S6 Chemistry Organic Chemistry (Functional Groups), Paper 3: saponification, ester yields and isomer counting
- 2026-11-02
S6 Chemistry Organic Chemistry (Functional Groups), Paper 3: saponification, ester yields and isomer counting
Paper 3 covers the molar mass and combustion of propan-1-ol, the homologous series of carboxylic acids, neutralisation with alkali, the isomers of C4H8O2 and C4H10O, the saponification of ethyl ethanoate with a mass balance, a percentage yield and an alcohol combustion analysis.
Focus Saponification and esterification both run one to one, a yield compares actual with theoretical moles, and isomer questions should fix the functional group before splitting the carbons.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 25 |
| Difficulty mix | Foundation 4 · Intermediate 3 · Challenge 2 · Error redo 1 |
| Chapter | Chapter: S6 Chemistry: Organic Chemistry (Functional Groups) |
| Level | Four basic, three advanced, two challenge and one error-repair item |
02 Questions (10 items)
- FoundationWhat is the molar mass of propan-1-ol, C3H7OH, and how many moles are in 6.0 g?
- FoundationPropan-1-ol burns as C3H7OH plus 4.5O2 giving 3CO2 and 4H2O. For 0.10 mol, find the moles and mass of oxygen and the mass of carbon dioxide.
- FoundationMethanoic acid and ethanoic acid belong to the same carboxylic acid series. Give their molar masses and the difference.
- Foundation0.10 mol of ethanoic acid is neutralised with sodium hydroxide. How many moles and grams of alkali are needed?
- IntermediateHow many structural isomers does the ester C4H8O2 have, and what are they?
- IntermediateHow many structural isomers does the alcohol C4H10O have?
- Intermediate0.20 mol of ethyl ethanoate is boiled with sodium hydroxide until saponification is complete. Find the mass of alkali used and the masses of ethanol and sodium ethanoate formed.
- Challenge0.20 mol of ethanol and 0.20 mol of ethanoic acid give 0.15 mol of ester. Find the mass of ester and the percentage yield.
- ChallengeBurning 0.05 mol of a saturated alcohol gives 0.15 mol of carbon dioxide and 0.20 mol of water. Find the formula, a possible name and the mass of 0.05 mol.
- Error redoError repair: a student says the concentrated sulphuric acid in esterification serves only as a drying agent. What is wrong?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | 60.0 g per mole, so 0.10 mol | Three carbons give 36.0, eight hydrogens 8.0 and one oxygen 16.0, so 60.0 in total, and 6.0 g is 0.10 mol. |
| 2 | 0.45 mol or 14.4 g of oxygen, giving 13.2 g of carbon dioxide | The ratio is one to 4.5, so 0.45 mol of oxygen weighing 14.4 g is needed, and 0.30 mol of carbon dioxide weighs 13.2 g. |
| 3 | 46.0 and 60.0 g per mole, a difference of 14.0 | Neighbouring members differ by one CH2 unit, which weighs 12.0 plus 2.0, or 14.0. |
| 4 | 0.10 mol, which is 4.0 g | A monobasic acid neutralises a monobasic base one to one, and sodium hydroxide at 40.0 g per mole gives 4.0 g for 0.10 mol. |
| 5 | Three: propyl methanoate, ethyl ethanoate and methyl propanoate | An ester must contain a COO link, so the four carbons split into one plus three, two plus two and three plus one, giving three esters. |
| 6 | Four | The skeleton is either a straight chain, giving butan-1-ol and butan-2-ol, or branched, giving two methylpropanols, so there are four. |
| 7 | 8.0 g of sodium hydroxide, giving 9.2 g of ethanol and 16.4 g of sodium ethanoate | Saponification runs one to one, so 0.20 mol needs 8.0 g of alkali and gives 9.2 g of ethanol and 16.4 g of sodium ethanoate. |
| 8 | 13.2 g, a yield of 75 per cent | 0.15 times 88.0 gives 13.2 g, and 0.15 over the theoretical 0.20 is a yield of 75 per cent. |
| 9 | C3H8O, propan-1-ol or propan-2-ol, with a mass of 3.0 g | Carbon is three, hydrogen is eight and the alcohol adds one oxygen, so the molar mass is 60.0 and 0.05 mol weighs 3.0 g. |
| 10 | Wrong: it is the catalyst that makes the reaction fast enough, and removing water only helps to push the equilibrium right | Without proton catalysis the reaction is far too slow even with heat, so the acid acts first as a catalyst and only secondly as a dehydrating agent that shifts the equilibrium. |
04 What to do after marking
- After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
- Repeat the same item two days later; move to the next paper only when both attempts are correct.
- Parents only need the answer and note columns; there is no need to explain every item.
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