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DUMAS METHOD OF DETERMINING MOLAR MASS OF A VOLATILE LIQUID

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DUMAS METHOD OF DETERMINING MOLAR MASS OF A VOLATILE LIQUID

Postby Darwin420 » Mon Oct 20, 2008 6:22 pm

Ladies and Gents, I am having a bit of a difficulty writing this lab report and I would like some insight!

Ok, so I conducted this experiment where I was given methanol. I placed methanol in a flask, put tine foil over the end, poked to holes in it and placed it in a hot bath. I measured the flask before the methanol added and after. I also did three trials.

My results:

Textbook molar mass: 32.04 g/mol

M.M. = (0.1 grams)(0.08206 L atm/K mol)(371.35 K)/(91.023 atm)(0.139 L)
= 21.43 g/mol
% relative error: x-a/x = 32.04-21.43/32.04 = 33.11% error

Calculation for trial # 2:
M.M. = (0.1364 grams)(0.08206 L atm/K mole)(370.15 K)/(1.023 atm)(0.159L)
= 25.34 g/mol
% relative error: 20.91 %

Calculation for trails # 3:
M.M. = (0.1433 grams)(0.08206 L atm/K mole)(370.75 K)/(1.023 atm)(0.153 L)
=27.86 g/mol
% relative error: 13.05 %
Average relative error %: 22.35 %

As you can see my relative errors are somewhat consistant, any ideas why I am getting this gap?

I was thinking this:

Air contaminated the results which contains roughly 78% nitrogen, 21 % oxygen and 1 % trace elements. But I would assume that would add more weight as oppose to gain more weight.

Another possible reason would be that that I didn't let the flask cool down enough so all the methanol didn't condense thus making my results for molar mass less then they apparently should be. I let the the flask cool down relatively the same amount of time so that may explain the somewhat consisten relative % error.

Thanks for your time and I look forward to your ideas/insight
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Postby Darwin420 » Mon Oct 20, 2008 6:23 pm

I also to forgot that the sample of methanol I used for all trials was 4 mL.
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Postby alextemplet » Mon Oct 20, 2008 6:42 pm

If the sample of methanol is 4mL (0.004L), why did you place a different volume in your calculations? All of your calculations seem to be accurate, but I can't see where you get the volume from in each one. You should have also taken more precise measurements of the methanol's volume. If the mass was slightly different each time, then your volume should have been also, assuming you were using the same solution for each trial. It might also be helpful to know the molarity of the solution; I'm assuming you were using an aqueous methanol solution?

I'm also a bit confused as to exactly what you were trying to do with the methanol. I see you're working with temperatures slightly below the boiling point of water (373K); were you trying to denature the methanol into methane and oxygen? If some of your methanol boiled off as gas, that would be why your final sample was light-weight. I'm assuming this is what you were doing, since I can see no other reason why the ideal gas constant (0.08206Latm/Kmol) would be in your equation.
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Postby MichaelXY » Mon Oct 20, 2008 7:55 pm

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Re: DUMAS METHOD OF DETERMINING MOLAR MASS OF A VOLATILE LIQUID

Postby Darwin420 » Mon Oct 20, 2008 8:56 pm

Yes, the whole point was to convert the methanol into a gas, that is why we are using this equation in the first place. The volume I am using in these equations are the volume of the flasks we have been using not the volume of the methanol. When it comes to molarity we weren't given it in this experiment.

thanks for your input, much appreciated.
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Postby alextemplet » Mon Oct 20, 2008 9:54 pm

Okay, here's what I think you need to do. If you're converting it into a gas, then methanol decays per this equation:

2CH3OH -> 2CH4 + O2

As you can see, two moles of methanol produces three moles of gas. Here's where I think your calculations went wrong. The equation you used is mRT/PV, but it should just be PV/RT; this would give you the number of moles of gas produced. Since three moles of gas are produced by two moles of methanol, multiply that number of moles of gas by 2/3. This gives you the original number of moles of methanol. Then divide the mass of your original sample (which you determined by weighing the flask before and after adding the methanol, and subtracting the difference) by the number of moles that you calculated, and you should have the molar mass.
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Postby MichaelXY » Mon Oct 20, 2008 10:24 pm

Alex, take a look at this link. better than the one above.

http://web.centre.edu/miles/che135labs/ ... Method.pdf
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Postby alextemplet » Tue Oct 21, 2008 3:11 am

Okay so I might have been slightly mistaken in some assumptions. The procedure I was describing still works; we did this in my lab a few weeks a ago to measure quantities of gas produced by displacing water. I had assumed this experiment was similar to the one we conducted; might have been a bad assumption for me to make, as I can see there were a few differences.
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Postby AstusAleator » Tue Oct 21, 2008 6:01 pm

It looks like you're getting smaller-than-expected amounts of liquid in your condensation. This could be, as you suggested, because the condensation is incomplete. It could also be due to problems with your stopper (tinfoil). If the hole you poked is too big, air currents passing by could disturb the contents of the flask. Also, make sure you take the flask out of the bath as soon as you don't detect anymore vapor escaping. Finally, it could have something to do with the equipment you're using to measure.
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Re: DUMAS METHOD OF DETERMINING MOLAR MASS OF A VOLATILE LIQUID

Postby Darwin420 » Wed Oct 22, 2008 5:47 pm

Thanks for all your input guys. I am just finishing my lab report right now.

This is off topic, but do you guys know a good website where I could get some lab gear, more specifically a light microscope?

Thanks.
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Postby Darwin420 » Wed Oct 22, 2008 6:03 pm

Oh, one more question. In regards to Alex, so heating methanol forms methane?

so it is CH3OH --> CH4 + monotomic Oxygen? Seems strange
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Postby alextemplet » Wed Oct 22, 2008 6:13 pm

I am pretty sure it does, unless I've got my chemistry wrong. Those carbon compounds are tricky. You're not forming atomic oxygen (O), however; it's molecular (O2) oxygen, also called diatomic oxygen. This is important since any calculation of the number of moles of gas will reflect the amount of molecular oxygen, which is half the number of oxygen atoms present.
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