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### Topic: Combustion Analysis Question CxHyOz  (Read 16385 times)

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#### SoleSky

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##### Combustion Analysis Question CxHyOz
« on: September 24, 2011, 12:30:49 PM »
Hey I've attempted this problem twice and only get three attempts on it so I was wondering if I could get some help with answering this problem.

An unknown compound contains only carbon, hydrogen, and oxygen (CxHyOz). Combustion of 2.50 g of this compound produced 3.67 g of carbon dioxide and 1.50 g of water.

How many Mols of C and H were in the original sample?

So I started like this:

(.083 mol C)x(12.0 g C)/(1 mol C)x(3.67gCO2)x(1 mol CO2)/(44.01gCO2) = .083 moles CO2 which I thought would equal 1 mol of C because there is only one mol of C in the compound.

I did the same thing for H but I divided by two because there are 2 mols of H in H2O.

I got the wrong answer apparently, I'm just confused as to what I'm doing wrong. Then they gave me a hint saying, How many mols of C are in 8.33x10^-2 mol CO2. still not sure what I'm doing wrong.

#### Borek

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##### Re: Combustion Analysis Question CxHyOz
« Reply #1 on: September 24, 2011, 05:03:49 PM »
.083 moles CO2 which I thought would equal 1 mol of C

No idea what you mean by 0.083=1.
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#### SoleSky

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##### Re: Combustion Analysis Question CxHyOz
« Reply #2 on: September 24, 2011, 07:43:54 PM »
Yes I was going over my work and I'm not sure where I got that number either. I think it is the mols of C but still I have no idea how to set up this equation

#### SoleSky

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##### Re: Combustion Analysis Question CxHyOz
« Reply #3 on: September 25, 2011, 02:37:12 PM »
I figured it out. It was .0834 Moles of C .1667 mols of H and .0833 mols of O