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Topic: URGENT!!! Empirical Formula PLEASE !!  (Read 5258 times)

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Offline bec036

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URGENT!!! Empirical Formula PLEASE !!
« on: May 30, 2009, 09:56:59 PM »
URGENT

Hey if anyone could help me out if would be greatly appreciated!!   :D

I have a chem exam coming up soon and im doing some practice questions and was just wondering if anyone knew how to work this out. The question is:

"A hydrocarbon contains 16.3% hydrogen by mass. The empirical formula of the compound is...?"  ??? ??? ??? ???

I tried to work it out myself but my answer was C2H12 which is not correct apparently.  :-\

Please show how you worked this out as well (as best you can by typing it)

Thank-you so much  :D

Offline UG

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Re: URGENT!!! Empirical Formula PLEASE !!
« Reply #1 on: May 30, 2009, 10:59:56 PM »
Let's assume you have 100g of the compound. 16.3g of it would be hydrogen and therefore 83.7 grams of it will be carbon. Number of moles of hydrogen in 16.3 grams = 16.3/1 = 16.3 moles
Number of moles of carbon = 83.7/12 = 6.975 moles
So the ratio of hydrogen to carbon is, 16.3 : 6.975 which can be simplified to 2.33 : 1
Make the ratios whole numbers and you'll have your emperical formula  :)

Offline bec036

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Re: URGENT!!! Empirical Formula PLEASE !!
« Reply #2 on: May 30, 2009, 11:31:28 PM »
Let's assume you have 100g of the compound. 16.3g of it would be hydrogen and therefore 83.7 grams of it will be carbon. Number of moles of hydrogen in 16.3 grams = 16.3/1 = 16.3 moles
Number of moles of carbon = 83.7/12 = 6.975 moles
So the ratio of hydrogen to carbon is, 16.3 : 6.975 which can be simplified to 2.33 : 1
Make the ratios whole numbers and you'll have your emperical formula  :)

THANK-YOU so much!!  :D would you be able to explain how to make them whole numbers i dont get molecular formulas at all!!  ??? i'm sorry  :-[

Offline UG

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Re: URGENT!!! Empirical Formula PLEASE !!
« Reply #3 on: May 30, 2009, 11:41:36 PM »
Multiply the ratio by three so you get, 7 : 3
The answer should be C3H7


Offline bec036

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Re: URGENT!!! Empirical Formula PLEASE !!
« Reply #4 on: May 31, 2009, 12:21:10 AM »
Multiply the ratio by three so you get, 7 : 3
The answer should be C3H7



yes that is the correct answer. how did you know to multiply by 3??

Offline UG

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Re: URGENT!!! Empirical Formula PLEASE !!
« Reply #5 on: May 31, 2009, 01:47:36 AM »
2.33 written as a fraction is 7/3. So the ratio is 7/3 : 1
In order to get rid of the three you muiltiply the entire thing by three to get 7 : 3

Offline bec036

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Re: URGENT!!! Empirical Formula PLEASE !!
« Reply #6 on: May 31, 2009, 02:25:00 AM »
ah k. i get it!!
thank-you so much  :D

Offline UG

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Re: URGENT!!! Empirical Formula PLEASE !!
« Reply #7 on: May 31, 2009, 02:30:06 AM »
No problemo  :)
Or should I say, N2O4 Pr2O6B3Li5Mo8 

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