10 mL of gaseous organic compound contain C, H
and O only was mixed with 100 mL of o2 and
exploded under identical conditions and then cooled.
The volume left after cooling was 90 mL. On
treatment with KOH a contraction of 20 mL was
observed. If vapour density of compound is 23,
derive molecular formula of the compound.
Answers
Answer:
C2H60
Explanation:
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× vapour density =2×23=46 g/mol. The atomic weight of C is 12 g/mol. Two moles of C will weigh 2×12=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
Answer:
C2H6O
Explanation:
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× vapour density =2×23=46 g/mol. The atomic weight of C is 12 g/mol. Two moles of C will weigh 2×12=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 C2H6O