How can a saponification answer be checked?

The moles on both sides must match, so the mass of ester plus alkali equals the mass of alcohol plus carboxylate: 15.3 g plus 6.0 g and 6.9 g plus 14.4 g are both 21.3 g.

In what order should alcohol isomers be counted?

Group by the length of the main chain, then move the hydroxyl position along it, and add branches last, checking each structure is chemically possible.

S6 Chemistry Organic Chemistry (Functional Groups), Paper 4: dehydration, saponification balance and yields

2026-11-02

Tutor Prep Guide · Primary English

Paper 4 covers the molar mass and combustion of butan-1-ol, the homologous series of carboxylic acids, neutralisation with alkali, the eight isomers of C5H12O, the saponification of ethyl propanoate with a mass balance, a 72 per cent ester yield, an alcohol combustion analysis and the dehydration of butanol.

Focus　Saponification, esterification and dehydration all run one to one, a yield compares actual with theoretical moles, and isomer questions should fix the functional group before splitting the carbon skeleton.

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 butan-1-ol, C4H9OH, and how many moles are in 7.4 g?
FoundationButan-1-ol burns as C4H9OH plus 6O2 giving 4CO2 and 5H2O. For 0.10 mol, find the moles and mass of oxygen and the mass of carbon dioxide.
FoundationPropanoic acid and butanoic acid are neighbouring carboxylic acids. Give their molar masses and the difference.
Foundation0.25 mol of propanoic acid is fully neutralised with sodium hydroxide. How many moles and grams of alkali are needed?
IntermediateHow many structural isomers does the saturated alcohol C5H12O have, and what are they?
Intermediate0.15 mol of ethyl propanoate is boiled with sodium hydroxide until saponification is complete. Find the mass of alkali and the masses of ethanol and sodium propanoate formed.
Intermediate0.25 mol of propanoic acid and 0.25 mol of ethanol give 0.18 mol of ester. Find the mass of ester and the percentage yield.
ChallengeBurning 0.10 mol of a saturated alcohol gives 0.40 mol of carbon dioxide and 0.50 mol of water. Find the formula, a possible name and the mass of 0.10 mol.
Challenge0.25 mol of butan-1-ol is dehydrated to butene with a yield of 80 per cent. Find the theoretical and actual masses of butene.
Error redoError repair: a student says the carboxylic acid supplies the hydrogen and the alcohol supplies the hydroxyl group in esterification. What is wrong?

03　Answers and one-line notes

#	Answer	One-line note
1	74.0 g per mole, so 0.10 mol	Four carbons give 48.0, ten hydrogens 10.0 and one oxygen 16.0, so 74.0 in total, and 7.4 g is 0.10 mol.
2	0.60 mol of oxygen, which is 19.2 g, giving 17.6 g of carbon dioxide	The ratio is one to six, so 0.60 mol of oxygen weighing 19.2 g is needed, and 0.40 mol of carbon dioxide weighs 17.6 g.
3	74.0 and 88.0 g per mole, a difference of 14.0	Propanoic acid has three carbons, six hydrogens and two oxygens, giving 74.0, while butanoic acid has one extra CH2 group weighing 14.0, giving 88.0.
4	0.25 mol, which is 10.0 g	A monobasic acid neutralises a monobasic base one to one, and sodium hydroxide at 40.0 g per mole gives 10.0 g for 0.25 mol.
5	Eight isomers, counting the three straight-chain pentanols, four methylbutanols and 2,2-dimethylpropan-1-ol	A five-carbon chain gives three alcohols, a four-carbon chain with one methyl branch gives four, and a three-carbon chain with two methyls on one carbon gives the eighth.
6	6.0 g of sodium hydroxide, giving 6.9 g of ethanol and 14.4 g of sodium propanoate	Saponification runs one to one, so 0.15 mol needs 6.0 g of alkali and gives 6.9 g of ethanol and 14.4 g of sodium propanoate, both sides weighing 21.3 g.
7	18.36 g, a yield of 72 per cent	0.18 times 102.0 gives 18.36 g, and 0.18 over the theoretical 0.25 is a yield of 72 per cent.
8	C4H10O, butan-1-ol or butan-2-ol, with a mass of 7.4 g	Carbon is four and hydrogen is ten, and a saturated alcohol adds one oxygen, so the molar mass is 74.0 and 0.10 mol weighs 7.4 g.
9	14.0 g in theory and 11.2 g in practice	Dehydration is one to one, so 0.25 mol of butene at 56.0 g per mole is 14.0 g in theory, and eighty per cent of that is 11.2 g.
10	Wrong: the acid loses OH and the alcohol loses H, so the water is assembled from both and the ester oxygen comes from the acid	If the acid lost only a hydrogen, the ester oxygen would come from the alcohol, which contradicts isotope labelling; the acid loses OH and the alcohol loses H.

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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