Mass of monomer needed to obtain a polymer of a given molecular weight

Question

In a condensation polymerization between two different monomers, the average molecular weight of the product depends on the ratio of the monomers used. This question illustrates the effect of using different ratios of ethylenediamine ($\ce{A}$) and 1,4-butanoic acid chloride ($\ce{B}$):

The reaction of $\ce{A}$ ($\pu{2 mol}$) with $\ce{B}$ ($\pu{1 mol}$) gives a compound with molecular weight $\pu{202.3 g mol-1}$ as the major product.

The reaction of $\ce{A}$ ($\pu{1 mol}$) with $\ce{B}$ ($\pu{2 mol}$) gives a compound with molecular weight $\pu{297.1 g mol-1}$ as the major product.

What mass of $\ce{B}$ (in grams) should be combined with $\pu{400 g}$ of pure $\ce{A}$ in order to obtain a high molecular weight polymer?

This question is part of a sample test for my course. I calculated the answer as follows.
For the higher molecular weight polymer, we'll need a $1 : 2$ molar ratio between $\ce{A}$ and $\ce{B}$. That means we'll need $\pu{2067 g}$ of $\ce{B}$. However, this doesn't seem to be the right answer, which is given as $\pu{1032 g}$.
I'm really not sure what I'm doing wrong at this point.

Answer 1

You have a $2:1$ ratio. You need a $1:1$.

The $\bar{X}_m$ resulting (the number average degree of polymerisation) is derived from the Carothers equation:

$$\bar{X}_m = \frac{1+p}{1-p}$$

where $p$ is the mole ratio. For a $2:1$ this is $0.5$, and the effective $\bar{X}_m = 3$. If you mix at $2:1$, you'll have the $\ce{BAB}$ trimer as the major product with a $M_m$ of $\pu{297.1 g/mol}$.

If you have one component in $1\%$ excess ($1.01:1$) the $\bar{X}_m$ will be $199$, at $.1\%$ excess the $\bar{X}_m$ will be $1,999$ etc.