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Slightbreach of Etiquette
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Discussion Starter #1
My brain is fried from doing lab report and other chemistry stuff for the past 6 or so hours, and for some reason this is escaping me:

Complete combustion of 1.110g of a gaseous Hydrocarbon yields 3.613g of Carbon dioxide and 1.109g of water. A 0.288g sample of the hydrocarbon occupies a volume of 131mL at 24.8c and 753mm Hg. What is the molecular formula of the hydrocarbon?

So here's what I have so far:
(some hydrocarbon) + O2 -> CO2 + H20

.08211 Moles of CO2
.06161 Moles of H2O

Using PV=nRT I get that the Molecular weight of the hydrocarbon is 209g so there's .005311moles of that, but for some reason I'm having a hard time figuring out the molecular formula!
Any help is appreciated
 

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Slightbreach of Etiquette
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Discussion Starter #5 (Edited)
Got it! The Molecular weight of the hydrocarbon should be based off of the 0.288g sample not the 1.110g sample. I can't believe this took me so long haha... It works out to (1)H6C4 + (11/2) O2 --> (4)CO2 + (3) H2O
 

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Custom User Title
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.08211 Moles of CO2
.06161 Moles of H2O
It's been years since I had chem, but I'd assume your goal is to find the ratio of Hydrogen and Carbon.

Using just the info above, you should be able to solve for:


.005311 HxCy + O2 -> .08211 CO2 + .06161 H20


You could also approach it using conservation of mass:

1.110g of a gaseous Hydrocarbon yields 3.613g of Carbon dioxide and 1.109g of water
(meaning 3.612g Oxygen)

1.110g HxCy + 3.612g O2 -> 3.613 CO2 + 1.109g H20
 

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Killing Owls 2 stop O Rly
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I am getting 4, 8, 15, 16, 23, 42... though I am not sure all of those are correct.
 

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Slightbreach of Etiquette
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Discussion Starter #9
haha I took a break to watch lost and figured it out almost right when I got back.

It ended up being (1)H6C4 + (11/2) O2 --> (4)CO2 + (3) H2O

Thanks!
 

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haha I took a break to watch lost and figured it out almost right when I got back.

It ended up being (1)H6C4 + (11/2) O2 --> (4)CO2 + (3) H2O

Thanks!
Not sure how anal your chem prof is, but mine would mark points off for putting "H6C4" instead of "2 H3C2"
 

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Slightbreach of Etiquette
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Discussion Starter #12
She wants the molecular not the empirical, but thanks for looking out! (Plus the CO2 would have to be 8 and the H2O 6 in order for the proportions to work out that way)
 
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