Question

A compound is found to contain 65.44% C, 5.49% H and 29.07% O .What is its empirical formula.

Answer 1

Element	%	At mass	% / at mass	Simple ratio	Whole no ratio
C	65.44	12	65.44 / 12 = 5.45	5.45 / 1.82 = 2.99	3
H	5.49	1	5.49 / 1 = 5.49	5.49 / 1.82 = 3.02	3
O	29.07	16	29.07 / 16 = 1.82	1.82 / 1.82 = 1	1

Thus the empirical formula is C₃H₃O