Complete combustion of 2.10 g of a hydrocarbon produced 6.43 g of CO2 and 3.07 g of H2O. What is the empirical formula for the hydrocarbon?

Question

aaronq · Answer

convert all the masses given to moles

then divide all by the one with the least amount of moles

anonymous · Answer

That did not work..

aaronq · Answer

show me what you did

anonymous · Answer

2.10/(1.008+12)=0.161439114/0.146136364=1.104715552
6.43/(12+(16*2))=0.146136364/0.146136364=1
3.07/((1.008*2)+16)=0.170404085/0.146136364=1.166062167

anonymous · Answer

That would mean the formula would be just CH. Right?

anonymous · Answer

Obviously not because my homework told me I was wrong..

anonymous · Answer

Okay.. What exactly does that mean? How do I get my answer?

aaronq · Answer

i'm having a mental blank give me a few mins

anonymous · Answer

Okay. Take all the time you need.. It's due at 11:30 pm central

aaronq · Answer

Okay, so we have
$ C_xH_y + O_2 = CO_2 + H_2O$

C in the CO2 = 6.43 g/(12+(16*2)g/mol) = 0.146136363 mol

H in H2O = 3.07 g/(1.008*2)+16)g/mol*(2) =0.3408081705150976 mol


$H:C\rightarrow \dfrac{0.34080}{0.1461}:1\rightarrow 2.33:1$


now all you need to do is multiply to get whole numbers

anonymous · Answer

does : mean multiply?

aaronq · Answer

nope H:C means the ratio of H to C

anonymous · Answer

so what do i need to multiply?

aaronq · Answer

multiply 2.33 by a number to get a whole number (integer)
whatever number you multiply it by you need to multiply 1 by it as well.

anonymous · Answer

so 2.33*100=233? then 100*1=100?

anonymous · Answer

then what?

aaronq · Answer

no you wanna find the lowest possible number

2.33* 3= 7
1*3=3

$C_3H_7$

anonymous · Answer

ohhhh.. wow.. sorry, i'm not very good at this.. if you couldn't tell..
thank you so much though! Now i have to figure out how to post a second question..

aaronq · Answer

wait, that formula seems a little off, let me double check it.

anonymous · Answer

it's right

aaronq · Answer

yeah, i drew it and it didn;t make sense, but then i doubled it $C_6H_{14}$ and it worked.

aaronq · Answer

and no worries, we all gotta learn someday. No ones born a genius. Btw, you can close this question, as open a new one, just like you did with this one.