CIE iGCSE Co-Ordinated Science C11.7 Polymers Exam Style Questions Paper 4
Question
Polymers are made from small molecules called monomers.
(a) The structure of a polymer is shown.
Draw the structure of its monomer.
▶️Answer/Explanation
Monomer identification:
1. Identify repeating unit in polymer chain
2. Remove extension bonds on either side
3. Example for polyethene: H2C=CH2 (ethene)
Key features:
– Double bond breaks to form single bonds with adjacent monomers
– Must show correct functional group (e.g., C=C for addition polymers)
– For condensation polymers, include two reactive groups (e.g., -OH and -COOH)
(b) Poly(ethene) is an addition polymer. Nylon is a condensation polymer.
Describe the differences between addition polymerisation and condensation polymerisation.
▶️Answer/Explanation
| Feature | Addition Polymerisation | Condensation Polymerisation |
|---|---|---|
| Monomer requirement | Unsaturated monomers (C=C) | Two functional groups per monomer |
| Byproducts | None | Small molecules (H2O, HCl) |
| Examples | Polyethene, PVC | Nylon, polyester |
| Process | Chain reaction | Step-growth |
Key distinction: Addition keeps all atoms from monomers, condensation loses atoms to byproducts.
(c) A mixture containing 3.9 g of ethene and 4.0 g of steam is allowed to react. Ethanol, \(C_{2}H_{6}O\), is made.
\(C_{2} H_{4} + H_{2}O \to C_{2}H_{6}O\)
Determine the limiting reactant in this reaction. Show your working and explain your answer.
[Ar : C, 12; H, 1; O, 16]
▶️Answer/Explanation
Calculation steps:
1. Moles of ethene: 3.9g ÷ 28 g/mol = 0.139 mol
2. Moles of water: 4.0g ÷ 18 g/mol = 0.222 mol
3. 1:1 stoichiometry (C2H4 + H2O → C2H5OH)
4. Ethene is limiting (fewer moles)
Theoretical yield:
– Max ethanol = 0.139 mol × 46 g/mol = 6.4g
– 0.083 mol H2O remains unreacted