---|---|
| 1 | 79.89% Cu in CuO (accept 79.9%) |
| 1 | Identifies Q and uses the fact that its values are lower than the pure value |
| 1 | Uses the variation between Q samples, linked to the idea that a pure compound has a fixed composition |
| 1 | 94.1% CuO |
Do not award the points in (b) for “Q is a mixture” with no evidence from the data. The variation in bottle Q also suggests the impurity is not spread evenly through the bottle.
Question 7 (explanation · stretch)
A sample sold as “potassium chloride, KCl” is analysed. The results give a mole ratio of potassium atoms to chlorine atoms of 1 : 1.12.
(a) Explain how this result shows that the sample is not pure KCl. (b) A student suggests the impurity is sand, which contains neither potassium nor chlorine. Evaluate this suggestion. (c) Describe what kind of substance the impurity could be.
Model answer and suggested Marlbridge rubric
(a) In pure KCl every formula unit contains one K and one Cl, so the ratio must be exactly 1 : 1 in any pure sample. A ratio of 1 : 1.12 means there is extra chlorine that did not come from KCl, so a second substance is present.
(b) The suggestion is not consistent with the data. Sand adds mass but contains neither K nor Cl, so it would leave the K : Cl ratio at 1 : 1. (Sand could lower the mass percent of each element, but it cannot change their mole ratio.) The sample may also contain sand, but sand alone cannot explain the result.
(c) The impurity must contain chlorine but less than one potassium per chlorine (usually no potassium at all), for example a chloride of another metal such as MgCl₂ or CaCl₂. Elemental analysis of K and Cl alone cannot say which one; analysing for other elements, such as Mg or Ca, would identify it.
| Point | What earns it |
|---|---|
| 1 | States that pure KCl has a fixed 1 : 1 ratio, so 1 : 1.12 needs a second chlorine-containing substance |
| 1 | Explains that sand contains neither element and so cannot change the K : Cl ratio |
| 1 | Identifies the impurity as a chlorine-containing substance without matching potassium, with a valid example or a valid follow-up analysis |
How did you do?
- Q1 wrong: re-read “Pure substances and mixtures at the particle level” and Figure 1 in the study guide.
- Q2 or Q5(a) wrong: check the “between the two pure values” idea in Figure 2 of the study guide.
- Q3, Q4(a) or Q6(c) wrong: practise Worked example 1; remember to divide the element mass by its mass fraction.
- Q4(b) or Q5(b) wrong: practise Worked example 2, where both components contain the element.
- Q6(b) or Q7 incomplete: your answer needs the why: a pure compound has a fixed composition and fixed atom ratios.
Then tick off the topic checklist.