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04-BS-11 · May 2017

Question 4 of 7: Molecular Weights of a Polypropylene Sample

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exam 04-BS-11, Properties of Materials — May 2017. 3 hours, closed-book examination (approved Casio or Sharp calculator only). Any five of the seven questions constitute a complete paper (only the first five in the answer book are marked); all questions are of equal value. All seven questions are solved below for completeness.

Reference texts: Callister & Rethwisch, Materials Science and Engineering: An Introduction, 9th ed. (dislocations and slip / Schmid’s law, mechanical behaviour, ionic crystal geometry and the radius-ratio rule, X-ray diffraction, diffusion and Fick’s laws, polymer molecular weight, ASTM grain size, strengthening and annealing, steel heat treatment and hardenability, casting defects, ceramics, glasses and composites).

Question 4: Molecular Weights of a Polypropylene Sample (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 groups $(N_i,M_i)$ as tabulated; polypropylene repeat unit $\text{C}_3\text{H}_6$ with $M_0=3(12.0)+6(1.0)=42$ g·mol$^{-1}$.

Find. Number-average $\bar M_n$, weight-average $\bar M_w$, and the degree of polymerization from $\bar M_w$.

Approach. Form the sums $\sum N_i$, $\sum N_iM_i$, and $\sum N_iM_i^2$, then apply the number- and weight-average definitions and divide $\bar M_w$ by the repeat-unit mass.

  1. Totals. $\sum N_i=69{,}000$; $\sum N_iM_i=6.51\times10^{8}$; $\sum N_iM_i^2=7.155\times10^{12}$.
  2. Number-average molecular weight. $$\bar M_n=\frac{\sum N_iM_i}{\sum N_i}=\frac{6.51\times10^{8}}{69{,}000}=\boxed{9435\ \text{g}\cdot\text{mol}^{-1}}.$$
  3. Weight-average molecular weight. $$\bar M_w=\frac{\sum N_iM_i^2}{\sum N_iM_i}=\frac{7.155\times10^{12}}{6.51\times10^{8}}=\boxed{10{,}991\ \text{g}\cdot\text{mol}^{-1}}.$$
  4. Degree of polymerization. Based on $\bar M_w$, $$\text{DP}=\frac{\bar M_w}{M_0}=\frac{10{,}991}{42}=\boxed{262}.$$

The two averages are worth reading against each other. Because $\bar M_w$ weights every chain by its own mass, it is pulled toward the long-chain groups: the 3,000 chains of 18,000 g·mol$^{-1}$ material are only 4.3% of the population but carry 8.3% of the mass, while the 5,000 shortest chains are 7.2% of the population and just 2.3% of the mass. The ratio $\bar M_w/\bar M_n=10{,}991/9435=1.16$ is the polydispersity index, and a value this close to unity says the distribution is narrow — a fairly uniform commercial polypropylene rather than a broadly distributed one. The distinction matters in practice because melt viscosity and processability track $\bar M_w$ while colligative properties (and the end-group count) track $\bar M_n$, so a resin specification normally quotes both. Note also that the degree of polymerization inherits whichever average it is computed from: the same sample gives $\text{DP}_n=9435/42=225$ on a number-average basis against the $262$ reported here, so the question’s instruction to work from $\bar M_w$ has to be followed literally.

QuantityResult
Number-average $\bar M_n$9,435 g·mol$^{-1}$
Weight-average $\bar M_w$10,991 g·mol$^{-1}$
Polydispersity $\bar M_w/\bar M_n$1.16
Degree of polymerization (from $\bar M_w$)262