Question

The percentage composition of a compound having the molecular formula weight $122.5$ is $\mathrm{K}=31.84\%,\mathrm{Cl}=28.98\%$ and $\mathrm{O}=39.18\%$. Calculate the molecular formula of the compound.

Answer 1

Step 1: Given information

• The percentage composition of Potassium $\left(\mathrm{K}\right)=31.84\%$
• Atomic weight of $\mathrm{K}=39\mathrm{g}$.
• The percentage composition of Chlorine $\left(\mathrm{Cl}\right)=28.98\%$
• Atomic weight of $\mathrm{Cl}=35.5\mathrm{g}$.
• The percentage composition of Oxygen $\left(\mathrm{O}\right)=39.18\%$
• Atomic weight of $\mathrm{O}=16\mathrm{g}$.
• Molecular formula weight is $122.5$.

Step 2: The formula used to calculate the atomic ratio and simplest ratio:

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

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

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

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

Step 3: Determination of the empirical formula:

Element	Percentage composition	Atomic weight	Atomic ratio	Simplest ratio
K	$31.84$	$39$	$\frac{31.84}{39}=0.82$	$\frac{0.82}{0.82}=1$
Cl	$28.98$	$35.5$	$\frac{28.98}{35.5}=0.82$	$\frac{0.82}{0.82}=1$
O	$39.18$	$16$	$\frac{39.18}{16}=2.45$	$\frac{2.45}{0.82}=3$

• The simplest ratio of $\mathrm{K}:\mathrm{Cl}:\mathrm{O}=1:1:3$ is a whole number ratio.
• Thus, the empirical formula of the compound is ${\mathrm{KClO}}_3$.

Step 4: Calculation of value of $\mathit{n}$:

• The empirical formula weight is calculated as follows:

$$\begin{gathered} \mathrm{Empirical}\mathrm{formula}\mathrm{weight}=36+35.5+\left(16\times 3\right) \\ =122.5 \end{gathered}$$

• Determine the value of $n$ as follows:

$$\begin{gathered} n=\frac{\mathrm{Molecular}\mathrm{weight}}{\mathrm{Empirical}\mathrm{formula}\mathrm{weight}} \\ \mathit{}\mathit{}\mathit{}\mathrm{=}\frac{122.5}{122.5} \\ =1 \end{gathered}$$

Step 5: Determination of the molecular formula:

• The molecular formula is determined as follows:

$$\begin{gathered} \text{Molecular formula}={\left(\mathrm{Empirical}\mathrm{formula}\right)}_n \\ ={\left({\mathrm{KClO}}_3\right)}_1 \\ ={\mathrm{KClO}}_3 \end{gathered}$$

Therefore, the molecular formula of the compound is ${\mathrm{KClO}}_3$.