[Krystalskye]

I believe I'm doing the calculations correctly, however my yield percent is smaller than I expected, so I just wanted to double check here.  In this lab I don't see where my sources of error could have came from to make the %yield so small.

The reaction is between adipoyl chloride (10mL) and hexamethylenediamine (10mL) to yield nylon.  (with OH-)
My  yield of nylon was 2.299g

Let A = Adipoyl Chloride
Let B = hexamethylenediamine
Let C = 6,6 Nylon

MW of A = 183.03(g/mol)
Density of A = 1.25(g/mL)

MW of B = 116.21(g/mol)
Density of B = 0.84(g/mL)

First I calculated the grams of each by using g = mL x (g/ML)
A(grams)= (10)(1.25) = 12.5g
B(grams)= (10)(0.84) = 8.4g

From there I did g to mol.
A=(12.5g)/(183.03g/mol) = .0683mol
B=(8.4g)/(116.21g/mol) = .0723mol
the ratio is 1:1 therefore A is our limiting reactant.

Then I figured out the MW of 6,6 Nylon
Its 116.21 + 183.03 -(Cl2) -(h2) = 116.21 + 183.03 - 72.9 = 226.34 g/mol


.0683mol---------->.0683mol               (.0683g)(226.34g/mol) = 15.46g
     A     +      B      =    C
                               15.46g Nylon

(2.229)/(15.46) X 100 = 14.42% yield

  Help?

[Quinoline]

Actually the molecular weight of your nylon-6,6 is not what you've said. Polymer reactions generate a variety of different sized chains (think spaghetti) which have molecular weights in kilo-daltons (10,000-100,000 g/mol). To characterize these one must use sophisticated chromatography techniques like gel permeated chromatography (GPC) or by mass spectrometry.

What you can report is a percent recovery of the polymer, which would be the mass of the polymer divided by the mass of the starting materials.

[zolarpwr]

What you can report is a percent recovery of the polymer, which would be the mass of the polymer divided by the mass of the starting materials.

Yes. Don't forget to account for the loss of one HCl molecule for each amide bond formed.

How are you recovering your polymer? Do you just distill off the remaining monomers or are you precipitating the polymer from the excess monomer mixture? If you do the second method you may lose a lot of dimers and small oligomers which dissolve instead of precipitate.

Your math so far is good, but this is far more complicated than an ordinary chemical reaction. In step polymerization, it is very important to have an equal stoichiometry ratio of monomers to get high MW. From the Carother's equation, Xn = (1+r)/(1-r), where r = (0.0683 mol/0.0723 mol) = 0.94 in your case, the average degree of polymerization for your reaction is, at best, 32. If the actual conversion (meaning the percentage of amide bonds that did form compared to the amount that could form) is low, then you will get even smaller chains, on average. Depending on your recovery method, you could have lost a lot of the small oligomers you made.

http://en.wikipedia.org/wiki/Step-growth_polymerization
http://en.wikipedia.org/wiki/Carother's_equation