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### Topic: Graham's law/emperical formula problem  (Read 5149 times)

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#### optimusprimer

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##### Graham's law/emperical formula problem
« on: December 05, 2007, 09:48:09 PM »
Here's the Problem: A compound contains only C,H, and N, It is 58.51% C and 7.37% H by mass.  Helium effuses through a porous frit 3.2 times as fast as the compound does. Determinine the emperical and molecular formulas of this compound.

What I have so far:
.5851 g C = .04876 moles
.0737 g H = .0737 moles
.3412 g N = .02437 moles
emperical formula = C3H2N6
emperical mass    = 122 g/mole

3.2R/R=sqrt(mm/4)
This gives me a molar mass of only 41 g/mole.
I know this can't be possible because it is less than the emperical molar mass.  Can someone try to help me and see where I went wrong?

#### Sev

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##### Re: Graham's law/emperical formula problem
« Reply #1 on: December 05, 2007, 10:27:59 PM »

#### optimusprimer

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##### Re: Graham's law/emperical formula problem
« Reply #2 on: December 05, 2007, 10:37:33 PM »
Oh, I see now.  The emperical formula should be C2H3N, correct?  This would make the molecular formula C2H3N too I guess.  I don't really know how I screwed that up, but thanks for pointing out the problem.

#### Sev

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##### Re: Graham's law/emperical formula problem
« Reply #3 on: December 05, 2007, 10:40:57 PM »
Quote
Oh, I see now.  The emperical formula should be C2H3N, correct?

Yep