Ideal gas equation worksheet answers for A Level chemistry (free PDF download)

Problem 1 Moles of salt

1. n = 117 ÷ 58.5 = 2.00 mol.
2. N = 2.00 × 6.02 × 10²³ = 1.20 × 10²⁴ formula units.
3. m = 0.250 × 58.5 = 14.6 g.

Problem 2 Sodium hydroxide solution

1. n = c × V = 0.400 × 0.250 = 0.100 mol.
2. m = 0.100 × 40.0 = 4.00 g.

Problem 3 Unit conversions

1. 250 × 10⁻⁶ = 2.50 × 10⁻⁴ m³.
2. 101 × 10³ = 1.01 × 10⁵ Pa.
3. 27 + 273 = 300 K.
4. 2.40 × 10⁻³ = 2.40 × 10⁻³ m³.

Problem 4 Formula of an iron oxide

1. n(Fe) = 5.60 ÷ 56.0 = 0.100 mol; n(O) = 2.40 ÷ 16.0 = 0.150 mol.
2. Ratio Fe : O = 0.100 : 0.150 = 2 : 3.
3. The formula is Fe₂O₃.

Problem 5 Vitamin C in a tablet

1. M = 6 × 12.0 + 8 × 1.0 + 6 × 16.0 = 72.0 + 8.0 + 96.0 = 176.0 g mol⁻¹.
2. n = 0.500 g ÷ 176.0 g mol⁻¹ = 2.84 × 10⁻³ mol.

Problem 6 Why is atom economy 100 %?

1. Atom economy = 100 %.
2. There is only one product, so every reactant atom ends up in the ethanol. The mass of the desired product equals the total mass of products.

Problem 7 Formula of sodium sulfate

1. n(Na) = 32.4 ÷ 23.0 = 1.41 mol; n(S) = 22.6 ÷ 32.1 = 0.704 mol; n(O) = 45.0 ÷ 16.0 = 2.81 mol.
2. Dividing by 0.704: 2.00 : 1.00 : 4.00. The empirical formula is Na₂SO₄.

Problem 8 Carbon dioxide at room conditions

T = 298 K and p = 1.00 × 10⁵ Pa. V = nRT ÷ p = (0.0800 × 8.31 × 298) ÷ 100 000 = 1.98 × 10⁻³ m³ = 1.98 dm³.

Problem 9 Gas in a cylinder

1. T = 293 K; V = 2.40 × 10⁻³ m³. n = pV ÷ RT = (1.50 × 10⁵ × 2.40 × 10⁻³) ÷ (8.31 × 293) = 360 ÷ 2435 = 0.148 mol.
2. m = 0.1479 × 32.0 = 4.73 g.

Problem 10 Molar mass of a volatile liquid

1. V = 62.0 × 10⁻⁶ = 6.20 × 10⁻⁵ m³; p = 1.01 × 10⁵ Pa; T = 373 K.
2. n = pV ÷ RT = (1.01 × 10⁵ × 6.20 × 10⁻⁵) ÷ (8.31 × 373) = 6.262 ÷ 3099.6 = 2.02 × 10⁻³ mol.
3. M = m ÷ n = 0.150 ÷ 0.00202 = 74 g mol⁻¹.

Problem 11 Phosphorus oxide

1. 2 × 31.0 + 5 × 16.0 = 62.0 + 80.0 = 142.0.
2. 284 ÷ 142.0 = 2, so the molecular formula is P₄O₁₀.

Problem 12 Burning an organic liquid

1. n(CO₂) = 1.76 ÷ 44.0 = 0.0400 mol, so n(C) = 0.0400 mol. n(H₂O) = 0.900 ÷ 18.0 = 0.0500 mol, so n(H) = 0.100 mol.
2. m(C) = 0.0400 × 12.0 = 0.480 g; m(H) = 0.100 × 1.0 = 0.100 g; m(O) = 0.740 − 0.480 − 0.100 = 0.160 g, so n(O) = 0.0100 mol.
3. C : H : O = 0.0400 : 0.100 : 0.0100 = 4 : 10 : 1. The empirical formula is C₄H₁₀O (mass 74.0), and M = 74.0, so the molecular formula is C₄H₁₀O.

Problem 13 Atom economy in iron making

1. 2 × 55.8 = 111.6; 3 × 44.0 = 132.0; total = 243.6.
2. Atom economy = 111.6 ÷ 243.6 × 100 = 45.8 %.

Problem 14 Percentage yield of ethyl ethanoate

1. n(ethanol) = 4.60 ÷ 46.0 = 0.100 mol, so the maximum mass = 0.100 × 88.0 = 8.80 g.
2. Yield = 3.10 ÷ 8.80 × 100 = 35.2 %.

Problem 15 Hydrated copper sulfate

1. m(H₂O) = 2.50 − 1.60 = 0.90 g; n = 0.90 ÷ 18.0 = 0.0500 mol.
2. n(CuSO₄) = 1.60 ÷ 159.6 = 0.0100 mol.
3. 0.0500 ÷ 0.0100 = x = 5.

Problem 16 Reading a pressure graph

1. Gradient = 150 ÷ 2.0 = 75 kPa dm³. Since p = nRT × (1/V), the slope is nRT; 1 kPa × 1 dm³ = 10³ Pa × 10⁻³ m³ = 1 J, so it is 75 J.
2. n = 75 ÷ (8.31 × 300) = 75 ÷ 2493 = 0.0301 mol.
3. The gradient increases to nR × 350 = 87.5 J (it is proportional to T).

Problem 17 Identifying a gas from density

1. From pV = nRT with n = m/M and ρ = m/V: pV = (m/M)RT, so M = ρRT / p.
2. ρ = 1.94 g dm⁻³ = 1940 g m⁻³. M = (1940 × 8.31 × 298) ÷ 100 000 = 48.0 g mol⁻¹.
3. O₂ is 32.0, O₃ is 48.0 and N₂O is 44.0. The gas is ozone, O₃.

Problem 18 Gas from baking soda

1. n(NaHCO₃) = 4.20 ÷ 84.0 = 0.0500 mol.
2. 2 mol NaHCO₃ give 1 mol H₂O and 1 mol CO₂, so 0.0500 mol gives 0.0250 + 0.0250 = 0.0500 mol gas.
3. T = 473 K. V = nRT ÷ p = (0.0500 × 8.31 × 473) ÷ 100 000 = 1.97 × 10⁻³ m³ = 1.97 dm³.

Problem 19 Two routes to hydrogen

1. Route A: 3 × 2.0 = 6.0; total products = 28.0 + 6.0 = 34.0; economy = 6.0 ÷ 34.0 × 100 = 17.6 %. Route B: 4 × 2.0 = 8.0; total = 44.0 + 8.0 = 52.0; economy = 8.0 ÷ 52.0 × 100 = 15.4 %.
2. Route A has the higher atom economy because the waste product CO is lighter than CO₂. Atom economy ignores the yield, energy use, rate and the hazard of the by-products, so the better route also depends on those.

Problem 20 Hydrocarbon from gas data

1. n = pV ÷ RT = (1.00 × 10⁵ × 1.55 × 10⁻⁴) ÷ (8.31 × 373) = 5.00 × 10⁻³ mol; M = 0.280 ÷ 0.00500 = 56 g mol⁻¹.
2. In 100 g: C = 85.7 ÷ 12.0 = 7.14 mol; H = 14.3 ÷ 1.0 = 14.3 mol. Ratio 1 : 2, so the empirical formula is CH₂ (14.0).
3. 56.0 ÷ 14.0 = 4, so the molecular formula is C₄H₈.