Question

An organic compound contains $40$% $C$, $6.66$% $H$ and $53.34$% $O$. The empirical formula of the compound is:

• A. $C{H}_2$
• B. $C{H}_{2}O$
• C. $CHO$
• D. $CH{O}_2$

Answer 1

The correct option is B $C{H}_{2}O$

Let us assume the mass of the compound = 100 gm.

Mass of carbon = 40 gm

Moles of carbon $=\frac{40}{12}$ moles

Mass of hydrogen = 6.66 gm

Moles of hydrogen $=\frac{6.66}{1}$ moles

Mass of oxygen = 53.34 gm

Moles of oxygen $=\frac{53.34}{16}$ moles

Ratio of moles $=$ $C:H:O=\frac{40}{12}:\frac{6.66}{1}:\frac{53.34}{16}=1:2:1$

Therefore, the empirical formula is $C{H}_{2}O$.