I synthesized Nylon in my lab for organic chem.

I need help figuring out the theoretical yield so I can compare it to my product that I synthesized.

I looked up the reaction online on Google by typing in "Synthesis of adipoyl chloride and nylon 6.6."
I clicked on the first link on the page and it has the rxn (scrolled down to part C. figure 5)

The problem is that I'm not sure which is the limiting reagent (currently confused as to the ratio) I'm thinking it's a 1:1 ratio of 1 mol adipoyl chloride to 1 mol hexamethylenediamine but I'm not sure. (based on the picture)

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)

to a beaker.

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  1. I went to a different site. This gives a pretty good discussion. About half way down, under "chemistry". I have highlighted in bold below where the article talks about a 1:1 ratio. I presume this is 1:1 MOLE ratio. From your directions, you appear to have 1:1 gram ratio since both solutions were 10 mL of 5% solution.

    This article comes from Wikipedia and I found it by googling on nylon.
    In the laboratory, nylon 6,6 can also be made using adipoyl chloride instead of adipic It is difficult to get the proportions exactly correct, and deviations can lead to chain termination at molecular weights less than a desirable 10,000 daltons (u). To overcome this problem, a crystalline, solid "nylon salt" can be formed at room temperature, using an exact 1:1 ratio of the acid and the base to neutralize each other. Heated to 285 °C, the salt reacts to form nylon polymer. Above 20,000 daltons, it is impossible to spin the chains into yarn, so to combat this, some acetic acid is added to react with a free amine end group during polymer elongation to limit the molecular weight. In practice, and especially for 6,6, the monomers are often combined in a water solution. The water used to make the solution is evaporated under controlled conditions, and the increasing concentration of "salt" is polymerized to the final molecular weight.

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  2. Hm..I actually looked at that page but I got confused as to if it was the same rxn since it has "Heated to 285 deg celcius" which I didn't do.
    I now looked it up and found that all I produced was the nylon salt but to purify it it is usually heated, thus that explains the heating part of the Wikipedia article.

    Well since that is clarified.
    I am now wondering if I have to factor into the calculations for theoretical yield..
    the fact that I used 5% solutions of each to calculate the moles? Or do I just use the density?

    I was thinking that (using density to find theoretical yield)
    for Hexamethylenediamine:
    10ml(0.84g/ml)= 8.4g

    8.4g(1mol nylon/1mol hexamethylenediamine)= 8.4g of nylon

    while (adipoyl chloride)

    10ml x (1.26g/ml)= 12.6g

    12.6g(1mol nylon/ 1mol adipoyl chloride)= 12.6g nylon

    Thus if this is correct, the hexamethylenediamine would be the limiting reagent..

    Problem Is that...I had 0.814g of product so I really think the 5% solution would come into the calculations.

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  3. Okay That was ALL WRONG...I illegally went and did this above: 8.4g(1mol nylon/1mol hexamethylenediamine)= 8.4g of nylon which cannot be true of course.

    My second attempt at this since you didn't say anything about the above Dr.Bob.

    Here's what I came up with (taking into consideration the dilution of the solutions [not sure if that's right though])

    A10ml 5% hexamethylenediamine (116.21g/mol)
    B10ml 5% adipoyl chloride (183.03g/mol)

    M= [(%sol)(10)] / Mwt
    based on this formula

    A [5%(10)]/ 116.21= 0.430M Hexamethylenediamine
    B [5%(10)/ 183.03 = 0.273M Adipoyl chloride

    A 0.10L(0.430M)= 0.00430mol
    B 0.10L(0.273M)= 0.00273mol

    Thus according to this the limiting reagent is the Adipoyl chloride with 0.00273mol
    based on that the theoretical yield should be:

    0.00273mol Adipoyl chloride (1mol Nylon/ 1mol Adipoyl chloride)(226.31868g/mol Nylon 6.6)= 0.617849g Nylon 6.6

    Problem is that when I weighed the product, I had 0.818g of Nylon 6.6 total.

    Thanks for your help Dr.Bob =)
    I hope you can help me find what's wrong with my calculations.

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  4. My brain is cooked today. Without going through your calcns, I would do this.
    5% means 5 g/100 g soln = 0.5 g/10 mL.
    Therefore, you had 0.5 g hexa material.
    you had 0.5 g adi material.
    Convert each to mols, convert mols to mols nylon 6,6 formed and go from there.

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  5. I replied to the post labled Dr.Bob with what I came up with which is the same answer I derived from my own thought process.

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  6. Note: it's on pg 1 currently

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