Question

What is the empirical formula of ${\mathrm{C}}_2{\mathrm{H}}_2$ and ${\mathrm{C}}_6{\mathrm{H}}_6$?

Answer 1

Part 1: Defining Empirical and molecular formula

Empirical formula:

• An empirical formula is a formula in which the atoms in a molecule are present in their lowest ratio.
• An empirical formula gives the simplest ratio of atoms in a molecule or compound.

Molecular formula:

• The molecular formula is the formula in which the atoms in a molecule is present in fixed ratio.
• The molecular formula gives the exact number of atoms in a molecule or compound.

Relation between molecular formula and empirical formula:

• $\mathrm{n}=\frac{\text{Molecular formula}}{\text{Empirical formula}}$
• Where, $n$ is the whole number multiple that is $1,2,3,4\dots .$so on.

Part 2: Determining the Empirical formula of ${\mathrm{C}}_2{\mathrm{H}}_2$

• The molecular formula of the compound is ${\mathrm{C}}_2{\mathrm{H}}_2$.
• In this compound, the value of $‘n’$ is $2$ that means the subscript of Carbon and Hydrogen are divided by the whole number $2$ and we get an empirical formula for ${\mathrm{C}}_2{\mathrm{H}}_2$.

$$\begin{gathered} 2=\frac{{\text{C}}_2{\text{H}}_2}{\text{Empirical formula}} \\ \text{Empirical formula =}\frac{{\text{C}}_2{\text{H}}_2}{2} \\ \text{Empirical formula = CH} \end{gathered}$$

Part 3: Determining the Empirical formula of ${\mathrm{C}}_6{\mathrm{H}}_6$

• The molecular formula of the compound is ${\mathrm{C}}_6{\mathrm{H}}_6$.
• In this compound, the value of $‘n’$ is $6$ that means the subscript of Carbon and Hydrogen are divided by the whole number $6$ and we get an empirical formula for compound $2^{nd}$.

$$\begin{gathered} 6=\frac{{\text{C}}_6{\text{H}}_6}{\text{Empirical formula}} \\ \text{Empirical formula =}\frac{{\text{C}}_6{\text{H}}_6}{6} \\ \text{Empirical formula = CH} \end{gathered}$$

In both compounds, the ratio of atoms of Carbon and Hydrogen are $1:1$.

Thus, the empirical formula of both ${\mathrm{C}}_2{\mathrm{H}}_2$ and ${\mathrm{C}}_6{\mathrm{H}}_6$ is $\mathrm{CH}$.