Question

A hydrocarbon has the following percentage composition:

Hydrogen $=2.2\%$, Carbon $=26.6\%$ and oxygen $=71.2\%$.

Calculate the empirical formula of the compound. If its molecular mass is $90$, find its molecular formula.

$\left[\mathrm{C}=12,\mathrm{H}=1,\mathrm{O}=16\right]$

Answer 1

Step 1: Given information

• The percentage composition of Hydrogen$\left(\mathrm{H}\right)=2.2\%$
• Atomic weight of Hydrogen is $1\mathrm{g}$.
• The percentage of Carbon $\left(\mathrm{C}\right)=26.6\%$
• Atomic weight of Carbon is $12\mathrm{g}$.
• The percentage of Oxygen $\left(\mathrm{O}\right)=71.2\%$
• Atomic weight of Oxygen is $16\mathrm{g}$.
• Molecular weight is $90$.

Step 2: Formula used to calculate 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
C	$26.6$	$12$	$\frac{26.6}{12}=2.22$	$\frac{2.22}{2.2}=1$
H	$2.2$	$1$	$\frac{2.2}{1}=2.2$	$\frac{2.2}{2.2}=1$
O	$71.2$	$16$	$\frac{71.2}{16}=4.45$	$\frac{4.45}{2.2}=2$

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

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

• The empirical formula weight is calculated as follows:

$$\begin{gathered} \mathrm{Empirical}\mathrm{formula}\mathrm{weight}=12+1+\left(16\times 2\right) \\ =45 \end{gathered}$$

• Determine the value of $n$ as follows:

$$\begin{gathered} n=\frac{\mathrm{Molecular}\mathrm{weight}}{\mathrm{Empirical}\mathrm{formula}\mathrm{weight}} \\ \mathit{}\mathit{}\mathit{}\mathrm{=}\frac{90}{45} \\ =2 \end{gathered}$$

Step 5: Determination of the molecular formula

• The molecular formula is determined as follows:

$$\begin{gathered} \text{Molecular formula}={\left(\mathrm{Empirical}\mathrm{formula}\right)}_n \\ ={\left({\mathrm{CHO}}_2\right)}_2 \\ ={\mathrm{C}}_2{\mathrm{H}}_2{\mathrm{O}}_4 \end{gathered}$$

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