new in bold
(copied in normal text)
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: 0.61807g Nylon
What I had when I weighed it:
0.818g Nylon
You said:
"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"
Hm...well I did read somewhere that nylon absorbs moisture even from the air so that would account for it. As for the way I calculated the g of nylon. I based it on that site I tried to get you to go to in the first place. It's actually a pdf lab manual of the experiment and it says, I quote "(Theoretical yield can be calculated in grams. To do this, you do not need to know the molecular mass of the polymer – just calculate the molecular mass of the repetitive unit and multiply this by the number of moles of limiting reagent.)"
The repetative unit I calculated for nylon 6.6 was 226.319g. So I just multiplied it by the moles of limiting reagent since it was 1:1 molar ration btwn the nylon and the adi.
Thanks Dr.Bob =)
Um this should be bolded.
Hm...well I did read somewhere that nylon absorbs moisture even from the air so that would account for it. As for the way I calculated the g of nylon. I based it on that site I tried to get you to go to in the first place. It's actually a pdf lab manual of the experiment and it says, I quote "(Theoretical yield can be calculated in grams. To do this, you do not need to know the molecular mass of the polymer – just calculate the molecular mass of the repetitive unit and multiply this by the number of moles of limiting reagent.)"
The repetative unit I calculated for nylon 6.6 was 226.319g. So I just multiplied it by the moles of limiting reagent since it was 1:1 molar ration btwn the nylon and the adi.
Personally, I think your sample had a little water in it. At least that would account for the discrepancy. As for the molar mass, I've not worked with polymers like this before. That's a new way of doing it so I learned something. As for the pdf site, my computer won't let me look at a pdf site unless I turn off my fire wall and I don't like to do that.
Okay, well I think I should also mention that when it was dried or so I thought, I pulled it off the paper towel but I think by the sound of it some paper fibers may have been stuck to it also accounting for the additional mass.
Hm, it's quite interesting as I have my firewall turned on but it allows me to see pdf files. It just asks me about downloads. I just checked this and my firewall security has a option to "block all incoming connections" but I have that not checked.
~
another thing is you would you know how to make plastics? (I notice you say you've worked with polymers before)
My hwk Q that I was trying to find the answer to yesterday is this below but I don't know exactly how the placticizing process works. I compiled the chemicals I think I need but other than that I'm stumped as to how to put them together or how.
Q:
A water pipe has burst in your home. Being a poor college student and a good organic laboratory student you decide to fix the pipe yourself keeping in mind that a simple patch will not work. You will have to construct your own piping. your are able to order any chemicals needed. Describe what polymer you would make, why, and how.
I don't even know why they ask us these things...anyway. I came up with polyethylene as the plastic I think is suitable for producing pipes.
I think I'd need
ethylene, Ziegler-Natta catalyst (TiCl4)+ alluminum based co-catalyst Al(C2H5)3
I read somewhere that commercially they put it in a fluidized bed but I'm not sure if I could just say that I would use the fluidized bed at this imaginary school..(I don't think any school has one of these in reality)
statement taken from site:
"ethylene is put into a fluidized bed, where a polymerization reaction takes place, converting the ethylene into polyethylene. This process needs very high pressures and a catalyst."
-The catalyst is the zielgler natta catalyst with the alluminium compound
but other than that..I don't know how it works or how I can describe it.
Thanks alot Dr.Bob =D
Okay now I've basically given up on this.(I have to go do physics now) The last thing I found is this:
I looked up "making plastic at home"
I came up with:
Making Polystyrene
by using acetone and styrofoam and allowing the styrofoam to be dissolved by the acetone. Then the styrofoam is molded into whatever you want.
problem: it's not standardly used for pipes and I'm not sure it would be strong enough.
Well that's all I came up with.
Hope you can help me out with this.
Thanks Dr.Bob
It seems like you have a question about the calculation of the grams of nylon in your experiment. Let me explain it to you.
First, you calculated the moles of adipic acid (adi) and hexamethylenediamine (hex) based on their respective masses. You used the molar masses of adi (183.03 g/mol) and hex (116.21 g/mol) to convert the grams to moles.
Next, you determined the limiting reagent, which in this case was adi. To find the grams of nylon formed, you used the molar ratio between adi and nylon. The lab manual you mentioned states that you can calculate the theoretical yield of nylon in grams by multiplying the molar mass of the repetitive unit by the number of moles of the limiting reagent.
To find the molar mass of the repetitive unit of nylon 6.6, you used the value of 226.319 g/mol. This is the sum of the molar masses of adipic acid (hexane-1,6-dioic acid) and hexamethylenediamine. Since the molar ratio between adi and nylon is 1:1, you multiplied the moles of adi (0.002731 mol) by the repetitive unit's molar mass to obtain the grams of nylon.
Now, moving on to the issue of the difference between the theoretically calculated grams of nylon (0.61807 g) and the actual weight measured (0.818 g).
You mentioned that nylon absorbs moisture from the air, which could explain the additional weight. Nylon is known to be hygroscopic, meaning it readily absorbs water. In this case, it's possible that the nylon sample absorbed some moisture, leading to the higher measured weight. It's important to ensure that nylon is thoroughly dried and weighed quickly to minimize moisture absorption.
I hope this explanation helps clarify the calculation and the difference in the measured weight. If you have any further questions, feel free to ask!