Polyester Calculation

Hi,

Need your help once more:

What is the OH number of the resin below:

Adipic Acid - 552,2grams Trimethilopropane TMP - 35,5 grams DEG - 412,3 grams

I=B4m not sure, but I think it should be 56, isn=B4t ?

Please, help me, it=B4s urgent.

Regards,

Rodrigo Toledo

Reply to
toledo11
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Tells us the number of moles of

Adipic Acid - Trimethilopropane - DEG -

and the number of equivalents of carboxylic acid, and total hydroxyls.

After that, I will tell you if 56 is correct answer. But, I suspect that at that point you will not need any more help.

Ernie

Need your help once more:

What is the OH number of the resin below:

Adipic Acid - 552,2grams Trimethilopropane TMP - 35,5 grams DEG - 412,3 grams

I´m not sure, but I think it should be 56, isn´t ?

Please, help me, it´s urgent.

Regards,

Rodrigo Toledo

Reply to
Ernie

Ernie,

Adipic Acid - 552,3 grams - 3,777 moles - 7,554 equivalents

DEG - 412,3 grams - 3,882 moles - 7,765 equivalents TMP - 35,5 grams - 0,265 moles - 0,794 equivlantes

OH =3D (7,765+0,794)-7,554 =3D 1,005 equivalents of hydroxyls

1,005*56100/(412,3+35,5+552,3) =3D 56,38

Is that correct ?

regards,

Rodrigo

Ernie wrote:

Reply to
toledo11

Rodrigo,

Almost! What is eliminated (removed) when an ester (or polyester) is formed from a carboxylic acid and an alcohol (polyol)? How many moles of this molecule are formed and eliminated to make the polyester, and what is the total weight of the substance eliminated?

Correct the weight of the polymer formed by the weight of the eliminated substance and redo your calculation.

Good luck,

Ernie

Adipic Acid - 552,3 grams - 3,777 moles - 7,554 equivalents

DEG - 412,3 grams - 3,882 moles - 7,765 equivalents TMP - 35,5 grams - 0,265 moles - 0,794 equivlantes

OH = (7,765+0,794)-7,554 = 1,005 equivalents of hydroxyls

1,005*56100/(412,3+35,5+552,3) = 56,38

Is that correct ?

regards,

Rodrigo

Ernie wrote:

Reply to
Ernie

Ernie,

Ok ! Water is produced and must be removed (Le chatellier ). But how to use this information (amount of water produced/removed) to calculate the OH number ?

Can you hlep me ?

Regards, Rodrigo

Ernie wrote:

Reply to
toledo11

He's already told you this. Use the weight of water removed to correct the weight of the polymer in your calculation.

Colin

Ok ! Water is produced and must be removed (Le chatellier ). But how to use this information (amount of water produced/removed) to calculate the OH number ?

Can you hlep me ?

Regards, Rodrigo

Ernie wrote:

Reply to
Colin Reed

Please keep in mind that you are only looking at the difference between acid groups and OH group. At start of the reaction nothing has reacted and your OH is 480 (and AV=424). Remove water and both will go down. Without losses and decomposition you will end up with your OH=56! regards Koos

Reply to
kvb

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