Question

Write any two examples of compounds which have same molecular and empirical formulae.

Answer 1

Part 1: Empirical formula of a compound:-

1. It is the simplest formula, that shows the simplest whole-number ratio between the atoms of the elements in the compound.
2. The empirical formula of a compound represents only the atomic ratio of the various elements present in it's molecule.

Part 2: Relation between molecular formula and empirical formula:-

Since,

$$\begin{gathered} \mathrm{molecular}\mathrm{weight}=\mathrm{n}\times \mathrm{empirical}\mathrm{formula}\mathrm{weight} \\ \Rightarrow \mathrm{n}=\frac{\mathrm{Molecular}\mathrm{weight}}{\mathrm{Empirical}\mathrm{formula}\mathrm{weight}} \end{gathered}$$

Therefore, $\mathrm{Molecular}\mathrm{formula}={(\mathrm{Empirical}\mathrm{formula})}_{\mathrm{n}}$

If $\mathrm{n}=1$, the molecular formula and empirical formula are the same.

Part 3: Molecular and empirical formula of Formaldehyde (${\mathrm{CH}}_2\mathrm{O}$),:-

1. Empirical formula of formaldehyde is

Compound	Molecular Formula	Whole Numbers	Simplest Ratio of Empirical Formula
Formaldehyde	${\mathrm{CH}}_2\mathrm{O}$	$1:2:1$	${\mathrm{CH}}_2\mathrm{O}$

$$\begin{gathered} \therefore \mathrm{n}=\frac{\mathrm{Molecular}\mathrm{weight}\mathrm{of}\mathrm{formaldehyde}}{\mathrm{Empirical}\mathrm{formula}\mathrm{weight}\mathrm{of}\mathrm{formaldehyde}} \\ \Rightarrow \mathrm{n}=\frac{12+(2\times 1)+16}{12+(2\times 1)+16} \\ \Rightarrow \mathrm{n}=1 \end{gathered}$$

Since $\mathrm{n}=1$, as a result, the molecular formula of formaldehyde (${\mathrm{CH}}_2\mathrm{O}$) is the same as the empirical formula.

Part 4: Molecular and empirical formula of methane (${\mathrm{CH}}_4$),:-

1. Empirical formula of methane(${\mathrm{CH}}_4$) is

Compound	Molecular Formula	Whole Numbers	Simplest Ratio of Empirical Formula
Methane	${\mathrm{CH}}_4$	$1:4$	${\mathrm{CH}}_4$

$$\begin{gathered} \therefore \mathrm{n}=\frac{\mathrm{Molecular}\mathrm{weight}\mathrm{of}\mathrm{methane}}{\mathrm{Empirical}\mathrm{formula}\mathrm{weight}\mathrm{of}\mathrm{methane}} \\ \Rightarrow \mathrm{n}=\frac{12+(4\times 1)}{12+(4\times 1)} \\ \Rightarrow \mathrm{n}=1 \end{gathered}$$

Since $\mathrm{n}=1$, as a result, the molecular formula of formaldehyde (${\mathrm{CH}}_4$) is the same as the empirical formula.

Hence, two examples of compounds which have same molecular and empirical formulae are Formaldehyde(${\mathrm{CH}}_2\mathrm{O}$)and Methane(${\mathrm{CH}}_4$)