Polymers: Numerical Problems Set (1)

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Question

A sample of polyethylene has number-average molecular weight Mˉn=28000\bar{M}_n = 28000 g/mol. The repeating unit (CH2CH2-\text{CH}_2-\text{CH}_2-) has molar mass 2828 g/mol. Find the degree of polymerisation.

Solution — Step by Step

Degree of polymerisation, nn, is the number of repeating monomer units in a polymer chain.

n=MˉnMrepeating unitn = \frac{\bar{M}_n}{M_{\text{repeating unit}}}

n=2800028=1000n = \frac{28000}{28} = 1000

Final answer: Degree of polymerisation =1000= 1000.

Why This Works

The polymer chain is built by stringing together monomer units. The total molecular weight is approximately n×Mmonomern \times M_{\text{monomer}} (small end-group corrections matter for short chains but are negligible for n=1000n = 1000).

For polyethylene, each CH2CH2-\text{CH}_2-\text{CH}_2- comes from one ethylene molecule (CH2=CH2\text{CH}_2=\text{CH}_2, M=28M = 28). So nn also equals the number of ethylene molecules incorporated.

Alternative Method

If given the mass average Mˉw\bar{M}_w instead, divide by the same monomer mass to get the weight-average degree of polymerisation. The two are related by the polydispersity index Mˉw/Mˉn1\bar{M}_w / \bar{M}_n \geq 1.

Distinguish addition polymers (no atom loss, Mrepeating=MmonomerM_{\text{repeating}} = M_{\text{monomer}}) from condensation polymers (water or HCl lost, repeating unit smaller than monomer). Polyethylene is addition; nylon-6,6 is condensation.

Common Mistake

Students sometimes use molecular weight of ethylene gas including a leading CH2=CH2\text{CH}_2=\text{CH}_2 formula but compute it as 1414 instead of 2828. Each ethylene contributes 2 carbons and 4 hydrogens, total 24+4=2824 + 4 = 28 g/mol. Don’t undercount.

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