Question

A carbon compound containing only carbon and oxygen has a approximate molecule weight of $290\mathrm{g}$. On analysis, it is found to contain $50\%$ by weight of each element. What is the molecular formula of the compound?

Answer 1

Step 1: Given information

• Molecular weight of an organic$=290\mathrm{g}$
• Percentage composition of carbon $\left(\mathrm{C}\right)=50\%$
• Percentage composition of oxygen $\left(\mathrm{O}\right)=50\%$

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
C	$50$	$12$	$\frac{50}{12}=4.2$	$\frac{4.2}{3.1}=1.35$
O	$50$	$16$	$\frac{50}{16}=3.1$	$\frac{3.1}{3.1}=1$

• The simplest ratio of $\mathrm{C}:\mathrm{O}=1.35:1$, which is not a whole number ratio.
• Thus, multiplying by $3$, the simplest ratio of $\mathrm{C}:\mathrm{O}=4:3$
• Therefore, the empirical formula of the compound is ${\mathrm{C}}_4{\mathrm{O}}_3$.

Step 7: Calculation of value of $\mathit{n}$

• The empirical formula weight is calculated as follows:

$$\begin{gathered} \mathrm{Empirical}\mathrm{formula}\mathrm{weight}=\left(12\times 4\right)+\left(16\times 3\right) \\ =96 \end{gathered}$$

• Determine the value of $n$ as follows:

$$\begin{gathered} n=\frac{\mathrm{Molecular}\mathrm{weight}}{\mathrm{Empirical}\mathrm{formula}\mathrm{weight}} \\ \mathit{}\mathit{}\mathit{}\mathrm{=}\frac{290}{96} \\ =3 \end{gathered}$$

Step 7: Determination of the molecular formula

• The molecular formula is determined as follows:

$$\begin{gathered} \mathrm{Molecular}\mathrm{formula}={\left(\mathrm{Empirical}\mathrm{formula}\right)}_n \\ ={\left({\mathrm{C}}_4{\mathrm{O}}_3\right)}_3 \\ ={\mathrm{C}}_{12}{\mathrm{O}}_9 \end{gathered}$$

Therefore, the molecular formula of the compound is ${\mathrm{C}}_{12}{\mathrm{O}}_9$.