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Topic: xylene empirical formula?  (Read 7036 times)

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Offline tomgreenschic

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xylene empirical formula?
« on: November 05, 2008, 09:32:24 PM »
mass percent of carbon is 90.51% and mass percent of hydrogen is 9.49%
What is the empirical formula?
is it C5H6?
Thats what I get but when I looked it up online I couldn't find much info on it.
Any help thanks! :D
« Last Edit: November 05, 2008, 09:48:31 PM by tomgreenschic »

Offline Astrokel

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Re: xylene empirical formula?
« Reply #1 on: November 05, 2008, 11:00:13 PM »
no it is not. There is a standard approach to this type of question, all you need to do is to find its mole ratio assuming there is 100g of compound. As you are working with ratio, it doesn't matter you assume how many grams of compound to work with, result will still be the same.
No matters what results are waiting for us, it's nothing but the DESTINY!!!!!!!!!!!!

Offline tomgreenschic

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Re: xylene empirical formula?
« Reply #2 on: November 05, 2008, 11:21:40 PM »
I know.
Here is what I am getting.

90.51g C
9.49g H

90.51g C x 1 mol C/12g C = 7.536

9.49g H x 1 mol H/1g H = 9.49

I then divide both by the smaller number, 7.536.

ratios:
C = 1
H = 1.3

multiply to make them both whole numbers...
C = 1 x 3 =3
H = 1.3 x 3 = 4

So it should be C3H4

right? maybe? hmmm.
:D

Offline AWK

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Re: xylene empirical formula?
« Reply #3 on: November 06, 2008, 02:26:58 AM »
Do not round numbers before finding numbers very close to integers.
Divide 9.49 by 7.536 and then multiply the result by 2, 3, 4 ... up to the obtained number and the closest integer will differentiate less than (or close to) 1 %
But when you obtain this ratio you can easily see which number can fulfill this requirement without checking many integer coefficient.
AWK

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