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Say I'm given a mixture of batches of polydisperse polymers. I know the weight $w$, the number average $ \bar M_n$, and the weight average $ \bar M_w$ of each batch that was added to the final mixture. How would I go about calculating the $\bar M_w, \bar M_n$ and PD of the final mixture?

I'm thinking that the $\bar M_w$ value would simply be a weighted average of each of the batches, but I'm not quite sure how to get the number average.

In math, I'm thinking that $\bar M_{w,final} = \frac{\Sigma_{i} w_i \bar M_{w,i}}{\Sigma_i w_i} $

I haven't been able to reason what I should do to account for the differing numbers of molecules/moles of each batch. I think it should also be a weighted average, but using the numbers of each batch for weighting instead of the weight, but how would I calculate that?

Thank you

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  • $\begingroup$ What do you mean by "numbers of each batch"? $\endgroup$
    – Karl
    Apr 6 '20 at 20:22
  • $\begingroup$ The way I think of number weighted average is molecular weight averaged by weighting by the numbers. It could be number of molecules or moles or any similar unit, I think. In this case Moles probably makes the most sense since that's how the averages are given. $\endgroup$
    – Z. E.
    Apr 6 '20 at 20:30
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It should not be difficult to derive either of the number-averaged molecular weight or weight-averaged molecular weight by their definitions. To avoid ambiguity, I will use $i$ as the index of each actual length of chains and $b$ as the index of batches.

First, the definitions:

$\bar{M}_\mathrm{n,final}=\frac{\Sigma_in_iw_i}{\Sigma_in_i}$

$\bar{M}_\mathrm{w,final}=\frac{\Sigma_in_iw_i^2}{\Sigma_in_iw_i}$ where $n$ and $w$ are number and molecular weight of each chain length.

The numerator of $\bar{M}_\mathrm{n,final}$ is, in fact, the total mass of the mixture. Hence, $\Sigma_in_iw_i=\Sigma_bw_b$. The denominator is the total number of chains. Hence, $\Sigma_in_i=\Sigma_bn_b=\Sigma_b\frac{w_b}{\bar{M}_\mathrm{n,b}}$.

It is slightly more complicated in the case of $\bar{M}_\mathrm{w,final}$ but you got it correct. Feel free to ignore my derivation. The numerator does not have a straightforward definition. It has to be derived from the $\bar{M}_\mathrm{w,b}$ instead. $\Sigma_in_iw_i^2=\Sigma_b\Sigma_{i,b}n_{i,b}w_{i,b}^2=\Sigma_b\bar{M}_\mathrm{w,b}w_b$. Since we have already defined the $\Sigma_in_iw_i$ above, we can have the final equation.

$\bar{M}_\mathrm{n,final}=\frac{\Sigma_bw_b}{\Sigma_b\frac{w_b}{\bar{M}_\mathrm{n,b}}}$

$\bar{M}_\mathrm{w,final}=\frac{\Sigma_b\bar{M}_\mathrm{w,b}w_b}{\Sigma_bw_b}$

$\bar{M}_\mathrm{w,final}/\bar{M}_\mathrm{n,final}=\frac{\Sigma_b\frac{w_b}{\bar{M}_\mathrm{n,b}}\Sigma_b\bar{M}_\mathrm{w,b}w_b}{(\Sigma_bw_b)^2}$

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