Question

(a) (i) A compound has the following percentage composition by mass:

Carbon $14.4\%$, hydrogen $1.2\%$, and chlorine $84.5\%$. Determine the empirical formula of this compound. Work correct to $1$ decimal place.

$[\mathrm{H}=1,\mathrm{C}=12,\mathrm{Cl}=35.5]$

Answer 1

Given information.

• The percentage composition of Hydrogen $\left(\mathrm{H}\right)=1.2\%$
• Atomic weight of $\mathrm{H}=1\mathrm{g}$.
• The percentage composition of Carbon $\left(\mathrm{C}\right)=14.4\%$
• Atomic weight of $\mathrm{C}=12\mathrm{g}$.
• The percentage composition of Chlorine $\left(\mathrm{Cl}\right)=84.5\%$
• Atomic weight of $\mathrm{Cl}=35.5\mathrm{g}$

Formulas used for the calculation of atomic ratio and simplest ratio

• The atomic ratio is calculated by dividing the percentage composition of each element by its atomic weight.

$\mathrm{Atomic}\mathrm{ratio}=\frac{\mathrm{Percentage}\mathrm{composition}\mathrm{of}\mathrm{an}\mathrm{element}}{\mathrm{Atomic}\mathrm{weight}\mathrm{of}\mathrm{an}\mathrm{element}}$

• The simplest ratio is calculated by dividing the atomic ratio of each element by the smallest atomic ratio.

$\mathrm{Simplest}\mathrm{ratio}=\frac{\mathrm{Atomic}\mathrm{ratio}\mathrm{of}\mathrm{an}\mathrm{element}}{\mathrm{Smallest}\mathrm{atomic}\mathrm{ratio}}$

Determination of empirical formula

Element	Percentage composition	Atomic weight	Atomic ratio	Simplest ratio
$\mathrm{H}$	$1.2$	$1$	$\frac{1.2}{1}=1.2$	$\frac{1.2}{1.2}=1$
$\mathrm{C}$	$14.4$	$12$	$\frac{14.4}{12}=1.2$	$\frac{1.2}{1.2}=1$
$\mathrm{Cl}$	$84.5$	$35.5$	$\frac{84.5}{35.5}=2.38$	$\frac{2.38}{1.2}=2$

• The simplest ratio of $\mathrm{H}:\mathrm{C}:\mathrm{Cl}=1:1:2$, which is a whole number ratio.
• Thus, the empirical formula of the compound is ${\mathrm{HCCl}}_2$.