Condensation polymerisation (HT only) (chemistry only) - AQA GCSE Chemistry Exam Questions

Practise 22 exam-style questions on Condensation polymerisation (HT only) (chemistry only) for AQA GCSE Chemistry. Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.

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1–6 marks each
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20 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

A polyester is made from two different monomers.

Name the functional group present in each type of monomer, and state how many of these groups each monomer molecule has.

Model answer

One monomer is a diol, with –OH groups, and the other is a dicarboxylic acid, with –COOH groups. Each monomer molecule has two of its functional group, one at each end.

Explanation

The simplest condensation polymers are made from two different monomers, each with two of the same functional group. An –OH group reacts with a –COOH group to form an ester link, and because each monomer has two groups the chain can keep growing.

Mark scheme

1 mark
one monomer has –OH (alcohol / hydroxyl) groups and the other has –COOH (carboxylic acid) groups
both needed; allow diol and dicarboxylic acid
1 mark
each monomer has two of these groups (one at each end)

Question 2

2 marks

Explain why ethanediol, HO–CH₂CH₂–OH, cannot form a polyester on its own.

Model answer

Ethanediol has two functional groups, but both are –OH groups. An ester link only forms when an –OH group reacts with a –COOH group, so ethanediol needs a second monomer with two –COOH groups, such as hexanedioic acid.

Explanation

For condensation polymerisation the functional groups must be able to react with each other. Two identical alcohol groups cannot form an ester, which is why the simplest polyesters need two different monomers.

Mark scheme

1 mark
both of its functional groups are –OH (alcohol) groups
1 mark
an ester link needs an –OH group to react with a –COOH group / two –OH groups do not form an ester link
allow 'it needs a dicarboxylic acid to react with'

Question 3

3 marks

Lactic acid, CH₃CH(OH)COOH, can form a condensation polymer called poly(lactic acid). This polymer is used to make biodegradable packaging.

Explain how a single type of monomer, lactic acid, can form a condensation polymer.

Model answer

Each lactic acid molecule has two different functional groups: an –OH group and a –COOH group. The –OH group of one molecule can react with the –COOH group of another molecule, forming an ester link and releasing water. The product still has an –OH group at one end and a –COOH group at the other, so it can react with more lactic acid molecules, and a long polyester chain builds up.

Explanation

A condensation polymer needs monomers that can react at both ends. This can be two different monomers, each with two of the same group, or one monomer with two different groups that can react with each other. Amino acids polymerise in the same way.

Mark scheme

1 mark
lactic acid has two (different) functional groups: –OH and –COOH
1 mark
the –OH of one molecule reacts with the –COOH of another to form an ester link, losing water
1 mark
the product still has a free –OH and –COOH group, so it reacts with more molecules to form a long chain

Question 4

4 marks

PET is a polyester made from ethanediol, C₂H₆O₂, and benzene-1,4-dicarboxylic acid, C₈H₆O₄. Two molecules of water are lost for each repeating unit formed.

Calculate the relative formula mass (Mr) of the repeating unit of PET. Calculate the number of repeating units in a PET molecule with an Mr of 38 400.
Relative atomic masses (Ar): H = 1, C = 12, O = 16

Model answer

Mr of C₂H₆O₂ = (2 × 12) + (6 × 1) + (2 × 16) = 62
Mr of C₈H₆O₄ = (8 × 12) + (6 × 1) + (4 × 16) = 166
Mr of repeating unit = 62 + 166 − (2 × 18) = 192
number of repeating units = 38 400 ÷ 192 = 200

Explanation

The repeating unit is one of each monomer minus two water molecules (one for each ester link). Then divide the Mr of the whole polymer by the Mr of one repeating unit. Forgetting to subtract the water gives 228 and a wrong number of units.

Mark scheme

1 mark
Mr of C₂H₆O₂ = 62 and Mr of C₈H₆O₄ = 166
both needed
1 mark
Mr of repeating unit = 62 + 166 − (2 × 18)
allow ecf
1 mark
= 192
allow ecf
1 mark
number of repeating units = 38 400 ÷ 192 = 200
allow ecf; correct answers with no working score 4 marks

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