Question 6 of 8: Polyacrylonitrile Molecular Weight & Degree of Polymerization
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exam 04-BS-11, Properties of Materials — December 2015. 3 hours,
closed-book examination (approved Casio or Sharp calculator only). Candidates attempt five,
and only five, questions for a full paper: two from Section A, two from Section B, and the
fifth from either section. All eight questions are solved below for completeness.
Given. Six $(N_i,M_i)$ chain-count/mean-molecular-weight pairs (table above)
for polyacrylonitrile, repeat unit $-\text{CH}_2\text{-CH(CN)-}=\text{C}_3\text{H}_3\text{N}$.
Find. $\bar M_n$, $\bar M_w$, and the degree of polymerization $DP$ (based on
$\bar M_w$).
Approach
$\bar M_n$ weights each molecular-weight class by its number of chains (a simple
count-weighted average); $\bar M_w$ weights each class by the mass it
contributes, which biases the average toward the longer, heavier chains. The degree of
polymerization is the average molecular weight divided by the repeat unit's own molar mass.
Weight-average molecular weight.
$$\bar M_w=\frac{\sum N_iM_i^2}{\sum N_iM_i}.$$
$\sum N_iM_i^2=(10{,}000)(3000)^2+(18{,}000)(6000)^2+(17{,}000)(9000)^2+(15{,}000)(12{,}000)^2
+(9{,}000)(15{,}000)^2+(4{,}000)(18{,}000)^2=7.596\times10^{12}\ \text{g}^2$.
$$\bar M_w=\frac{7.596\times10^{12}}{678{,}000{,}000}=\boxed{11{,}204\ \text{g/mol}}.$$
(As expected, $\bar M_w>\bar M_n$, since the mass-weighting shifts the average toward the
higher-molecular-weight chains — the polydispersity index here is
$\bar M_w/\bar M_n=1.21$.)
Degree of polymerization. The repeat unit is
$-\text{CH}_2\text{-CH(CN)-}=\text{C}_3\text{H}_3\text{N}$:
$$M_0=3(12.01)+3(1.01)+1(14.01)=36.03+3.03+14.01=53.07\ \text{g/mol}.$$
$$DP=\frac{\bar M_w}{M_0}=\frac{11{,}204}{53.07}=\boxed{DP\approx211}.$$
