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

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.

• I cannot see anything wrong with the maths of it. Why do you think it is not the correct answer? It would be nice, if you could cite the source of the exercise. Dec 5, 2017 at 6:29
• @Martin-マーチン It's part of a sample test for my course. They give the answer as 1032g. I've been trying to figure out how they got that for a while. Dec 5, 2017 at 6:50
• I am not an expert on polymer chemistry, so there might be something very obvious I'm missing. I cannot see anything wrong with your reasoning. Your question is still young so hopefully someone else will enlighten us. Dec 5, 2017 at 6:54
• @Mithoron Duh, now that you say it... would you be so kind and add a few words and post as an answer. Dec 5, 2017 at 13:52
• @Mithoron Is that a property of polymerization? According to the info they give us in the question, it's completely different. Dec 5, 2017 at 18:20

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.