$0.262$g of a substance gave on combustion $0.361$g of $C{O}_2$ and $0.147$g Of $H_{2}O$. What is the empirical formula of the substance

$44$g of $C{O}_2$ contains $12$g of C. 

So,$0.361$g of $C{O}_2$ contains $12/ 44$x$0.361 = 0.098$g of C.

$18$g of $H_2{O}_{ }$ contains $2$g of H. 

So, $0.147$g of $H_2{O}_{ }$ contains $2/18 \times 0.147 = 0.016$g of $H$.

Amount of 'O' = $0.262 – 0.098 – 0.016 = 0.148$ g of O

% of $C$ = $0.098/ 0.262 x 100 = 37.4$

% of $H$ $=0.016/ 0.262 x 100 = 6.1$

% of $O$ $= 0.148 / 0.262 x 100 = 56.48$

$37.40/ 12 = 3.1$

$6.1/ 1= 6.1$

$56.48 / 16 = 3.5$

by dividing with lowest value(3.1)

Hence, $C$ : $H$ : $O$ = $1 : 2 : 1$. So, the formula is $C{H}_{2}O$.

Thus, the correct option is A.