Question 2 of 8: Identifying a polymer from its molecular weight and degree of polymerization

Question 2: Identifying a polymer from its molecular weight and degree of polymerization (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A single unknown polymer whose two measured chain statistics are to be matched against two candidate chemistries:

Find. Which of the two candidate polymers, polyvinyl chloride or polypropylene, is consistent with the measured pair of values.

Approach. The degree of polymerization is by definition the number of repeat units in an average chain, so dividing the average molecular weight by it returns the molar mass of one repeat unit; that single number is then compared with the repeat-unit formula weights of the two candidates.

1. Recover the repeat-unit molar mass from the definition of DP. The number-average molecular weight of a chain is the mass of its repeat unit multiplied by the number of repeat units it contains, $$M_n = \mathrm{DP}\times m_{\text{repeat}} \quad\Rightarrow\quad m_{\text{repeat}} = \frac{M_n}{\mathrm{DP}} = \frac{6250}{100}$$ $$\boxed{\ m_{\text{repeat}} = 62.50\ \text{g/mol}\ }$$Everything else in this question is arithmetic on the periodic table; this is the only measurement-derived quantity, and it is what the two candidates must be tested against.
2. Build the repeat unit of polyvinyl chloride. PVC polymerises from vinyl chloride, so its repeat unit is –CH₂–CHCl–, that is C₂H₃Cl. Summing the atomic weights supplied by the question, $$m_{\text{PVC}} = 2A_C + 3A_H + A_{Cl} = 2(12.01) + 3(1.01) + 35.45 = 24.02 + 3.03 + 35.45 = 62.50\ \text{g/mol}$$ which is exactly the value recovered in Step 1.
3. Build the repeat unit of polypropylene. PP polymerises from propylene, so its repeat unit is –CH₂–CH(CH₃)–, that is C₃H₆. Note that it contains no chlorine at all, which is why the question bothers to supply A_Cl: $$m_{\text{PP}} = 3A_C + 6A_H = 3(12.01) + 6(1.01) = 36.03 + 6.06 = 42.09\ \text{g/mol}$$
4. Compare and decide. The measured 62.50 g/mol agrees with PVC to the last digit and is 48.5 % heavier than a propylene repeat unit, a discrepancy far outside any measurement uncertainty. $$\boxed{\ \text{The polymer is polyvinyl chloride (PVC)}\ }$$
5. Cross-check the rejected candidate on its own terms. If the material really were polypropylene, a chain of molecular weight 6250 would have to contain $$\mathrm{DP}_{\text{PP}} = \frac{6250}{42.09} = 148.5\ \text{repeat units}$$ not the 100 that was measured. The two measurements are therefore mutually inconsistent for PP and mutually consistent for PVC, which is a stronger statement than matching one number alone.
6. Suggest an independent confirmation. Chlorine accounts for $$\frac{A_{Cl}}{m_{\text{PVC}}}\times 100 = \frac{35.45}{62.50}\times 100 = 56.72\ \%$$ of the mass of PVC and none of the mass of PP, so a simple elemental chlorine assay, or a Beilstein flame test, settles the identification without any molecular-weight measurement at all.