If a particular ether contains $$60\% $$  carbon and $$13.3\% $$  hydrogen, its formula could be

Solution :
Let the formula be $${C_x}{H_y}{O_z}$$
Molecular weight $$ = 12x + y + 16z;$$
$$\% $$  of oxygen $$ = 100\, - $$  $$\left[ {\,\,\% } \right.$$  of carbon $$ + \,\,\% $$  of hydrogen $$\left. {} \right]$$
$$\eqalign{ & = 100 - \left[ {60 + 13.3} \right] \cr & = 26.7\% \cr} $$
$$\% $$  of carbon $$ = \frac{{12x}}{{12x + y + 16z}} \times 100 = 60\% $$
$$\% $$  of hydrogen $$ = \frac{y}{{12x + y + 16z}} \times 100 = 13.3\% $$
$$\% $$  of oxygen $$ = \frac{{16z}}{{12x + y + 16z}} \times 100 = 26.7\% $$
On solving the above equations, we get the ratio of $$x:y:z = 3:8:1$$     (approximate)
∴ Formula of compound is $${C_3}{H_8}O.$$