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Form 6 Chemistry Practice Paper: Organic Chemistry, Polymers and Analytical Chemistry
Form 6 Chemistry – 12 questions – 100 marks – 75 minutes
Form 6 Chemistry practice paper: 12 questions, 100 marks, 75 minutes, on alkanes, isomerism, addition, oxidation, esterification, polymers, analysis and titration.
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Paper details Form 6 Chemistry practice paper with 12 structured questions on organic reactions, polymers and analytical chemistry; includes a full marking scheme and two printable PDFs.
Section A: Hydrocarbons and Organic Reactions (Q1-Q4)
Answer all parts; show all working, steps and units in calculations.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | A member of the alkane homologous series contains 7 carbon atoms. (a) Write its molecular formula. (b) Calculate its relative molecular mass (C = 12.0, H = 1.0). (c) Write the general formula of the alkane series and state the mass difference between successive members. |
(a) C7H16. (b) 7 x 12.0 + 16 x 1.0 = 100.0. (c) The general formula is CnH2n+2; successive members differ by one CH2 unit, which is 14.0. Marks: formula, calculation, general formula and difference. |
| 2 | The molecular formula C5H12 has several structural isomers. (a) State the number of structural isomers of C5H12 and name two of them. (b) State the number of structural isomers of C4H9Cl. (c) Explain why the isomers of C5H12 have different boiling points. |
(a) There are 3 isomers: pentane and 2-methylbutane are two of them, and the third is 2,2-dimethylpropane. (b) There are 4 isomers. (c) More branching makes the molecule more spherical, so the contact area between molecules is smaller, the van der Waals forces are weaker and the boiling point is lower. Marks: number of isomers, names, C4H9Cl count, branching explanation. |
| 3 | 0.200 mol of butane (C4H10) undergoes complete combustion in excess oxygen. (a) Write the balanced equation. (b) Calculate the number of moles of oxygen needed. (c) At room temperature and pressure (molar volume 24.0 dm3 mol-1), calculate the volume of oxygen needed and the volume of carbon dioxide formed. |
(a) C4H10 + 6.5 O2 → 4 CO2 + 5 H2O. (b) 0.200 x 6.5 = 1.30 mol. (c) Volume of oxygen = 1.30 x 24.0 = 31.2 dm3; moles of carbon dioxide = 0.200 x 4 = 0.800 mol, volume = 19.2 dm3. Marks: balanced equation, moles of oxygen, both volumes. |
| 4 | 14.0 g of ethene reacts with excess bromine in the dark. (a) State the type of reaction and the systematic name of the product. (b) Calculate the mass of bromine needed (Br = 80.0). (c) Use conservation of mass to check the mass of the product. |
(a) It is an addition reaction and the product is 1,2-dibromoethane. (b) Moles of ethene = 14.0 / 28.0 = 0.500 mol; the same number of moles of bromine is used, so the mass is 0.500 x 160.0 = 80.0 g. (c) Mass of product = 0.500 x 188.0 = 94.0 g, and 14.0 + 80.0 = 94.0 g, which agrees with conservation of mass. Marks: reaction type and name, mass of bromine, mass check. |
Section B: Functional Group Conversion and Addition Polymerisation (Q5-Q8)
Balance every equation and state the reaction conditions.
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | Compare the halogenation of methane and ethene. (a) Describe the reaction of methane with chlorine in ultraviolet light; give the equation and the reaction type. (b) Describe the type of reaction between ethene and chlorine in the dark, and name the product. (c) Explain why the two reactions have different atom economies. |
(a) CH4 + Cl2 → CH3Cl + HCl under ultraviolet light; this is a substitution reaction. (b) Ethene undergoes an addition reaction with chlorine to give 1,2-dichloroethane. (c) In an addition reaction every atom of the reactants ends up in the product, so the atom economy is 100%, whereas substitution releases HCl as a by-product and also gives a mixture of chlorinated products, so the atom economy is lower. Marks: equation and type, addition reaction, atom economy explanation. |
| 6 | 0.250 mol of ethanol is completely oxidised by acidified potassium dichromate. (a) State the product of each of the two oxidation steps and the reagent needed. (b) Describe the colour change observed. (c) Calculate the mass of ethanoic acid finally formed (C = 12.0, H = 1.0, O = 16.0). |
(a) In the first step ethanol is oxidised to ethanal by acidified potassium dichromate; in the second step ethanal is oxidised further to ethanoic acid. (b) The solution turns from orange to green because of chromium(III) ions. (c) 0.250 x 60.0 = 15.0 g. Marks: two products, reagent, colour change, mass. |
| 7 | 0.400 mol of ethanoic acid reacts with excess ethanol using concentrated sulphuric acid as catalyst; the yield of the esterification is 75%. (a) Write the equation and name the product. (b) Calculate the number of moles and the mass of ester obtained (molar mass of the ester = 88.0 g mol-1). (c) State two ways to raise the yield and explain each in terms of equilibrium shift. |
(a) CH3COOH + C2H5OH = CH3COOC2H5 + H2O (reversible) with concentrated sulphuric acid; the product is ethyl ethanoate. (b) 0.400 x 0.75 = 0.300 mol; mass = 0.300 x 88.0 = 26.4 g. (c) First, distil off the ester or the water formed so that the equilibrium shifts to the right; second, use excess ethanol so that the concentration of a reactant is higher, which also shifts the equilibrium to the right. Marks: equation and name, moles and mass, two methods with explanations. |
| 8 | Polyvinyl chloride (PVC) is made by the addition polymerisation of chloroethene. (a) Write the structure and molecular formula of the repeat unit. (b) A sample of PVC has a relative molecular mass of 62500 and the repeat unit has a molar mass of 62.5 g mol-1; calculate the degree of polymerisation. (c) State the difference between addition polymerisation and condensation polymerisation in terms of by-products. |
(a) The repeat unit is -CH2-CHCl-, with molecular formula C2H3Cl. (b) Degree of polymerisation = 62500 / 62.5 = 1000. (c) Addition polymerisation gives no small-molecule by-product and all atoms go into the polymer, while condensation polymerisation releases one small molecule such as water or hydrogen chloride for each bond formed. Marks: repeat unit, degree of polymerisation, difference in by-products. |
Section C: Condensation Polymerisation, Instrumental Analysis and Titration (Q9-Q12)
Explanations must link structure to properties; keep significant figures consistent with the data.
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | Nylon-6,6 is made by the condensation polymerisation of adipic acid (molar mass 146.0 g mol-1) and hexane-1,6-diamine (molar mass 116.0 g mol-1). (a) Name the functional group in each monomer and the bond formed during polymerisation. (b) Given that two water molecules (molar mass 18.0 g mol-1) are eliminated per repeat unit, calculate the molar mass of the repeat unit. (c) Explain why this type of polymerisation needs two different monomers. |
(a) Adipic acid contains carboxyl groups and hexane-1,6-diamine contains amino groups; an amide bond is formed. (b) 146.0 + 116.0 – 2 x 18.0 = 226.0 g mol-1. (c) Each monomer has only two functional groups, so one kind of monomer alone cannot keep extending the chain; the two monomers react alternately to build a long chain, and the acid-amine reaction does not terminate by itself. Marks: functional groups and bond, molar mass, explanation. |
| 10 | An unknown organic compound X gives a molecular ion at m/z = 46 in its mass spectrum. Its infrared spectrum shows a broad absorption at 2500-3300 cm-1 and an absorption at 1050 cm-1, but no absorption near 1700 cm-1. (a) Identify the functional group present in X. (b) Suggest the structure of X. (c) Describe a chemical test to distinguish X from ethanoic acid. |
(a) A hydroxyl group (-OH) is present and there is no carbonyl group. (b) X is ethanol, C2H5OH, with relative molecular mass 46. (c) Add sodium hydrogencarbonate solution: ethanoic acid gives bubbles of carbon dioxide while ethanol does not; alternatively, acidified potassium dichromate turns from orange to green with ethanol. Marks: functional group, structure, test and observation. |
| 11 | 10.0 cm3 of white vinegar is diluted to 250 cm3. A 25.0 cm3 portion of the diluted solution is titrated with 0.100 mol dm-3 sodium hydroxide and the mean titre is 20.00 cm3. (a) Calculate the number of moles of sodium hydroxide used in each titration. (b) Calculate the number of moles of ethanoic acid in the original vinegar sample. (c) Calculate the concentration of ethanoic acid in the vinegar in grams per 100 cm3 (molar mass of ethanoic acid = 60.0 g mol-1). |
(a) 0.100 x 20.00 / 1000 = 0.00200 mol. (b) The dilution factor is 250 / 25.0 = 10, so the moles of ethanoic acid = 0.00200 x 10 = 0.0200 mol. (c) Mass = 0.0200 x 60.0 = 1.20 g, so the concentration is 1.20 / 10.0 x 100 = 12.0 g per 100 cm3. Marks: moles of sodium hydroxide, dilution factor and moles of acid, concentration. |
| 12 | Dealing with plastic waste depends on polymer structure. (a) Explain why condensation polymers are generally easier to recycle back to monomers than addition polymers. (b) Name the two monomers obtained when poly(ethylene terephthalate), PET, is completely hydrolysed. (c) Suggest two ways to reduce plastic waste and explain the scientific principle behind each. |
(a) The backbone of a condensation polymer contains ester or amide bonds that can be broken by hydrolysis or alcoholysis to recover the monomers, whereas the backbone of an addition polymer is a chain of carbon-carbon single bonds with high bond energy and no hydrolysable functional group, so it is harder to break down. (b) Ethane-1,2-diol and benzene-1,4-dicarboxylic acid (terephthalic acid). (c) First, recycle PET bottles into fibres, which reduces the use of raw materials and the amount sent to landfill; second, replace conventional plastics with biodegradable polymers whose ester bonds can be broken by micro-organisms, speeding up degradation. Marks: hydrolysis reason, two monomers, two methods with principles. |
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