A 1.24g sample of a hydrocarbon, when completely burned in excess of O2(g), yields 4.04g CO2 and 1.24g H20. Determine the empirical formula.

Answer 1

CnHm + O2 → CO2 + H2O

4.04g of CO2*(1 mole of CO2/44.01g of CO2)=moles of CO2=0.0913g of CO2

1 mole of CO2=1 mole C, moles of CO2=moles of C

1.24 g of H20 *(1mole of H20/18 g of H20)=moles of H20=0.689

1 mole of H20=2 moles of H

2*moles of H20= moles of H=0.1376

divide the number of moles of H by moles of C

you should calculate C=1 H=2

Combustion reaction is
CH2 + O2 → CO2 + H2O

Answer 2

The reaction in the bottom of the post above is unbalanced

The balanced reaction is

2CH2 + 3O2--> 2CO2 + 2H20

Answer 3

I agree with 0.0918 mols C and 0.137 mols H but not the formula.
0.137/0.0918 = about 1.5; is that C2H3?
And the balanced equation is
4C2H3 + 11O2 ==> 8CO2 + 6H2O
I've not heard of C2H3?