Question
A compound contains $$54.55\% $$ carbon, $$9.09\% $$ hydrogen, $$36.36\% $$ oxygen. The empirical formula of this compound is :
A.
$${C_3}{H_5}O$$
B.
$${C_4}{H_8}{O_2}$$
C.
$${C_2}{H_4}{O_2}$$
D.
$${C_2}{H_4}O$$
Answer :
$${C_2}{H_4}O$$
Solution :
$$\eqalign{
& C\,\,\,\,\,\,54.55\,\,\,\,\,\,\frac{{54.55}}{{12}} = 4.5\,\,\,\,\,\,\,\frac{{4.5}}{{2.27}} = 2 \cr
& H\,\,\,\,\,9.09\,\,\,\,\,\,\,\,\frac{{9.09}}{1} = 9.09\,\,\,\,\,\,\,\,\frac{{9.09}}{{2.27}} = 4 \cr
& O\,\,\,\,\,36.36\,\,\,\,\,\frac{{36.36}}{{16}} = 2.27\,\,\,\,\,\frac{{2.27}}{{2.27}} = 1 \cr} $$
Hence empirical formula of the compound $$ = {C_2}{H_4}O$$