Question

Analysis of a sample of a compound shows that it contains 24.3% carbon, 4.1 % hydrogen and 71.6% chlorine. The empirical formula of the compound is:

• A. $C{H}_{2}Cl$
• B. $C_2{H}_{2}Cl$
• C. $C{H}_{2}C{l}_2$
• D. $CHC{l}_3$

Answer 1

The correct option is A $C{H}_{2}Cl$

Percentages of elements $=C\to 24.3$%, $H\to 4.1$%, $Cl\to 71.6$%

Dividing the percentages by atomic weights $=C=\frac{24.3}{12}=2$ $H=\frac{4.1}{1}=4$ $Cl=\frac{71.6}{35.5}=2$

Ratio of $C:H:Cl$ in the compound $=2:4:2=1:2:1$

$\therefore$ Emperical formula is ${CH}_{2}Cl$.