MW

You need some basis for calculating this. In cannot be done in a vacuum.

For example, if you are polymerizing it, you can make certain assumptions about the polymerization conditions and calculate a MW.

John Aspen Research, -

Opinions expressed herein are my own and may not represent those of my employer.

Dear John,

I=B4ve some basis indeed, problem is how to use them.

I charged the vessel with 1362 kg of phtalic anhidride + 1637,47 kg of DEG. The result was 2645 kg of polyester with OH number of 160 mgKOH/ g and acidity index of 1.

How to calculate de Mw ?

Regards,

rodrigo

Think of the mg KOH/g of polymer as g of KOH/1000 g of polymer.

The total number of ends then is 2.869 per 1000 g of polymer.

Linear polymers have two ends per chain, therefore there are 2.869/2=1.435 chains per 1000 g. Thus, on the average each chain has a mw of 1000/1.435 = 697 g/mole.

If one assumes that all of the mass loss during polymerization was water and DEG than your mass balance would imply a MW of around 2600 g/mole. That the end measurements indicate 697 g/mole mw, suggest that some phthalic compound was lost as well.

Ernie

I´ve some basis indeed, problem is how to use them.

I charged the vessel with 1362 kg of phtalic anhidride + 1637,47 kg of DEG. The result was 2645 kg of polyester with OH number of 160 mgKOH/ g and acidity index of 1.

How to calculate de Mw ?

Regards,

rodrigo

Dear Ernie, Thank You.

Why do you say that the average MW should be 2600 g/mole instead of

Other doubt: When checking the OH number, we tritate the excess of DEG, isn=B4t it ? So, the OH number we=B4re checking is the OH of the polymer, OH of the unreated DEG or both ?

In the example I sent above, We=B4ve an excess of 12,46 equivalents of OH (DEG) then (12,46*56100)/ 3000 - 165(water) =3D OH should be 246 but I get 160 so I lost DEG+water by vacuum distilattion.

Thank You very much

rodrigo