These eleven questions practise the whole chain in relative masses and amounts: formula mass, mass to moles, equation ratios, limiting quantities and unit checks. They are original and ordered from easier to harder. Cover each answer, write your working with units on every line, then open it.
Ar values used: H = 1, C = 12, N = 14, O = 16, Na = 23, Mg = 24, S = 32, Cl = 35.5, Ca = 40, Fe = 56, Zn = 65.
Section A: relative formula mass
1. Find the Mr of carbon dioxide, CO₂.
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C: 12. O: 2 × 16 = 32. Total: 12 + 32 = 44.
2. Find the Mr of calcium nitrate, Ca(NO₃)₂.
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The bracket NO₃ = 14 + 48 = 62, and two of them give 124. Then 40 + 124 = 164. Check: N = 2 × 14 = 28, O = 6 × 16 = 96, Ca = 40, and 28 + 96 + 40 = 164.
3. Find the Mr of hydrated magnesium sulfate, MgSO₄·7H₂O.
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MgSO₄ = 24 + 32 + 64 = 120. 7H₂O = 7 × 18 = 126. Total: 120 + 126 = 246.
Section B: mass and amount
4. How many moles are in 11.7 g of sodium chloride, NaCl?
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Mr = 23 + 35.5 = 58.5. n = 11.7 ÷ 58.5 = 0.200 mol. Check: 0.200 × 58.5 = 11.7 g.
5. What is the mass of 0.050 mol of sodium carbonate, Na₂CO₃?
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Mr = 46 + 12 + 48 = 106. m = 0.050 × 106 = 5.3 g.
6. A bag holds 1.5 kg of calcium carbonate, CaCO₃. How many moles is that?
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Convert: 1.5 kg = 1500 g. Mr = 40 + 12 + 48 = 100. n = 1500 ÷ 100 = 15 mol. Check: 15 × 100 = 1500 g.
Section C: equation ratios
7. In 2Mg + O₂ → 2MgO, what mass of MgO forms from 1.2 g of Mg? (MgO: Mr = 40)
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n(Mg) = 1.2 ÷ 24 = 0.050 mol. Ratio 2 : 2, so n(MgO) = 0.050 mol. Mass = 0.050 × 40 = 2.0 g.
8. In 2H₂O₂ → 2H₂O + O₂, how many moles of O₂ and what mass of O₂ form from 0.40 mol of H₂O₂ in the model?
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Ratio H₂O₂ : O₂ = 2 : 1, so n(O₂) = 0.40 ÷ 2 = 0.20 mol. Mr of O₂ = 32, so mass = 0.20 × 32 = 6.4 g.
Section D: limiting quantities
9. In 2Na + Cl₂ → 2NaCl, a model mixture has 0.30 mol of Na and 0.20 mol of Cl₂. Which is limiting? Find the mass of NaCl formed and the amount of excess reactant left.
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Na ÷ 2 = 0.15. Cl₂ ÷ 1 = 0.20. Na is limiting. n(NaCl) = 0.30 mol, Mr = 58.5, mass = 0.30 × 58.5 = 17.55 g, so 17.6 g. Cl₂ used = 0.15 mol, so 0.05 mol remains.
10. In Zn + 2HCl → ZnCl₂ + H₂, 6.5 g of Zn meets 0.20 mol of HCl in the model. Which is limiting, and what mass of ZnCl₂ forms? (ZnCl₂: Mr = 136)
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n(Zn) = 6.5 ÷ 65 = 0.10 mol. Zn ÷ 1 = 0.10. HCl ÷ 2 = 0.10. The values are equal, so neither is in excess: both are used up exactly. n(ZnCl₂) = 0.10 mol, mass = 0.10 × 136 = 13.6 g.
Section E: unit check
11. A student calculates the mass of 0.25 mol of CO₂ as 0.25 ÷ 44 = 0.0057 g. Find the error and give the correct answer.
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The operation is wrong: mol ÷ (g/mol) gives mol²/g, not g. The correct step is mol × (g/mol). m = 0.25 × 44 = 11 g. A size check agrees: 0.25 mol is a quarter of 44 g, which is 11 g.
If you got these wrong
| Error type | Questions | Go to |
|---|---|---|
| Brackets, hydrates, adding Ar | 1 to 3 | Calculate a relative formula mass |
| Dividing or multiplying the wrong way, kg and g | 4 to 6, 11 | Convert mass to amount |
| Ratio applied to masses, or the wrong coefficient | 7, 8 | Use an equation ratio |
| Comparing raw moles, missing excess | 9, 10 | Identify a limiting quantity |
| Answer has the wrong unit or size | 5, 6, 11 | Check by units |
Record the pattern, not just the question, in the mistake log and retest queue. The mole and equation-ratio tutor can write out the steps for a similar reaction, and the equation balance reasoning trainer helps with the equation itself.
When the same mistake keeps returning, it often helps to have a teacher ask you to explain your reasoning aloud. That is a routine part of online one-to-one Chemistry tuition.