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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: tomgreenschic on November 05, 2008, 09:32:24 PM

Title: xylene empirical formula?
Post by: tomgreenschic 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

Title: Re: xylene empirical formula?
Post by: Astrokel 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.

Title: Re: xylene empirical formula?
Post by: tomgreenschic 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

Title: Re: xylene empirical formula?
Post by: AWK 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.