IGCSE Chemistry 0620 Paper 4 Practice Paper: Organic Chemistry, Polymers and Fuels (Extended Theory, 75 marks)

Level: Year 10-11 (IGCSE) | Subject: Chemistry 0620 | Time: 75 minutes | Total: 75 marks | Parts: 31

Paper at a glance

Paper	Time	Marks	Calculator
Paper 4 (Extended, structured)	1 h 15 min	75	Not allowed

Questions

Question 1 (14 marks) – Alkanes and alkenes: combustion of methane, cracking and polymer feedstock

(a) Write the balanced chemical equation for the complete combustion of methane. (2 marks)

(b) Calculate the relative molecular mass of methane (Ar: C 12, H 1) and hence the percentage by mass of carbon in methane. (2 marks)

(c) 8.0 g of methane is burnt completely. Calculate the amount, in mol, of methane burnt. (2 marks)

(d) At the same temperature and pressure (molar volume 24 dm3/mol), calculate the volume of methane used. (3 marks)

(e) Calculate the mass of carbon dioxide formed (Mr of CO2 = 44). (2 marks)

(f) An oil refinery cracks decane: C10H22 -> C8H18 + C2H4. Explain two industrial advantages of this reaction. (3 marks)

Question 2 (12 marks) – Ethanol: fermentation of glucose compared with hydration of ethene

(a) Write the balanced equation for the fermentation of glucose to ethanol by yeast. (2 marks)

(b) Calculate the relative molecular mass of glucose (Ar: C 12, H 1, O 16). (2 marks)

(c) 180 g of glucose is fermented completely. Calculate the amount and mass of ethanol formed (Mr of ethanol = 46). (3 marks)

(d) The experiment gives only 41.4 g of ethanol. Calculate the percentage yield. (2 marks)

(e) Compare fermentation and ethene hydration as routes to ethanol: give one raw material, the reaction conditions and a difference in rate for each. (3 marks)

Question 3 (14 marks) – Polymers: addition and condensation polymerisation

(a) Write the equation for the addition polymerisation of ethene to poly(ethene) using the repeat unit, and state whether any other product is formed. (2 marks)

(b) A sample of poly(ethene) has a relative molecular mass of 28000. Calculate its degree of polymerisation. (3 marks)

(c) Describe the features of addition polymerisation and condensation polymerisation, giving one example of each. (3 marks)

(d) Nylon-6,6 is made by condensation of hexane-1,6-diamine with hexanedioic acid. Name the small molecule eliminated and state the type of force between the polymer chains. (3 marks)

(e) Explain why polyalkenes are slow to degrade in the environment and suggest one method of dealing with waste polyalkene, with one limitation. (3 marks)

Question 4 (12 marks) – Fractional distillation of petroleum, combustion and energy calculations

(a) State how the boiling range of a fraction is related to chain length and to its position in the fractionating column. (2 marks)

(b) Give two reasons why crude oil is separated into fractions industrially. (2 marks)

(c) 2.0 g of a fuel burns completely and raises the temperature of 500 g of water by 20 degrees C (specific heat capacity of water 4.2 J/(g degree C)). Calculate the energy released and the energy per gram of fuel in kJ/g. (3 marks)

(d) State the products of complete and of incomplete combustion of a hydrocarbon, and give one hazard of incomplete combustion. (2 marks)

(e) Suggest two ways of reducing pollutants from combustion and explain the principle of each. (3 marks)

Question 5 (13 marks) – Carboxylic acids and esters: esterification, atom economy and yield

(a) Name the functional group in ethanoic acid and calculate its relative molecular mass (Ar: C 12, H 1, O 16). (2 marks)

(b) 6.0 g of ethanoic acid reacts with excess ethanol using concentrated sulfuric acid as catalyst. Calculate the amount, in mol, of ethanoic acid. (3 marks)

(c) Calculate the theoretical mass of ethyl ethanoate (Mr 88) and the atom economy of the reaction. (3 marks)

(d) Name the functional group of an ester and state one use of low-molecular-mass esters in the food industry. (2 marks)

(e) The experiment gives 7.04 g of ethyl ethanoate. Calculate the percentage yield and suggest one way of increasing it, with the principle involved. (3 marks)

Question 6 (10 marks) – Methanol: fuel cells and biofuels

(a) Calculate the relative molecular mass of methanol, CH3OH (Ar: C 12, H 1, O 16). (2 marks)

(b) What is the amount, in mol, of 3.2 g of methanol? (2 marks)

(c) This amount of methanol burns completely. Calculate the mass of carbon dioxide formed (Mr of CO2 = 44). (2 marks)

(d) Compare a methanol fuel cell with burning methanol directly, and give two advantages of the fuel cell. (2 marks)

(e) State one feedstock for bio-methanol and one limitation of using biofuels. (2 marks)

Answers and marking notes

Question 1 – Alkanes and alkenes: combustion of methane, cracking and polymer feedstock

(a) CH4 + 2O2 -> CO2 + 2H2O [2 marks]

(b) Mr = 12 + 4 x 1 = 16; carbon = 12 / 16 x 100 = 75%. [2 marks]

(c) 8.0 / 16 = 0.50 mol. [2 marks]

(d) 0.50 mol x 24 dm3/mol = 12 dm3 (one mark amount, one mark multiplication, one mark unit). [3 marks]

(e) 0.50 mol CO2 x 44 = 22.0 g. [2 marks]

(f) (1) The alkene (ethene) is a feedstock for polymers and has high value. (2) Short-chain alkanes are also formed and can be used in petrol, matching demand. (3) The products have wider uses than the long-chain alkane, improving profitability. [3 marks]

Question 2 – Ethanol: fermentation of glucose compared with hydration of ethene

(a) C6H12O6 -> 2C2H5OH + 2CO2 [2 marks]

(b) 6 x 12 + 12 x 1 + 6 x 16 = 180. [2 marks]

(c) 180 / 180 = 1.0 mol glucose -> 2.0 mol ethanol -> 2.0 x 46 = 92 g (one mark each step). [3 marks]

(d) 41.4 / 46 = 0.90 mol; 0.90 / 1.0 x 100 = 90%. [2 marks]

(e) Fermentation: sugar or starch feedstock, 30-40 degrees C at ordinary pressure with yeast, slow. Hydration: ethene from petroleum, 300 degrees C, 60 atm with phosphoric acid catalyst, fast. One mark for the comparison of rate. [3 marks]

Question 3 – Polymers: addition and condensation polymerisation

(a) n CH2=CH2 -> a polymer with the repeat unit -CH2-CH2-; no other product is formed because all atoms of the monomer enter the chain. [2 marks]

(b) Repeat unit Mr = 2 x 12 + 4 x 1 = 28; 28000 / 28 = 1000 repeat units (one mark for the unit). [3 marks]

(c) Addition: unsaturated monomers join with no small molecule eliminated, e.g. poly(ethene). Condensation: two functional groups react and a small molecule such as water is eliminated, e.g. nylon or a polyester. [3 marks]

(d) Water, H2O, is eliminated. Hydrogen bonding between amide groups on neighbouring chains gives high strength and a relatively high melting point. [3 marks]

(e) The C-C and C-H bonds are non-polar and there are no functional groups for microbes to attack, so biodegradation is very slow. Methods include melting and remoulding for recycling, or incineration with energy recovery; recycling needs sorting and washing, and incineration releases CO2. [3 marks]

Question 4 – Fractional distillation of petroleum, combustion and energy calculations

(a) Longer chains have stronger forces between molecules and higher boiling points, and are collected near the bottom; shorter chains have lower boiling points and are collected near the top. [2 marks]

(b) (1) The fractions have different boiling ranges, so they can be separated. (2) Each fraction has a different use, for example petrol as fuel and bitumen for roads, so separation makes them usable. [2 marks]

(c) E = 500 x 4.2 x 20 = 42000 J = 42 kJ (two marks); per gram = 42000 / 2.0 = 42000 J/g = 21 kJ/g (one mark). [3 marks]

(d) Complete: CO2 and H2O. Incomplete: CO, soot and H2O; CO is toxic because it binds to haemoglobin. [2 marks]

(e) (1) A catalytic converter on vehicles oxidises CO to CO2 and reduces nitrogen oxides to N2. (2) Using biofuels or hydrogen reduces net fossil carbon emissions. [3 marks]

Question 5 – Carboxylic acids and esters: esterification, atom economy and yield

(a) Carboxyl group, -COOH (one mark); Mr = 2 x 12 + 4 x 1 + 2 x 16 = 60 (one mark). [2 marks]

(b) 6.0 / 60 = 0.10 mol (two marks for the value, one for the unit). [3 marks]

(c) 0.10 x 88 = 8.8 g; atom economy = 88 / (60 + 46) x 100 = 83.0%. [3 marks]

(d) Ester group, -COO-; used as flavourings or fruit-ester additives. [2 marks]

(e) 7.04 / 8.8 x 100 = 80%; distilling off the ester or removing water shifts the equilibrium towards the products. [3 marks]

Question 6 – Methanol: fuel cells and biofuels

(a) 12 + 4 x 1 + 16 = 32. [2 marks]

(b) 3.2 / 32 = 0.10 mol. [2 marks]

(c) 0.10 mol CH3OH -> 0.10 mol CO2 -> 0.10 x 44 = 4.4 g. [2 marks]

(d) (1) Higher energy conversion efficiency, less energy wasted as heat. (2) No flame, it runs at lower temperature and the main product is water, so fewer pollutants. [2 marks]

(e) Feedstock: crop waste or woody biomass that is gasified. Limitation: it needs large areas of land and water and may compete with food production. [2 marks]

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