Find the empirical formula of a compound found to contain 63.15 percent carbon 5.30 percent hydrogen and 31.55 percent oxygen

Question

I got C27H27O

Answer 1

That is a very odd answer.
assume you had 100grams.
63.15g c
5.3g H
31.55g O
in moles, that is
C 63.15/12.011= 5.26Mol
H 5.3/1.008 =5.26 mol H
O 31.55/15.999=1.971 mol O
Now divide each by the lowest number in the mole ratio:
C 5.26/1.971=2.69
H = 2.69
O=1
Now, to covert to whole numbers, note that 2.69*3=8
empirical is C8H8O3
Which is again, weird, but check my work.

Answer 2

Oh I got C and H = to 2.71