Question

The compound A has the following percentage composition by mass:

Carbon $26.7\%$, oxygen $71.1\%$, hydrogen $2.2\%$.

(b) If the relative molecular mass of A is $90$, what is the molecular formula of A?

Answer 1

Given information

• The percentage composition of Carbon$\left(\mathrm{C}\right)=26.7\%$
• Atomic weight of $\mathrm{C}=12\mathrm{g}$.
• The percentage composition of Oxygen$\left(\mathrm{O}\right)=71.1\%$
• Atomic weight of $\mathrm{O}=16\mathrm{g}$.
• The percentage composition of Hydrogen$\left(\mathrm{H}\right)=2.2\%$
• Atomic weight of $\mathrm{H}=1\mathrm{g}$.
• The relative molecular mass or molecular weight of compound A = $90$.

Formulas 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
C	$26.7$	$12$	$\frac{26.7}{12}=2.2$	$\frac{2.2}{2.2}=1$
O	$71.1$	$16$	$\frac{71.1}{16}=4.4$	$\frac{4.4}{2.2}=2$
H	$2.2$	$1$	$\frac{2.2}{1}=2.2$	$\frac{2.2}{2.2}=1$

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

Calculation of empirical formula weight of ${\mathrm{CO}}_2\mathrm{H}$

• The empirical formula weight is calculated as follows:

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

Calculation for the value of $\mathit{n}$

• The value of $n$ is calculated as follows:

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

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 ${\mathrm{C}}_2{\mathrm{H}}_2{\mathrm{O}}_4$.