Question 3 of 6: Molecular-Weight Averages and Variance from Fractionation

Question 3: Molecular-Weight Averages and Variance from Fractionation (20 points)

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 fractions with molecular weights M_i and mole (number) fractions x_i as above; monomer M₀ = 25 g/mol.

Find. M̄_n, M̄_w, the number-average variance σ_n², and PDI.

Approach. The mole fractions are number fractions, so M̄_n is their weighted mean, M̄_w weights by mass (i.e. by x_iM_i), the variance is the second central moment of the number distribution, and PDI is their ratio.

1. Number-average molecular weight is the mole-fraction-weighted mean:$$\bar M_n=\sum_i x_iM_i=0.1(10^4)+0.2(1.5\times10^4)+0.4(2\times10^4)+0.15(2.5\times10^4)+0.1(3\times10^4)+0.05(3.5\times10^4)=\boxed{20\,500\ \text{g/mol}}$$
2. Weight-average molecular weight weights each fraction by its mass, i.e. by x_iM_i:$$\bar M_w=\frac{\sum_i x_iM_i^2}{\sum_i x_iM_i}=\frac{4.60\times10^{8}}{20\,500}=\boxed{22\,439\ \text{g/mol}}$$
3. Number-average variance is the second central moment of the number distribution:$$\sigma_n^2=\sum_i x_i(M_i-\bar M_n)^2=\sum_i x_iM_i^2-\bar M_n^2=4.60\times10^{8}-(20\,500)^2=\boxed{3.975\times10^{7}\ (\text{g/mol})^2}$$i.e. a standard deviation σ_n = 6305 g/mol.
4. Polydispersity follows directly, and satisfies the exact identity $\bar M_w=\bar M_n+\sigma_n^2/\bar M_n$ (a useful check):$$\text{PDI}=\frac{\bar M_w}{\bar M_n}=\frac{22\,439}{20\,500}=\boxed{1.095}\qquad(20\,500+\tfrac{3.975\times10^7}{20\,500}=22\,439\ \checkmark)$$For reference the degree of polymerization is X̄_n = M̄_n/M₀ = 20 500/25 = 820.