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### Topic: Hard stoic problem  (Read 6206 times)

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#### Korokian

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##### Hard stoic problem
« on: September 25, 2006, 03:41:25 PM »
A 1.25-g sample of a compound containing only C,H, and O is analyzed by complete combustion of CO2 and H2O and is found to form 1.45g of CO2 and 0.592g H2O.  Find the empirical formula of the compound.  In a seperate measurement the molecular weight is found to be 153.  Find the molecular formula of the compound.
A) amount of C in sample_____
B) amount of H in sampe______
C) mass of O in sample_______
D) amount of O in sample_______
E) emperical formula_______
F) molecular formula_______

can i get a clue on where to start please?
CxHyOz + O2 ------> CO2 + H2O
do i use the mass composition formula for C?

#### sdekivit

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##### Re: Hard stoic problem
« Reply #1 on: September 25, 2006, 04:53:37 PM »
Hint:

when you have CxHyOz is completely combusted to H2O and CO2, all the C from CxHyOZ ends up in CO2 and all H ends up as H2O.

--> so when you know the mass of CO2 and H2O formed, you also know the amount of mol C and H in CxHyOz and all the rest of mass will be O in CxHyOz.

#### Korokian

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##### Re: Hard stoic problem
« Reply #2 on: September 25, 2006, 05:07:02 PM »
so i converted 1.45 g CO2 to 0.033 mol CO2
and 0.529g H2O to .033 mol H2O
so is it .0329 mol C
and .0329 mol H2?

#### Dan

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##### Re: Hard stoic problem
« Reply #3 on: September 25, 2006, 06:05:27 PM »
Yes.
My research: Google Scholar and Researchgate

#### Korokian

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##### Re: Hard stoic problem
« Reply #4 on: September 25, 2006, 06:47:32 PM »
how do you get the mass of the O?

#### BaO

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##### Re: Hard stoic problem
« Reply #5 on: September 26, 2006, 01:56:47 AM »
there is a previous post related to your question: http://www.chemicalforums.com/index.php?topic=10022.msg47591#msg47591