10 mL of a gaseous organic compound containing C, H and O only was mixed with 100 mL of oxygen and exploded under conditions which allowed the water formed to condense. The volume of the gas after explosion wa 90 mL. On treatment with potash solution, a further contraction of 20 mL in volume was observed. Given that the vapour density of the compound is 23, deduce the molecular formula. All volume measurements were carried out under the same conditions.

Explanation

Volume of unreacted `O_(2)+` vol. of `CO_(2)=90mL`.
(`H_(2)O` vapour condensed)
Volume of `CO_(2)` (absorbed by KOH) = 20 mL.
`therefore` volume of unreacted `O_(2)=(90-20)mL=70mL`
Volume of `O_(2)` reacted with 10 mL of compound
`=(100-70)mL=30mL`.
Thus,
`{:(,C_(x)H_(y)O_(z)"(suppose)"+O_(2)rarrCO_(2)+H_(2)O),(,"10 mL 30 mL 20 mL"),(or,"10 moles 30 moles 20 moles"):}`
Applying POAC for C atoms,
`x xx` moles of `C_(x)H_(y)O_(z)=1xx` moles of `CO_(2)`
`x xx10=1xx20, x=2`
Again applying POAC for H and O atoms, we get respectively,
`yxx` moles of `C_(x)H_(y)O_(z)=2xx` moles of `H_(2)O`
`10y=2xx` moles of `H_(2)O....(1)`
and `zxx` moles of `C_(x)H_(y)O_(z)+2xx` moles of `O_(2)`
`=2xx` moles of `CO_(2)+1xx` moles of `H_(2)O`
`10z+2xx30=2xx20+` moles of `H_(2)O`.
Eliminating moles of `H_(2)O` from eqns. (1) and (2), we get
`y-2z=4`.
Now, molecular wt. of the compound `=2xx23=46`.
`therefore` for the compound `C_(x)H_(y)O_(z)`,
`2xx12+yxx1+z+16=46`
`y+16z=22....(4)`
From eqns. (3) and (4), we get,
`y=6 and z=1`.
`therefore` formula of the compound is `C_(2)H_(6)O`.