Empirical formula question?

Start with Hydrogen, because all of the hydrogen from the compound will end up in water. You'll want to find the mass of hydrogen (2 * 1.01) / (16.00 + 2* 1.01) that is, the mass percentage of hydrogen in water, so that you can covert that to moles using its molar mass. Given that you can find the number of moles of hydrogen in the compound. Repeat the same process for carbon, and then use those two masses that you found in the process of getting the moles to subtract from .519 to find the mass of oxygen. Convert that to moles. You have the ratios, so divide all three separately by the smallest so that the smallest subscript will be one. You should end up with the correct formula. Send a reply to the answer if it doesn't work. Good luck!

(1.24 g CO2) / (44.00964 g CO2/mol) x (1 mol C / 1 mol CO2) x (12.01078 g C/mol) = 0.33841 g C

(0.255 g H2O) / (18.01532 g H2O/mol) x (2 mol H / 1 mol H2O) x (1.007947 g H/mol) = 0.028534 g H

(0.519 g total) - (0.33841 g C) - (0.028534 g H) = 0.152056 g O

(0.33841 g C) / (12.01078 g C/mol) = 0.0281755 mol C

(0.028534 g H) / (1.007947 g H/mol) = 0.0283090 mol H

(0.152056 g O) / (15.99943 g O/mol) = 0.00950384 mol O

Divide by the smallest number of moles:

(0.0281755 mol C) / 0.00950384 mol = 2.96

(0.0283090 mol H) / 0.00950384 mol = 2.98

(0.00950384 mol O) / 0.00950384 mol = 1.00

Round to the nearest whole numbers to find the empirical formula:

C3H3O