Question

An organic compound contains Carbon, Hydrogen and Oxygen. Its elemental analysis gave $\mathrm{C}$,$38.71\%$ and $\mathrm{H}$, $9.67\%$.The empirical formula of the compound would be?

• A. $\mathrm{CHO}$
• B. ${\mathrm{CH}}_2\mathrm{O}$
• C. ${\mathrm{CH}}_3\mathrm{O}$
• D. ${\mathrm{CH}}_4\mathrm{O}$

Answer 1

The correct option is C

${\mathrm{CH}}_3\mathrm{O}$

The correct option is (C):

Explanation for the correct answer:

${\mathrm{CH}}_3\mathrm{O}$:

Step 1: Mole ratio for Carbon (C), Hydrogen (H) and Oxygen (O)

Element Carbon:

The percentage composition of Carbon : $38.71\%$

Atomic mass of Carbon: $12$

So, $\mathrm{Mole}\mathrm{ratio}=\frac{\%\mathrm{composition}}{\mathrm{Atomic}\mathrm{mass}}$

$\mathrm{Mole}\mathrm{ratio}=\frac{38.71}{12}=3.22$

Element Hydrogen:

The percentage composition of Hydrogen : $9.67\%$

Atomic mass of Hydrogen: $1$

So, $\mathrm{Mole}\mathrm{ratio}=\frac{\%\mathrm{composition}}{\mathrm{Atomic}\mathrm{mass}}$

$\mathrm{Mole}\mathrm{ratio}=\frac{9.67}{1}=9.67$

Element Oxygen

The percentage composition of Oxygen: $100-(38.71+9.67)=51.62\%$

Atomic mass of Oxygen:$16$

So, $\mathrm{Mole}\mathrm{ratio}=\frac{\%\mathrm{composition}}{\mathrm{Atomic}\mathrm{mass}}$

$\mathrm{Mole}\mathrm{ratio}=\frac{51.62}{16}=3.22$

Step 2: Simple ration for C, H and O.

For carbon (C)

$\mathrm{Simple}\mathrm{ratio}:\frac{\mathrm{mole}\mathrm{ratio}}{\mathrm{smallest}\mathrm{mole}\mathrm{ratio}}=\frac{3.22}{3.22}=1$

Hence, Carbon: 1

For Hydrogen (H)

$\mathrm{Simple}\mathrm{ratio}:\frac{\mathrm{mole}\mathrm{ratio}}{\mathrm{smallest}\mathrm{mole}\mathrm{ratio}}=\frac{9.67}{3.22}=3$

Hence, Hydrogen: 3

For Oxygen (O)

$\mathrm{Simple}\mathrm{ratio}:\frac{\mathrm{mole}\mathrm{ratio}}{\mathrm{smallest}\mathrm{mole}\mathrm{ratio}}=\frac{3.22}{3.22}=1$

Hence, Oxygen: 1

Therefore, the empirical formula of the compound is Option (C): ${\mathrm{CH}}_3\mathrm{O}$