The problems are on the crude oil and hydrocarbons worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Crude oil is a mixture of hydrocarbons separated by fractional distillation into fractions with different chain lengths and boiling points. These worked answers correct the 20 problems of the Chemistry 9-10 worksheet on crude oil and hydrocarbons, from hydrocarbon vocabulary to find the faulty statements: each solution shows the formula, the substitution and the final answer with its unit.
Every answer shows the method: the formula or rule, the substitution, and the final result in bold with its unit. Compare each step with your own working before checking the final number, because marks in an exam usually go to the method as well as to the answer.
Level: Chemistry grades 9-10 (GCSE · IGCSE · O Level · Class 9-10) · Curriculum: GCSE C7
Problem 1 Hydrocarbon vocabulary
1. Carbon and hydrogen.2. Saturated, single bonds.
3. CnH2n+2.
4. Fractional.
Problem 2 Alkane formulae
Use CnH2n+2.1. n = 6: H = 14. C₆H₁₄.
2. n = 11: H = 24. C₁₁H₂₄.
3. 2n + 2 = 42, so 2n = 40 and n = 20.
Problem 3 Spot the alkenes
1. An alkene has H = 2 × C: C₄H₈, C₃H₆ and C₆H₁₂.2. The formula fits CnH2n. The others fit CnH2n+2 and are alkanes.
3. C₈H₁₆.
Problem 4 Test for unsaturation
1. The mixture stays orange: alkanes do not react with bromine water.2. The orange colour disappears and the liquid becomes colourless: bromine adds across C=C.
3. The bottle that decolourised the water contains hex-1-ene.
Problem 5 Reading the column
1. A: refinery gases, the smallest molecules.2. F, the hottest part at the bottom.
3. E, fuel oil.
4. Petrol has shorter chains, so it is more flammable than diesel.
Problem 6 Complete combustion equations
1. Oxygen atoms needed: 8 × 2 + 10 = 26, so 13 O₂. 2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O.2. Oxygen atoms: 10 + 6 = 16, so 8 O₂. C₅H₁₂ + 8O₂ → 5CO₂ + 6H₂O.
3. One butane molecule needs 6.5 O₂, and half a molecule cannot be written in an equation, so every coefficient is doubled to whole numbers.
Problem 7 Trends along the series
1. Diesel has a higher boiling point, a higher viscosity and is harder to ignite.2. Longer molecules have stronger forces between them, so more energy is needed to separate them: the boiling point rises with chain length.
Problem 8 Balance the cracking
1. Carbon left 14 − 10 = 4, hydrogen left 30 − 22 = 8. The product is C₄H₈.2. It is an alkene (it fits CnH2n).
3. Carbon left 8, hydrogen left 16: C₁₆H₃₄ → C₈H₁₈ + C₈H₁₆.
Problem 9 Mass of carbon dioxide
1. C₃H₈ + 5O₂ → 3CO₂ + 4H₂O.2. M = 3 × 12 + 8 = 44 g/mol. n = 44 ÷ 44 = 1.0 mol.
3. CO₂ = 3 × 1.0 = 3.0 mol. M(CO₂) = 44 g/mol. Mass = 3.0 × 44 = 132 g.
Problem 10 Reading a molecule
1. Three carbons and six hydrogens: C₃H₆.2. Unsaturated: it contains a C=C double bond between carbons 2 and 3.
3. The bromine water changes from orange to colourless.
Problem 11 Carbon percentage and soot
1. M(CH₄) = 16. 12 ÷ 16 × 100 = 75.0 %.2. M(C₂H₆) = 30: 24 ÷ 30 × 100 = 80.0 %. M(C₂H₄) = 28: 24 ÷ 28 × 100 = 85.7 %.
3. Ethene has the highest percentage of carbon, so in limited oxygen more carbon is left over as soot.
Problem 12 Boiling points of alkanes
1. About −42 °C.2. Pentane boils at 36 °C, which is above 25 °C, so at 25 °C it is a liquid.
3. The boiling point rises as the chain gets longer, in smaller steps each time. Hexane: 36 + 33 = 69 °C (accept 65 to 70 °C).
Problem 13 Fuel tank conversion
1. 1 dm³ = 1000 cm³, so 45 dm³ = 45 000 cm³.2. Mass = density × volume = 0.74 × 45 000 = 33 300 g.
3. 33 300 ÷ 1000 = 33.3 kg.
Problem 14 Why fractions separate
1. Petrol vapour travels up until the temperature falls below its boiling point, then it condenses and is drawn off.2. Bitumen has a very high boiling point and very long molecules, so it condenses almost at once at the bottom.
3. The temperature gradient (hot at the bottom, cool at the top) makes each fraction condense at its own height.
Problem 15 Supply and demand
1. 18 % of 5000 = 900 t.2. Fuel oil = 0.24 × 5000 = 1200 t. Cracked = 600 t. Extra petrol = 0.40 × 600 = 240 t.
3. Increase = 240 ÷ 900 × 100 = 26.7 %.
Problem 16 Ethene from a big alkane
1. Carbon: 16 = 8 + 8. Hydrogen: 34 = 18 + 16. Balanced.2. M = 16 × 12 + 34 = 226.
3. 1000 g ÷ 226 = 4.425 mol. Ethene = 4 × 4.425 = 17.70 mol. Mass = 17.70 × 28 = 495.6 g, about 496 g.
Problem 17 Bromine addition by mass
1. M(C₂H₄) = 28; n = 2.8 ÷ 28 = 0.10 mol.2. 1 : 1 ratio, so 0.10 mol Br₂; M = 160; mass = 16.0 g.
3. M(C₂H₄Br₂) = 24 + 4 + 160 = 188; mass = 0.10 × 188 = 18.8 g. Reactants 2.8 + 16.0 = 18.8 g, equal to the product.
Problem 18 Design an identification test
1. Add a few drops of bromine water to a sample of each and shake. The sample that makes the orange colour disappear is hex-1-ene.2. Hexane does not mix with water: it stays as a separate layer on top of the orange water, while water mixes completely with the bromine water.
3. Hex-1-ene has a C=C double bond where bromine can add; hexane has only single bonds and is saturated.
Problem 19 Air needed for burning
1. C₃H₈ + 5O₂ → 3CO₂ + 4H₂O.2. Gas volumes are in the mole ratio 1 : 5, so O₂ = 5 × 2.0 = 10.0 dm³.
3. Air = 10.0 ÷ 0.21 = 47.6 dm³. CO₂ = 3 × 2.0 = 6.0 dm³.
Problem 20 Find the faulty statements
1. Statements (a), (c) and (d) are wrong; (b) is correct.2. (a) Alkenes are unsaturated because they contain a C=C double bond and fewer hydrogen atoms than the alkane. (c) Fractional distillation is a physical separation that breaks no bonds; cracking breaks them. (d) The fraction at the top has the lowest viscosity; bitumen at the bottom has the highest.
