A10.00 g sample of a compound containing only carbon, hydrogen, and oxygen forms 23.98 g co2 and 4.91 g h2o upon complete combustion. what is the empirical formula of the compound?

There are three major steps for finding an empirical formula from a combustion reaction.
1.) Determine the grams of carbon and hydrogen from the given data.

C: 23.98 g x (12.011 g / 44.01 g) = 6.54 g C
H: 4.91 g x (2.0158 g / 18.02 g) = 0.55 g H
Determine the grams of oxygen in the sample by subtracting the mass of the compound given from the total mass solved earlier.
O: 10.0 - (6.54 + 0.55) = 2.91 g O

2) Convert the values in step 1 to moles.

C: 6.54 g / 12.01 g / mol = 0.54 mol
H: 0.55 g / 1.01 g/mol = 0.54 mol
O: 2.91 g / 16.00 g/mol = 0.18 mol

3) Divide each  by the lowest value calculated in step 2

C:  0.54 mol / 0.18 mol = 3
H:  0.54 mol / 0.18 mol = 3
O: 0.18 mol / 0.18 mol = 1

Thus, the empirical formula is C3H3O.

Ithink it's the third answer