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?

We can determine the empirical formula by first converting each of the grams to moles. remember to do this, first, we need the molar mass of the molecules which can be calculated by adding the mass of the atoms from the periodic table. 

molar mass of CO2= 44.0 g/mol
molar mass of H2O= 18.02 g/mol

now, lets determine the grams of each atom

Carbon: 23.98 g x (12.011 g / 44.01 g) = 6.54 g C

Hydrogen: 4.91 g x (2.0158 g / 18.02 g) = 0.55 g H

Oxygen: 10.0 - (6.54 + 0.55) = 2.91 g O

Now let's convert each mass 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

now that we have the moles of each atom, we need to divide them by the smallest value to find the ration. If you do not get the whole number, you need to multiply until to get a whole number.

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

empirical formula--> C₃H₃O