# Combination of polydisperse polymer mixtures

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

• What do you mean by "numbers of each batch"? – Karl Apr 6 '20 at 20:22
• 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. – Z. E. Apr 6 '20 at 20:30

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.

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

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

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