A 1.082 g sample of a component of the light petroleum distillate called naptha is found to yield 3.317 g CO2 and 1.584 g H2O on complete combustion

Question

This particular compound is also found to be an alkane with one methyl group attached to a longer carbon chain and to have a molecular formula twice its empirical formula

The compound also has the following properties:

melting point: -154 C
boiling point: 60.3 C
density: 0.6532 g/mL at 20.0 C
specific heat: 2.25 J
DeltaH f = -204.6 kJ/mol

Use the masses of CO2 and H2O to determine the empirical formula of the alkane component.

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First i calculated the number of moles of Carbon and Hydrogen present in the unknown:

Carbon:
Molar mass = 43.99
CO2 mass - 3.317 g

moles= 0.0754
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Hydrogen:
Molar mass: 18.01
Mass of H20- 1.584 g

moles = 0.08795

Since there are 2 hydrogens in H2O
the moles are 0.01760.

And then i divide the moles by the lowest number:

Carbon: 0.0754/ 0.0754 = 1
Hydrogen: 0.0879/ 0.0754 = 1.16

I am not sure what to do next...

Answer 1

You're on the right track but you didn't start quite right. You haven't taken into account the mass of the sample. I would have determined the %C and %H, then converted those to mols and found the ratio from that.

Answer 2

i figured that the answer was C3H7

I found the mass percentage of C and H. Then I calculated the mass of C and H with the percentage.

Then I used the calculated masses to determine the number of moles of C and H. With that, I divided out the moles with the smallest number of moles. You should have gotten 1:2.33333333. In which you multiply by 3, to get the answer