ratio question

[asterio]

hey guys, having some problems with the chem question here.

10g of a hydrocarbon undergoes complete combustion leading to the formation of 30.5 g of CO2. What is the % of carbon in the hydrocarbon?

PS: I'm not exactly sure about what the formula of hydrocarbon is.

[fady_22]

n(c in hydrocarbon)=n(CO2)
That should get you started.

EDIT: Is this from the ILT?

[bomb]

A hydrocarbon a molecule with Carbon and Hydrogen in it.
So we don't know the formula of it exactly...and we don't need it (I'll show you how to find it later on though).

Now the best place to start off is with the info we have: the 30.5 grams of CO2.

The formula for combustion is:

nHydrocarbon + nO2 -> nCO2 + nH2O

Where n is just the number of each molecule needed. Now, all the carbons in CO2 come from the carbons in the hydrocarbon (where else can they come from?) so the number of mole of C in CO2 will be the same as in the Hydrocarbon.

#1, find n(CO2) ...now each mole of CO2 has 1 mole of C, so we're really finding the mole of C here, as that's what we need.
n= m/Mr
n(C)= 30.5/(12+32)
n(C)= 0.693mol

So now we know there was 0.693mol of Carbon in the 10g of Hydrocarbon.

#2 Let's find the mass of 0.693mol of Carbon
m = Mr x N
m = 12 x 0.693
m = 8.32 grams.

#3 Find %mass

8.32/10.0 *100
= 83.2%

Find the formula (a step further than you need to go)
Now we know that we have 0.693mol of Carbon, but how many mol of H?
Well 10.0-8.32=1.68grams
We have 1.68 grams of Hydrogen, so find the mole of Hydrogen:
n= m/Mr
n= 1.68/1
n=1.68

Now divide both number of mol by the LOWEST number of mole to find the ratio;
H: 1.68/0.693 = ~2.5 (2.43)
C: 0.693/0.693 = 1

Now we can see that the ratio between C and H is roughly 1:2.5 - we can't have this so we double both:
H: 5
C: 2

Formula is C2H5. However, I am mistake prone and would like someone to check this