HKDSE Chemistry Practice Paper: Organic Chemistry And Polymers

HKDSE Chemistry Practice Paper: Organic Chemistry and Polymers

2026-11-03

HKDSE Chemistry: Organic Chemistry and Polymers

DSE Chemistry practice paper on organic chemistry and polymers: 12 questions, 100 marks, 75 minutes, with full answers and marking points for S4-S6 revision.

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Paper details　Level: S4-S6 | Subject: Chemistry | Time: 75 minutes | Marks: 100 | Questions: 12

Part A: Short questions (6 marks each)

Part A: Write formulae and equations, and balance every equation; show substituted values with units.

#	Question	Answer and marking notes
1	An alkane molecule contains 8 carbon atoms. (a) Write its molecular formula. (b) Calculate its relative molecular mass (C = 12, H = 1). (c) Write the balanced equation for its complete combustion. (6 marks)	(a) Alkanes follow CnH2n+2 with n = 8, so the formula is C8H18 (2). (b) Mr = 8 × 12 + 18 × 1 = 96 + 18 = 114 (2). (c) C8H18 + 12.5O2 → 8CO2 + 9H2O; the oxygen coefficient may also be written as 25/2, and atoms must balance on both sides (2).
2	0.44 g of a hydrocarbon burns completely to give 1.32 g of carbon dioxide and 0.72 g of water (C = 12, H = 1, O = 16). (a) Find the amount and mass of carbon in the compound. (b) Find the amount and mass of hydrogen. (c) Hence find its empirical formula. (6 marks)	(a) Amount of carbon dioxide = 1.32 ÷ 44 = 0.03 mol, so carbon is 0.03 mol and its mass is 0.03 × 12 = 0.36 g (2). (b) Amount of water = 0.72 ÷ 18 = 0.04 mol, and each water molecule has 2 hydrogen atoms, so hydrogen is 0.08 mol with mass 0.08 g (2). (c) The mole ratio of carbon to hydrogen is 0.03 : 0.08 = 3 : 8, so the empirical formula is C3H8; 0.36 + 0.08 = 0.44 g matches the sample mass, showing the compound contains only carbon and hydrogen (2).
3	2.8 g of ethene (C2H4) reacts completely with bromine water (C = 12, H = 1, Br = 80). (a) Find the amount of ethene. (b) State the type of reaction and write its equation. (c) Find the mass of bromine required. (6 marks)	(a) Amount of ethene = 2.8 ÷ 28 = 0.1 mol (2). (b) This is an addition reaction: C2H4 + Br2 → C2H4Br2, in which the carbon-carbon double bond opens and one bromine atom adds to each carbon (2). (c) The ratio is 1 : 1, so 0.1 mol of bromine is needed, giving 0.1 × 160 = 16 g (2).
4	0.1 mol of ethanol and 0.1 mol of ethanoic acid are heated with concentrated sulphuric acid catalyst, giving ethyl ethanoate in 68% yield (Mr of ethyl ethanoate = 88). (a) Find the amount expected from the theoretical yield. (b) Find the amount actually formed. (c) Find the mass of ethyl ethanoate formed. (6 marks)	(a) Ethanol and ethanoic acid react in a 1 : 1 ratio, so the theoretical yield is 0.1 mol of ethyl ethanoate (2). (b) The amount actually formed is 0.1 × 68% = 0.068 mol (2). (c) Mass = 0.068 × 88 = 5.984 g, which rounds to 5.98 g (2). Note that esterification is reversible, so the yield is never 100%.

Part B: Calculations (8 marks each)

Part B: List the known quantities and mole ratios, work step by step, and mind significant figures.

#	Question	Answer and marking notes
5	Polypropene is made by addition polymerisation of propene (C3H6). (a) Write the formula of the repeating unit and calculate its relative molecular mass. (b) A sample has a relative molecular mass of 126000; find its degree of polymerisation. (c) State whether this is addition or condensation polymerisation and give one reason. (8 marks)	(a) The repeating unit is C3H6, so Mr = 3 × 12 + 6 × 1 = 42 (3). (b) Degree of polymerisation = 126000 ÷ 42 = 3000 (3). (c) It is addition polymerisation: the carbon-carbon double bonds of propene open and link into a long chain, no small molecule such as water is released, and the repeating unit has the same atoms as the monomer (2).
6	The molar volume of a gas at room temperature and pressure is taken as 24 dm³/mol (C = 12, O = 16). (a) Find the amount of carbon dioxide in 4.4 g. (b) Find the volume it occupies at room temperature and pressure in dm³. (c) What would the volume be at 0 °C and 1 atm, where the molar volume is 22.4 dm³/mol? (8 marks)	(a) Amount = 4.4 ÷ 44 = 0.1 mol (3). (b) Volume = 0.1 × 24 = 2.4 dm³, which is 2400 cm³ (3). (c) At 0 °C and 1 atm the volume is 0.1 × 22.4 = 2.24 dm³, which is smaller because the lower temperature reduces the average kinetic energy of the molecules (2).
7	23 g of ethanol is oxidised by air over a catalyst to ethanoic acid (C = 12, H = 1, O = 16). (a) Find the amount of ethanol. (b) Write the relationship for the oxidation and state the mole ratio. (c) Assuming complete reaction, find the theoretical mass of ethanoic acid in g. (8 marks)	(a) Amount of ethanol = 23 ÷ 46 = 0.5 mol (3). (b) C2H5OH + [O] → CH3COOH, so the mole ratio of ethanol to ethanoic acid is 1 : 1 (2). (c) Theoretical mass = 0.5 × 60 = 30 g (3). A lower actual yield would indicate incomplete oxidation or ethanal not being oxidised further.
8	Butane (C4H10) has two structural isomers. (a) Name the two isomers and describe their structural difference, stating the number of isomers. (b) State which one has the higher boiling point and explain this in terms of intermolecular forces. (8 marks)	(a) The two isomers are butane, with a straight chain of four carbon atoms, and 2-methylpropane, with a three-carbon chain and a methyl group on the middle carbon; the number of isomers is 2 (3). (b) Butane has the higher boiling point (1) because its straight chain gives a larger contact area between molecules, so the van der Waals forces are stronger and more energy is needed to overcome them (4).

Part C: Extended questions (11 marks each)

Part C: Show the reasoning and the equations; answers without working score no marks.

#	Question	Answer and marking notes
9	Polyvinyl chloride (PVC) is made by polymerising chloroethene (C2H3Cl) (C = 12, H = 1, Cl = 35.5). (a) Write the formula of the repeating unit and calculate its relative molecular mass. (b) A PVC sample has a relative molecular mass of 125000; find its degree of polymerisation. (c) State whether the polymerisation is addition or condensation and whether any small molecule is released. (11 marks)	(a) The repeating unit is C2H3Cl, so Mr = 2 × 12 + 3 × 1 + 35.5 = 24 + 3 + 35.5 = 62.5 (4). (b) Degree of polymerisation = 125000 ÷ 62.5 = 2000 (4). (c) It is addition polymerisation with no small molecule released, and the repeating unit contains exactly the same atoms as the chloroethene monomer (3).
10	89 g of a fat of average relative molecular mass 890 is boiled with sodium hydroxide solution (Na = 23, O = 16, H = 1). (a) Find the amount of fat. (b) Find the amount and mass of sodium hydroxide needed. (c) Find the amount and mass of glycerol formed (Mr of glycerol = 92). (11 marks)	(a) Amount of fat = 89 ÷ 890 = 0.1 mol (3). (b) Each fat molecule has three ester linkages and needs 3 mol of sodium hydroxide, so 0.3 mol is required (2); mass = 0.3 × 40 = 12.0 g (2). (c) Glycerol formed is 0.1 mol (1), with mass 0.1 × 92 = 9.2 g (3). Saponification is the alkaline hydrolysis of an ester, giving soap and glycerol.
11	The standard enthalpy of combustion of ethanol is −1367 kJ/mol (C = 12, H = 1, O = 16). (a) Given that 4.6 g of ethanol releases 136.7 kJ on complete combustion, find its enthalpy of combustion. (b) 0.46 g of ethanol burns completely and only 80% of the heat is absorbed by 500 g of water; find the temperature rise (specific heat capacity of water = 4.2 J g⁻¹ K⁻¹). (11 marks)	(a) ΔHc = −136.7 ÷ 0.1 = −1367 kJ/mol, where the negative sign shows heat is released (4). (b) 0.46 g of ethanol is 0.01 mol and releases 13.67 kJ (2); the heat absorbed by the water is 0.8 × 13.67 = 10.94 kJ (2); from q = mcΔT, ΔT = 10936 ÷ (500 × 4.2) ≈ 5.21 K (3).
12	Three unlabelled colourless liquids are hexane, hex-1-ene and ethanol. (a) Describe how bromine water can identify hex-1-ene, giving the observation and the type of reaction. (b) Describe how acidified potassium dichromate solution can identify ethanol, giving the observation. (c) Describe one further test that confirms ethanol and its observation. (11 marks)	(a) Add bromine water and shake: hex-1-ene decolourises the orange solution while the other two do not; this is an addition reaction at the carbon-carbon double bond (4). (b) Add acidified potassium dichromate solution and warm gently: ethanol turns the orange solution green, which is the chromium(III) ion, while hexane shows no change; ethanol is oxidised to ethanal and then to ethanoic acid (4). (c) Add a small piece of sodium: ethanol reacts giving bubbles of hydrogen and sodium ethoxide, while hexane does not react (3). An infrared spectrum showing a broad O-H absorption would confirm ethanol further.

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