GC question

[dcao]

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Combustion analysis of a 2.88 mg sample of an oxygenated hydrocarbon affords 5.31 mg CO2. The mass %C in the compound undergoing analysis is..

A) 50.3% C
B) 64.6% C
C) 42.3% C
D) 37.8% C
E) 55.4% C

Can someone tell me how to work this problem? Thanks!

 

[herkulease]

is water not formed?

 

[UCB05]

In a complete combustion reaction, it's reasonable to say that all the carbon in the hydrocarbon is converted into CO2.

The mass% of carbon in CO2 is 12/44=27.3%. 27.3% of the mass of CO2 generated in this reaction is carbon, or (0.273)(5.31)=1.45mg

The mass% of carbon in the original hydrocarbon would be 1.45/2.88

 

[herkulease]

In a complete combustion reaction, it's reasonable to say that all the carbon in the hydrocarbon is converted into CO2.

The mass% of carbon in CO2 is 12/44=27.3%. 27.3% of the mass of CO2 generated in this reaction is carbon, or (0.273)(5.31)=1.45mg

The mass% of carbon in the original hydrocarbon would be 1.45/2.88

yeah that's right. I wasn't sure if he left it out or not. I'm so used to these type chem questions where I would have to do more than something this straightforward.

I remember coming across a question on DAT destroyer and was like really, he's throwing out something this easy. I need to read it again to be sure.

 

[predent7]

so we do not need to know exact formula of comp..

 

[UndergradGuy7]

so we do not need to know exact formula of comp..

No, the reaction is a combustion so C(x)H👍 + O2 <-> CO2 and H2O
The point of this problem is to determine the formula or the carbon at least.

Notice the carbon only comes from the hydrocarbon and only goes to co2. So if you find out the amount in co2, you know the amount in the hydrocarbon.

You could also do 5.31mg CO2/44 g/mol CO2 = millimoles of co2. Millimoles of co2 = moles of c in co2. Then do moles co2*12 m.w. of co2 = 1.448 mg of c.

Then find the percent by doing 1.45/2.88 * 100.