Question

Calculate the Mn and Mw of a polyester that would be produced from the polymerization of an equimolar mixture of dimethyl terephthalate and ethylene glycol with an extent of conversion of reaction of 0.970

Answer 1

Reaction of dimethyl terephthalate and ethylene glycol gives rise to transesterification reaction. The product obtained is hydroxy ethoxy.

Now, 1 mole of dimethyl terephthalate and 1 mole of ethylene glycol react to give 1 mole of hydroxy ethoxy.

Conversion of reaction is 0.97

So, moles of hydroxy ethoxy formed = 0.97 * 1 mole = 0.97 mole

Mw = Weight average molecular weight = sum (Mi * Ci)/sum(Ci)

where Mi is molecular weight of fraction i and Ci is concentration of fraction i

Here, Mi = molecular weight of hydroxy ethoxy = 224.25

Ci = 0.97

The other product formed is methyl alcohol with molecular weight 32

Therefore, Mw = (224.25 * 0.97 + 32 * 0.03)/(0.97 + 0.03)

= 218.4825

Mn = Number average molecular weight

= sum (Ci)/(sum(Ci/Mi)

= (0.97 +0.03)/(0.97/224.25 + 0.03/32)

= 190