AS Chemistry: Atoms, Molecules and Stoichiometry — Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Stoichiometry revision notes

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Section A

1. State the four expressions used to calculate an amount in moles. [4]

2. Define relative atomic mass. [2]

3. Explain why relative atomic masses are rarely whole numbers. [2]

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Section B

4. A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Its relative molecular mass is 180.

(a) Determine the empirical formula. [3]

(b) Determine the molecular formula. [2]

5. 4.00 g of calcium carbonate is heated until it fully decomposes. CaCO₃ → CaO + CO₂ (A_r: Ca = 40.1, C = 12.0, O = 16.0)

(a) Calculate the moles of calcium carbonate. [2]

(b) Calculate the mass of calcium oxide formed. [2]

(c) Calculate the volume of carbon dioxide produced at room temperature and pressure. [2]

(d) The reaction is carried out in an open crucible. Explain what happens to the mass and why. [2]

6. 3.20 g of methane is burned in 16.0 g of oxygen. CH₄ + 2O₂ → CO₂ + 2H₂O

(a) Determine which reagent is limiting, showing your reasoning. [3]

(b) Calculate the maximum mass of carbon dioxide that can form. [3]

7. A sample of gas occupies 250 cm³ at a pressure of 150 kPa and a temperature of 25 °C. (R = 8.31 J K⁻¹ mol⁻¹)

(a) Convert each quantity into the units required by the ideal gas equation. [3]

(b) Calculate the amount, in moles, of gas present. [2]

8. Ethanol can be made by fermenting glucose: C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂ (M_r: glucose = 180, ethanol = 46.0, CO₂ = 44.0)

(a) Calculate the atom economy of this reaction with respect to ethanol. [3]

(b) 45.0 g of glucose is fermented and 18.4 g of ethanol is obtained. Calculate the percentage yield. [4]

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Answers

1. n = m ÷ M [1]; n = c × V [1]; n = V ÷ 24 (gas at RTP, V in dm³) [1]; n = N ÷ N_A [1].

2. The weighted mean mass of an atom of an element [1] compared with 1/12 the mass of a carbon-12 atom [1].

3. Elements exist as a mixture of isotopes of different masses [1], and the relative atomic mass is the weighted average allowing for their abundances [1].

4. (a) C: 40.0 ÷ 12.0 = 3.33; H: 6.7 ÷ 1.0 = 6.7; O: 53.3 ÷ 16.0 = 3.33 [1] Divide by smallest: 1 : 2.01 : 1 [1] Empirical formula = CH₂O [1].

(b) Empirical mass = 30 [1]. 180 ÷ 30 = 6, so molecular formula = C₆H₁₂O₆ [1].

5. (a) M_r(CaCO₃) = 100.1 [1]; n = 4.00 ÷ 100.1 = 0.0400 mol [1].

(b) n(CaO) = 0.0400 mol; M_r = 56.1 [1] m = 0.0400 × 56.1 = 2.24 g [1].

(c) n(CO₂) = 0.0400 mol [1]; V = 0.0400 × 24 = 0.96 dm³ [1].

(d) The mass decreases [1], because carbon dioxide gas escapes from the open crucible [1].

6. (a) n(CH₄) = 3.20 ÷ 16.0 = 0.200 mol [1]; n(O₂) = 16.0 ÷ 32.0 = 0.500 mol [1]. Divide by coefficients: CH₄ 0.200 ÷ 1 = 0.200; O₂ 0.500 ÷ 2 = 0.250. Methane is limiting [1].

(b) n(CO₂) = n(CH₄) = 0.200 mol [1]; M_r(CO₂) = 44.0 [1] m = 0.200 × 44.0 = 8.80 g [1].

7. (a) V = 250 ÷ 10⁶ = 2.50 × 10⁻⁴ m³ [1]; p = 150 × 1000 = 150,000 Pa [1]; T = 25 + 273 = 298 K [1]. (b) n = pV ÷ RT = (150,000 × 2.50 × 10⁻⁴) ÷ (8.31 × 298) [1] = 37.5 ÷ 2476 = 0.0151 mol [1].

8. (a) M_r of all products = (2 × 46.0) + (2 × 44.0) = 92.0 + 88.0 = 180 [1]. Atom economy = (92.0 ÷ 180) × 100 [1] = 51.1% [1]. (b) n(glucose) = 45.0 ÷ 180 = 0.250 mol [1]; theoretical n(ethanol) = 2 × 0.250 = 0.500 mol, so theoretical mass = 0.500 × 46.0 = 23.0 g [1]. Percentage yield = (18.4 ÷ 23.0) × 100 [1] = 80.0% [1].

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Where marks are usually lost

• Not dividing by the smallest value in an empirical formula calculation.
• Working from mass ratios rather than converting to moles first.
• Failing to divide by the stoichiometric coefficient when identifying the limiting reagent.
• Forgetting that gas volume at RTP uses 24 dm³ mol⁻¹.
• Using pV = nRT with pressure still in kPa or volume still in cm³ — both must be converted to Pa and m³ before substituting, or the answer is out by a factor of a thousand or a million.
• Comparing atom economy and percentage yield as if they measure the same thing — a reaction can have a poor atom economy (lots of by-product) yet still achieve a high percentage yield of the desired product from what the equation predicts, and vice versa.
• Forgetting to double the ethanol moles when the balanced equation gives two moles of product per mole of glucose — a very common one-mark slip in yield calculations with a coefficient other than 1.

Questions 7 and 8 draw on the ideal gas equation and percentage yield/atom economy sections of the Stoichiometry revision notes, material the earlier questions on this page don’t reach.