An organic compound contains carbon, hydrogen and oxygen. Its elemental analysis gave C, 38.71% and H, 9.67%. The empirical formula of the compound would be

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Element	%	Atomic mass	mole ratio	simple ratio
$C$	38.71	$12$	$\frac{38.71}{12} = 3.22$	$\frac{3.22}{3.22} = 1$
$H$	9.67	$1$	$\frac{9.67}{1} = 9.67$	$\frac{9.67}{3.22} = 3$
$O$	51.62	$16$	$\frac{51.62}{16} = 3.22$	$\frac{3.22}{3.22} = 1$

Hence empirical formula of the compound would be $C H_{3} O$ .

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