Question

An organic compound on analysis gave the following results C=54.5%, O=36.4%, H=91%. The Empirical formula of the compound is

Answer 1

I assume you made a typo and that is 9.1% H.
Take a 100 g sample to give you 54.5 g C, 36.4 g O, 9.1 g H
mols C = 54.5/12 = approx 4.5
mols O = 36.4/16 = approx 2.3
mols H = 9.1/1 = approx 9.1
Divide all by the smallest to get
4.5/2.3 = about 2
2.3/2.3 = 1
9.1/2.3 = about 4
ratio is C2H4O