Calculate Moles of Terephthalic Acid from PET Depolymerization

[PoisonCupcake]

Homework Statement

Hi, I recently did a lab where PET plastic was depolymerized into terephthalic acid and ethylene glycol. I was wondering how to figure out the moles of terephthalic acid (in mol) that was reacted. I'm not really sure how to calculate this since terephthalic acid isn't a reactant itself and is just mostly present in the form of a copolymer.
PET reacted = 2.482g or 0.01291605 mol
Mass of the copolymer unit = 192.164 g/mol
Mass of terephthalic acid = 166.128 g/mol
Mass of ethylene glycol = 62.068 g/mol

The Attempt at a Solution

If terephathalic acid was a reactant I would think it would mean that since 0.01291605 mol of PET reacted, 0.01291605 mol of terephathalic acid reacted, but don't know if this is correct.

Thanks!

 

[Borek]

You are doing some strange things. How do you know number of moles of PET? And what do you mean by "acid that reacted" - acid that was produced?

 

[PoisonCupcake]

In the experiment we used 3g of PET plastic and after it had been in the reflux condenser I weighed out how much PET plastic hadn't reacted, which when subtracted from the 3g, gave me the mass (2.482g). I converted that into the amount of moles of PET that reacted with this equation : (1 mol PET copolymer unit/192.164g)*2.482g = 0.01291605 mol.

What's meant about the amount of acid reacted is the theoretical yield as I need to do % yield calculations. It's definitely not the amount of acid produced though, as that was weighed out at the end.

 

[Borek]

So when you state "moles of PET" you really mean "moles of copolymer unit". That's not exactly how a mole is understood in polymer chemistry.

But I agree - seems like 1 mole of acid is produced per 1 mole of copolymer units.

 

[DeeNos]

Hello,

Thank you for sharing your lab results with me. To calculate the moles of terephthalic acid (TA) present in the PET depolymerization reaction, we can use the following equation:

Moles of TA = (Mass of PET reacted in grams) x (1 mol PET/192.164 g PET) x (2 mol TA/1 mol PET) x (166.128 g TA/1 mol TA)

Plugging in your values, we get:

Moles of TA = (2.482 g) x (1 mol PET/192.164 g PET) x (2 mol TA/1 mol PET) x (166.128 g TA/1 mol TA) = 0.01291605 mol TA

So, your initial assumption was correct. Since PET and TA are in a 2:2 ratio in the copolymer, the moles of PET reacted is equal to the moles of TA reacted. I hope this helps clarify the calculation for you. Let me know if you have any further questions. Good luck with your lab!