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S6 Chemistry Organic Chemistry (Functional Groups), Paper 3: saponification, ester yields and isomer counting

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)

  1. FoundationWhat is the molar mass of propan-1-ol, C3H7OH, and how many moles are in 6.0 g?
  2. 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.
  3. FoundationMethanoic acid and ethanoic acid belong to the same carboxylic acid series. Give their molar masses and the difference.
  4. Foundation0.10 mol of ethanoic acid is neutralised with sodium hydroxide. How many moles and grams of alkali are needed?
  5. IntermediateHow many structural isomers does the ester C4H8O2 have, and what are they?
  6. IntermediateHow many structural isomers does the alcohol C4H10O have?
  7. 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.
  8. 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.
  9. 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.
  10. 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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