Question 6 of 8: Polypropylene Molecular Weight Averages and Degree of Polymerization

Question 6: Polypropylene Molecular Weight Averages 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. Six chain-length groups (number of chains $N_i$, mean chain molecular weight $M_i$):

Find. Number-average $\bar M_n$; weight-average $\bar M_w$; degree of polymerization $DP$ (from $\bar M_w$).

Approach

Apply the standard defining sums for number-average and weight-average molecular weight directly to the tabulated groups, then divide the weight-average by the repeat-unit molar mass to get the (weight-average) degree of polymerization.

1. Sums. $\sum N_i=69{,}000$; $\sum N_iM_i=651{,}000{,}000$; $\sum N_iM_i^2=7.155\times10^{12}$ (all in g/mol, chain-count units).
2. Number-average molecular weight. $$\bar M_n=\frac{\sum N_iM_i}{\sum N_i}=\frac{651{,}000{,}000}{69{,}000}=\boxed{9435\ \text{g/mol}}.$$
3. Weight-average molecular weight. $$\bar M_w=\frac{\sum N_iM_i^2}{\sum N_iM_i}=\frac{7.155\times10^{12}}{651{,}000{,}000}=\boxed{10{,}991\ \text{g/mol}}.$$
4. Degree of polymerization. The propylene repeat unit is C$_3$H$_6$: $M_0=3(12.0)+6(1.0)=42.0$ g/mol, so $$DP_w=\frac{\bar M_w}{M_0}=\frac{10{,}991}{42.0}=\boxed{261.7}.$$