Calculate Moles of Terephthalic Acid from PET Depolymerization

  • Thread starter Thread starter PoisonCupcake
  • Start date Start date
  • Tags Tags
    pet
AI Thread Summary
To calculate the moles of terephthalic acid produced from PET depolymerization, it is essential to understand that terephthalic acid is a product rather than a reactant. The lab involved 2.482g of PET, which corresponds to 0.01291605 moles based on its copolymer unit mass of 192.164 g/mol. The discussion clarifies that while the moles of PET reacted can be calculated, the focus should be on the theoretical yield of terephthalic acid for yield percentage calculations. It is generally accepted that one mole of terephthalic acid is produced per mole of copolymer unit. Accurate understanding of these relationships is crucial for proper calculations in polymer chemistry.
PoisonCupcake
Messages
6
Reaction score
0

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!
 
Physics news on Phys.org
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?
 
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.
 
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.
 
Thread 'Confusion regarding a chemical kinetics problem'
TL;DR Summary: cannot find out error in solution proposed. [![question with rate laws][1]][1] Now the rate law for the reaction (i.e reaction rate) can be written as: $$ R= k[N_2O_5] $$ my main question is, WHAT is this reaction equal to? what I mean here is, whether $$k[N_2O_5]= -d[N_2O_5]/dt$$ or is it $$k[N_2O_5]= -1/2 \frac{d}{dt} [N_2O_5] $$ ? The latter seems to be more apt, as the reaction rate must be -1/2 (disappearance rate of N2O5), which adheres to the stoichiometry of the...
Back
Top