Question

A organic compound having carbon, hydrogen, oxygen and nitrogen was found to contain $\mathrm{C}=41.37\%,\mathrm{H}=5.75\%,\mathrm{O}=36.8\%$ and the rest nitrogen. If the vapor density of the compound is $43.3$, calculate the molecular formula of the compound.

Answer 1

Step 1: Given information

• A organic compound contains carbon, hydrogen, oxygen and nitrogen.
• The percentage of Carbon $\left(\mathrm{C}\right)=41.37\%$
• The percentage of Hydrogen $\left(\mathrm{H}\right)=5.75\%$
• The percentage of Oxygen $\left(\mathrm{O}\right)=36.8\%$
• The vapor density of the compound is $43.3$.

Step 2: Calculation of percentage of nitrogen in the compound

• The percentage of nitrogen is calculated as follows:

$$\begin{gathered} \mathrm{Percentage}\mathrm{of}\mathrm{nitrogen}=\left(100-41.37-5.75-36.8\right)\% \\ =16.08\% \end{gathered}$$

Step 3: Formula used for the determination of the atomic ratio and simplest ratio

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

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

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

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

Step 4: Determination of the empirical formula

Element	Percentage composition	Atomic weight	Atomic ratio	Simplest ratio
C	$41.37$	$12$	$\frac{41.37}{12}=3.45$	$\frac{3.45}{1.15}=3$
H	$5.75$	$1$	$\frac{5.75}{1}=5.75$	$\frac{5.75}{1.15}=5$
N	$16.08$	$14$	$\frac{16.08}{14}=1.15$	$\frac{1.15}{1.15}=1$
O	$36.8$	$16$	$\frac{36.8}{16}=2.3$	$\frac{2.3}{1.15}=2$

• The simplest ratio of $\mathrm{C}:\mathrm{H}:\mathrm{N}:\mathrm{O}=3:5:1:2$ is a whole number ratio.
• Thus, the empirical formula of the compound is ${\mathrm{C}}_3{\mathrm{H}}_5{\mathrm{NO}}_2$.

Step 5: Calculate relative molecular mass or molecular weight

• The calculation for molecular weight is as follows:

$$\begin{gathered} \text{Molecular weight=2×Vapor density} \\ \text{=2×43.3} \\ \text{=86.6} \end{gathered}$$

Step 6: Calculation of value of $\mathit{n}$

• The empirical formula weight is calculated as follows:

$$\begin{gathered} \mathrm{Empirical}\mathrm{formula}\mathrm{weight}=\left(12\times 3\right)+\left(1\times 5\right)+14+\left(16\times 2\right) \\ =87 \end{gathered}$$

• Determine the value of $n$ as follows:

$$\begin{gathered} n=\frac{\mathrm{Molecular}\mathrm{weight}}{\mathrm{Empirical}\mathrm{formula}\mathrm{weight}} \\ \mathit{}\mathit{}\mathit{}\mathrm{=}\frac{86.6}{87} \\ =0.99\simeq 1 \end{gathered}$$

Step 7: Determination of the molecular formula

• The molecular formula is determined as follows:

$$\begin{gathered} \mathrm{Molecular}\mathrm{formula}={\left(\mathrm{Empirical}\mathrm{formula}\right)}_n \\ ={\left({\mathrm{C}}_3{\mathrm{H}}_5{\mathrm{NO}}_2\right)}_1 \\ ={\mathrm{C}}_3{\mathrm{H}}_5{\mathrm{NO}}_2 \end{gathered}$$

Therefore, the molecular formula of the compound is ${\mathrm{C}}_3{\mathrm{H}}_5{\mathrm{NO}}_2$.