I desperately need help with this "chemical combustion" equation O_O

Question

vitamin E is an antioxidant that plays an especially important role protecting cellular structures in the lungs. Combustion of a 0.497-g sample of vitamin E produced 1.47g of carbon dioxide and 0.518g of water. Determine the empirical formula of vitamin E.

God I desperately need help on this question! O_O i've been trying to figure it out for about an hour... I'm ready to pull my hair out

Answer 1

C13H23O

Answer 2

1.47g CO2 is 0.03341mol and has 0.03341mol C 0.4009g C

0.518g H2O is 0.02877mol and has 0.05755mol H 0.05801g H

oxygen in vitamin is:

0.497g-0.4009g-0.05801g=0.0381 g

0.002380mol O

form proportions:

C:0.03341/0.002380=14.04

H:0.05755/0.002380=24.18

C14.04H24.18O real:(C14.5H25O)*2

it is very common for empirical formula to contain some  irational proportions,dew to practical errors.