If a compound on analysis was found to contain `C=18.5%, h=1.55%,Cl=55.04%` and `O=25.81%`, then its empirical formul is

Options

A. `CHClO`
B. `CH_(2)ClO`
C. `C_(2)_H_(2)OCl`
D. `ClCH_(2)O`

Answer

a

Explanation

Element `% ` amount At. wt relative number simple ratio
`C=18.5%" "12" "(18.5)/(12)=1.54" "1`
`H=1.55%" "1" "(1.55)/(1)=1.55" "1`
`O=24.81%" "6" "(24.81)/(16)" "1`
`Cl=55.04% " "35.5" "(55.04)/(35.5)=1.55" "1`
Empirical formula is `CHClO`