Question

A gaseous compound of nitrogen and hydrogen contains $12.5\%$ hydrogen by mass. Find the molecular formula of the compound if its relative molecular mass is $37$. $\left[\mathrm{N}=14,\mathrm{H}=1\right]$

Answer 1

Step 1: Given information

• The percentage of hydrogen in the compound is $12.5\%$
• Relative molecular mass is $37\mathrm{g}$.

Step 2: Calculation of percentage of nitrogen in the compound

• Since the compound is composed of nitrogen and hydrogen only, the percentage of nitrogen can be calculated by subtracting the %composition of hydrogen from 100%.
• The calculations are done as follows

$$\begin{gathered} \%\mathrm{Composition}\mathrm{of}\mathrm{Nitrogen}=100\%-\%\mathrm{Composition}\mathrm{of}\mathrm{Hydrogen} \\ =100\%-12.5\% \\ =87.5\% \end{gathered}$$

Step 3: Determination of empirical formula

Element	Percentage composition	Atomic weight	Atomic ratio	Simplest ratio
H	12.5	1	$\frac{12.5}{1}=12.5$	$\frac{12.5}{6.25}=2$
N	87.5	14	$\frac{87.5}{14}=6.25$	$\frac{6.25}{6.25}=1$

Hence, the empirical formula of the gaseous compound is ${\mathrm{NH}}_2$.

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

• The empirical formula weight is calculated as follows

$\mathrm{Empirical}\mathrm{formula}\mathrm{weight}=14+\left(2\times 1\right)=16\mathrm{g}$

• Molecular weight is $37\mathrm{g}$.
• Determine the value of $n$ as follows

$n\mathit{=}\frac{\text{Molecular weight}}{\text{Empirical formula weight}}\mathit{\text{=}}\frac{37}{16}\text{=2.31≃2}$

Step 5: Determination of the molecular formula

• The molecular formula is determined as follows

$$\begin{gathered} \text{Molecular weight}={\left(\mathrm{Empirical}\mathrm{formula}\right)}_{\mathit{n}} \\ ={\left({\mathrm{NH}}_2\right)}_2 \\ ={\mathrm{N}}_2{\mathrm{H}}_4 \end{gathered}$$

Therefore, the molecular formula of the compound is ${\mathrm{N}}_2{\mathrm{H}}_4$.