could you check the next page where I replied to the post from yesterday pertaining to nylon calculations.

Question

Thanks
C

Answer 1

I partially answered the question.

Answer 2

Oh..I'm sorry that your brain is cooked but I greatly appreciate you helping me out.

Thanks Dr.Bob =)

Answer 3

Good grief...I actually did the problem correctly HOWEVER it doesn't figure in with what I found as the theoretical yield =(

Problem (copied)
I added in the lab:

10ml 5% solution of Hexamethylenediamine +

10 drops NaOH (20% solution) +

10ml 5% aq solution of adipoyl chloride (slowly pouring to form a top layer over the hexamethylenediamine and NaOH solution)

_______________________
what I did...based on what you said...even though it ended up the same as what I did on my own.

0.5g hex (1mol/116.21g)= 0.004302mol hex
0.5g adi (1mol/183.03g)= 0.002731mol adi

Limiting: 0.002731mol adi(1mol nylon/1mol adi)(226.319g/1mol nylon6.6)= 0.61807 g Nylon

Supposed to have theoretically: 0.61807 g Nylon

What I had when I weighed it:
0.818g Nylon

I think it was a little wet on one spot about 1/2 a inch but I don't think it was enough to account for the 0.2g difference.
Thus I am throughly confused and have a headache right now..
(I spent the whole day looking up what polymer would make the best pipe and how to make it and I really didn't get anywhere)

Thanks Dr.Bob

Answer 4

above

Answer 5

Remember it doesn't take very much water to weigh 200 mg; i.e., only about 0.2 mL (about four (4) big drops) and your product could easily absorb that much water if it wasn't completely dry. I didn't understand exactly how you converted the mols of nylon to grams; i.e., isn't the molar mass of nylon in the big numbers because it has polymerized.