Question

10 mL of gaseous organic compound contain C, H and O only was mixed with 100 mL of $O_2$ and exploded under identical conditions and then cooled. The volume left after cooling was 90 mL. On treatment with KOH a contraction of 20mL was observed. If vapour density of compound is 23, derive molecular formula of the compound.

• A. $C_2{H}_{6}O$
• B. $C_3{H}_{8}O$
• C. $C_4{H}_6{O}_2$
• D. $C_2{H}_6{O}_2$

Answer 1

The correct option is A $C_2{H}_{6}O$

On treatment with KOH a contraction of 20mL was observed. The contraction of 20 ml is due to the absorption of carbon dioxide. Thus 10 ml of compound gives 20 ml of carbon dioxide. The volume of carbon dioxide is twice the volume of compound.
Volume is directly proportional to number of moles.
Hence, 1 mole of compound contains 2 moles of C atoms.
Also, Molecular weight $=2\times$ vapour density $=2\times 23=46$ g/mol. The atomic weight of C is 12 g/mol. Two moles of C will weigh $2\times 12=24$ g.
Out of 46 g for 1 mole of compound, 24 g corresponds to 2 moles of C.
The remaining $46-24=22$ g corresponds to oxygen and hydrogen.
The atomic masses of O and H are 16 g/mol and 1 g/mol respectively.
Hence, 22 g can have only one mole of O atoms and 6 moles of H atoms.
Thus, the molecular formula of the compound is $C_2{H}_{6}O$.