These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. All compounds and data are invented for practice unless a formula is named. Data for every question: H 1.008, C 12.01, N 14.01, O 16.00, S 32.06, Ca 40.08, Cu 63.55 g mol⁻¹.
Question 1 (multiple choice · foundation)
Which of these formulas is already an empirical formula?
- (A) C₂H₆
- (B) C₃H₈
- (C) N₂O₄
- (D) H₂O₂
Answer and explanation
Answer: (B). The subscripts 3 and 8 have no common factor other than 1, so C₃H₈ is already the lowest whole-number ratio.
- (A) C₂H₆ can be divided by 2, giving CH₃.
- (C) N₂O₄ can be divided by 2, giving NO₂.
- (D) H₂O₂ can be divided by 2, giving HO. A formula with equal subscripts is never empirical unless both are 1.
Question 2 (multiple choice · core)
A pure compound contains only elements X and Y. A 4.00 g sample of it contains 2.40 g of X. What mass of X is in a 10.0 g sample of the same pure compound?
- (A) 2.40 g
- (B) 4.00 g
- (C) 6.00 g
- (D) 15.0 g
Answer and explanation
Answer: (C). By the law of definite proportions, every pure sample has the same fraction of X by mass: 2.40 g ÷ 4.00 g = 0.600 (60.0%). In 10.0 g, the mass of X is 10.0 g × 0.600 = 6.00 g.
- (A) assumes the mass of X stays the same whatever the sample size.
- (B) is the mass of Y in 10.0 g (40.0% of 10.0 g), not the mass of X.
- (D) uses the X : Y ratio (2.40 ÷ 1.60 = 1.5) instead of the fraction of the whole. A part cannot have more mass than the 10.0 g sample.
Question 3 (multiple choice · core)
What is the percentage by mass of nitrogen in ammonium nitrate, NH₄NO₃?
- (A) 35.00%
- (B) 17.50%
- (C) 22.22%
- (D) 59.96%
Answer and explanation
Answer: (A). M(NH₄NO₃) = 2(14.01) + 4(1.008) + 3(16.00) = 80.052 g mol⁻¹. There are two N atoms in the formula (one in NH₄⁺, one in NO₃⁻), so % N = 28.02 ÷ 80.052 × 100 = 35.00%.
- (B) counts only one nitrogen atom.
- (C) is the fraction of atoms that are nitrogen (2 out of 9), not the fraction of mass.
- (D) is the percentage of oxygen.
Question 4 (calculation · core)
An invented food-acid analysis gives 40.91% carbon, 4.58% hydrogen and 54.51% oxygen by mass. Find the empirical formula. Show your working.
Worked solution
- Assume 100 g: 40.91 g C, 4.58 g H, 54.51 g O.
- Moles: C 40.91 ÷ 12.01 = 3.406 mol; H 4.58 ÷ 1.008 = 4.544 mol; O 54.51 ÷ 16.00 = 3.407 mol.
- Divide by the smallest (3.406): C 1.000; H 1.334; O 1.000.
- 1.334 is close to 1⅓, so multiply all three by 3: C 3.000, H 4.00, O 3.000.
Empirical formula: C₃H₄O₃.
Suggested mark points (3): 1 for converting all three masses to moles; 1 for dividing by the smallest to get 1 : 1.33 : 1; 1 for multiplying by 3 to reach C₃H₄O₃.
Common error: rounding 1.334 down to 1 gives CHO. Check it: CHO would be 41.39% C, which does not match. Accept equivalent methods, such as dividing by 3.407 instead of 3.406.
Question 5 (calculation · core)
A student heats 0.500 g of copper powder with excess sulfur. The unreacted sulfur is removed, leaving 0.626 g of a pure copper sulfide. Find the empirical formula of the copper sulfide.
Worked solution
- Mass of sulfur that combined: 0.626 g − 0.500 g = 0.126 g.
- Moles: Cu 0.500 ÷ 63.55 = 0.007868 mol; S 0.126 ÷ 32.06 = 0.003930 mol.
- Divide by the smallest (0.003930): Cu 2.002; S 1.000.
Empirical formula: Cu₂S.
Check: Cu₂S is 79.86% copper by mass; the sample is 0.500 ÷ 0.626 × 100 = 79.87% copper. The data fit.
Suggested mark points (3): 1 for the sulfur mass by subtraction; 1 for both mole values; 1 for the 2 : 1 ratio and the formula Cu₂S.
Common errors: using 0.626 g as the sulfur mass; dividing by the larger number of moles, which gives Cu : S = 1 : 0.4995, and then rounding 0.4995 up to 1 to write CuS. A ratio of 1 : 0.5 means multiply by 2, giving Cu₂S.
Question 6 (constructed response · core)
A lab receives two white powders, P and Q. Both are labelled “pure calcium carbonate, CaCO₃”. An analysis gives these invented results:
| Powder | Mass of sample | Mass of calcium found |
|---|---|---|
| P | 5.00 g | 2.00 g |
| Q | 4.00 g | 1.25 g |
(a) Calculate the percentage of calcium by mass in each powder. (b) Calculate the percentage of calcium by mass in pure CaCO₃. (c) Use the law of definite proportions to decide whether each powder could be pure CaCO₃. Justify your answer. (d) Write one testable scientific question that a follow-up experiment on powder Q could answer.
Worked solution and suggested Marlbridge rubric
(a) P: 2.00 ÷ 5.00 × 100 = 40.0%. Q: 1.25 ÷ 4.00 × 100 = 31.3% (31.25%).
(b) M(CaCO₃) = 40.08 + 12.01 + 3(16.00) = 100.09 g mol⁻¹. % Ca = 40.08 ÷ 100.09 × 100 = 40.04%.
(c) Every pure sample of CaCO₃ must have the same mass ratio of elements, so it must be 40.04% calcium whatever its size. Powder P (40.0%) matches within the precision of the data, so it could be pure CaCO₃. Powder Q (31.3%) does not match: 4.00 g of pure CaCO₃ would contain 1.60 g of calcium, not 1.25 g. So Q cannot be pure CaCO₃. It may be a different compound, or CaCO₃ mixed with something else.
(d) Any question that names something measurable, for example: “Does powder Q contain a second element that is not in CaCO₃?” or “Is the percentage of carbon in powder Q equal to 12.00%, the value for pure CaCO₃?”
| Point | What earns it |
|---|---|
| 1 | Both percentages in (a) correct |
| 1 | 40.04% for pure CaCO₃, with working |
| 1 | P could be pure CaCO₃ because its percentage matches the fixed value |
| 1 | Q cannot be pure CaCO₃, linked to the law: a pure compound always has the same composition |
| 1 | A question that could be answered by a measurement or observation |
Do not award the (d) point for a question that cannot be measured, such as “Is powder Q a good product?”. Accept carried-forward values from (a) and (b) in (c).
Question 7 (explanation · stretch)
Ethyne, C₂H₂, and benzene, C₆H₆, are different compounds.
(a) Show that both compounds have the same percentage of carbon by mass. (b) Explain why an analysis of percentage composition can give the empirical formula of a compound but cannot, by itself, tell you whether a sample is ethyne or benzene. (c) Sodium chloride is written NaCl. Explain why NaCl is described as a formula unit rather than a molecule.
Model answer and suggested Marlbridge rubric
(a) C₂H₂: M = 2(12.01) + 2(1.008) = 26.036 g mol⁻¹; % C = 24.02 ÷ 26.036 × 100 = 92.26%. C₆H₆: M = 78.108 g mol⁻¹; % C = 72.06 ÷ 78.108 × 100 = 92.26%. They are the same.
(b) Percentage composition is set by the ratio of atoms, not by how many atoms are in one molecule. Both compounds have a 1 : 1 ratio of C to H, so both have the empirical formula CH and give identical percentages. To tell them apart you need extra information that depends on the size of the molecule, such as the molar mass.
(c) Sodium chloride is a continuous lattice of Na⁺ and Cl⁻ ions. There is no separate NaCl particle. The formula only gives the fixed 1 : 1 ratio of the ions, so the smallest unit with that ratio is called a formula unit.
| Point | What earns it |
|---|---|
| 1 | Both percentages calculated correctly (92.26%) |
| 1 | Composition depends only on the atom ratio, which is the same (1 : 1, empirical formula CH) for both |
| 1 | States that extra data such as molar mass is needed to tell them apart |
| 1 | NaCl is a lattice of ions in a fixed ratio with no separate molecules |
How did you do?
- Q1 or Q7 wrong: re-read “The empirical formula” and Figure 1 in the study guide.
- Q2, Q3 or Q6 wrong: revisit “The law of definite proportions” and percentage by mass.
- Q4 or Q5 wrong: follow the four steps in Figure 2 and the whole-number table.
- Q6(d) unclear: see “Asking a testable question” in the study guide.
Then tick off the topic checklist.