The problems are on the alcohols and carboxylic acids worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Alcohols contain –OH and carboxylic acids contain –COOH. These worked answers correct the 20 problems of the Chemistry 9-10 worksheet on alcohols and carboxylic acids, from functional group vocabulary to two routes to ethanol: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows the rule or formula, the substitution and the final result in bold with its unit. Compare your method with the model, not just the final number, because exam marks reward the steps as well as the result.
Level: Chemistry grades 9-10 (GCSE · IGCSE · O Level · Class 9-10) · Curriculum: GCSE C7
Problem 1 Functional group vocabulary
1. Hydroxyl, –OH (an alcohol).2. CH₃COOH, which has –COOH.
3. CH₃COOC₂H₅, which has the –COO– link.
4. C₃H₇OH is propanol.
Problem 2 Formulae from names
1. C₄H₉OH.2. C₂H₅COOH (three carbons including the one in –COOH).
3. n = 5 gives C₅H₁₁OH.
Problem 3 Naming esters
The alcohol gives the first word and the acid the second.1. Ethyl propanoate.
2. Methyl ethanoate.
3. Butyl methanoate.
Problem 4 Fermentation apparatus
1. Sugar solution with yeast.2. C.
3. It turns cloudy (milky) because carbon dioxide is formed.
4. 30 to 40 °C.
Problem 5 Polymer words
1. False: a polymer is the long chain; monomers are the small units.2. True.
3. True.
4. False: it is made from ethene.
Problem 6 Fermenting glucose
1. C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂.2. M = 180; n = 36 ÷ 180 = 0.20 mol.
3. Ethanol 0.40 mol × 46 = 18.4 g; CO₂ 0.40 mol × 44 = 17.6 g. Check: 18.4 + 17.6 = 36.0 g.
Problem 7 Ethanol as a fuel
1. C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O.2. M = 46; n = 23 ÷ 46 = 0.50 mol.
3. CO₂ = 1.0 mol × 44 = 44 g. H₂O = 1.5 mol × 18 = 27 g.
Problem 8 Ethanoic acid reactions
1. Already balanced: C 3 = 3, H 5 = 5, O 5 = 5. CH₃COOH + NaHCO₃ → CH₃COONa + H₂O + CO₂.2. Sodium ethanoate and carbon dioxide.
3. 2CH₃COOH + Mg → (CH₃COO)₂Mg + H₂. The magnesium dissolves and bubbles of hydrogen appear.
Problem 9 Vinegar and pH
1. Microbes in the air oxidise ethanol to ethanoic acid.2. Each pH unit is a factor of 10. pH 1 means 0.1 mol/dm³ H⁺ and pH 3 means 0.001 mol/dm³, so ethanoic acid has 100 times fewer hydrogen ions.
3. Hydrochloric acid is a strong acid that ionises completely; ethanoic acid is weak: only a small fraction ionises.
Problem 10 Reading an ester
1. The ester group, –COO–.2. Three carbons, six hydrogens and two oxygens: C₃H₆O₂.
3. The alkyl part on oxygen has one carbon, so the alcohol is methanol; the other part has two carbons, so the acid is ethanoic acid. Q is methyl ethanoate.
Problem 11 Poly(chloroethene) chain
1. The repeat unit is –CH₂–CHCl–, so X is chlorine.2. The monomer is CH₂=CHCl, chloroethene: a C=C bond with one chlorine and three hydrogens.
3. Addition polymerisation.
Problem 12 Choosing a plastic
1. Softening above 100 °C: high-density poly(ethene) and poly(propene).2. Poly(propene), which softens at 150 °C.
3. PVC, density above 1.0 g/cm³.
4. m = 1.4 × 500 = 700 g.
Problem 13 Mass of a polymer chain
1. 2 × 12 + 4 = 28.2. 28 × 1000 = 28 000.
3. Repeat unit = 24 + 3 + 35.5 = 62.5; 62.5 × 800 = 50 000.
Problem 14 Ethanol petrol blend
1. 10 % of 40 dm³ = 4.0 dm³.2. Mass = 0.79 × 4.0 = 3.16 kg.
3. 3160 g ÷ 46 = 68.7 mol.
Problem 15 Plastic waste
1. Poly(ethene) is a long chain with strong C–C and C–H bonds, so it is unreactive. Microbes cannot break it down, so it is not biodegradable.2. Recycle or reuse bags, and use biodegradable materials or fewer bags.
Problem 16 Yield of an ester
1. M(CH₃COOH) = 60; n = 6.0 ÷ 60 = 0.10 mol.2. M(ester) = 88; 0.10 × 88 = 8.8 g.
3. 5.9 ÷ 8.8 × 100 = 67 %. The reaction is reversible, so it does not go to completion, and some ester is lost in separating it.
Problem 17 A polyester chain
1. 100 monomers in a line have 99 links.2. 99 water molecules.
3. 50 × 62 + 50 × 166 = 3100 + 8300 = 11 400; water lost 99 × 18 = 1782; M = 11 400 − 1782 = 9618.
Problem 18 Composition of PVC
1. M = 62.5; Cl = 35.5 ÷ 62.5 × 100 = 56.8 %.2. C = 24 ÷ 62.5 × 100 = 38.4 %; H = 3 ÷ 62.5 × 100 = 4.8 %.
3. 56.8 + 38.4 + 4.8 = 100.0 %.
Problem 19 Find the errors
1. All four are wrong.2. (a) Fermentation is anaerobic: it needs no oxygen. (b) The alcohol gives the first word, so methanol and ethanoic acid make methyl ethanoate; ethyl methanoate comes from ethanol and methanoic acid. (c) The C=C bonds open during polymerisation, so poly(propene) has only single bonds and does not decolourise bromine water. (d) They are weak acids because they ionise only partly.
Problem 20 Two routes to ethanol
1. Fermentation: 2 × 46 ÷ 180 × 100 = 51.1 %. Hydration: 46 ÷ (28 + 18) × 100 = 100 %.2. Fermentation uses renewable sugar at low temperature; hydration is fast and gives pure ethanol with no waste product.
3. Fermentation, because sugar cane is local and renewable while ethene would have to be imported as oil.
