If a compound, on analysis was found to contain C =18.5 %, H =1.55 %, Cl =55.04 % and O = 24.81 %, then its empirical formula is

Question

If a compound, on analysis was found to contain C =18.5 %, H =1.55 %, Cl =55.04 % and O = 24.81 %, then its empirical formula is (A) C 2 H 2 OCl (B) CH 2 ClO (C) CHClO (D) ClCH 2 O

Tardigrade Admin · Accepted Answer

Elements	%	At. mass	Relative no. of atom	Simplest ratio of atoms
C	18.5	12	$\frac{18.5}{12} = 1.542$	1
H	1.55	1	$\frac{1.55}{1} = 1.55$	1
Cl	55.04	35.5	$\frac{55.04}{35.5} = 1.55$	1
O	24.81	16	$\frac{24.16}{16} = 1.538$	1

Therefore, empirical formula of the compound is CHClO.

Q. If a compound, on analysis was found to contain $C = 18.5%, H = 1.55%, Cl = 55.04%$ and $O =$ $24.81%$ , then its empirical formula is

$C_{2} H_{2} OCl$

$C H_{2} ClO$

$C H ClO$

$ClC H_{2} O$

Elements % At. mass Relative no. of atom Simplest ratio
of atoms

C 18.5 12 $\frac{18.5}{12} = 1.542$ 1

H 1.55 1 $\frac{1.55}{1} = 1.55$ 1

Cl 55.04 35.5 $\frac{55.04}{35.5} = 1.55$ 1

O 24.81 16 $\frac{24.16}{16} = 1.538$ 1

Therefore, empirical formula of the compound is CHClO.

Q. If a compound, on analysis was found to contain C=18.5%,H=1.55%,Cl=55.04% and O= 24.81%, then its empirical formula is

C2​H2​OCl

CH2​ClO

CHClO

ClCH2​O