Question

An organic compound on analysis was found to contain $10.06\%$ carbon, $0.84\%$ hydrogen and $89.10\%$ chlorine. What will be the empirical formula of the substance?

• A. $C{H}_{2}C{l}_2$
• B. $CHC{l}_3$
• C. $CC{l}_4$
• D. $C{H}_{3}Cl$

Answer 1

Answer: (B)

Element	$\%$	Atomic mass	Molar ratio	Simple molar ratio
$C$	$10.06\%$	$12$	$\frac{10.06}{12}=0.84$	$\frac{0.84}{0.84}=1$
$H$	$0.84\%$	$1$	$\frac{0.84}{1}=0.84$	$\frac{0.84}{0.84}=1$
$Cl$	$89.10\%$	$35.5$	$\frac{89.10}{35.5}=2.5$	$\frac{2.5}{0.84}=3$

Thus, the emprical formula of the substance of $CHC{l}_3$.