20 cc of a hydrocarbon mixed with 66 cc of oxygen were exploded in a eudiometer tube. The residual gases after cooling occupied 56 cc. On treatment with KOH solution, the volume decreased to 16 cc. Find the formula of the hydrocarbon.

Explanation

Experimental Values : Volume of hydrocarbon taken = 20 cc
Volume of oxygen added = 66 cc
Volume after explosion and cooling
`"i.e., Volume of CO"_(2)" formed + unused oxygen = 56 cc"`
Volume after introducing KOH
`{:("i.e., ","Volume of unused oxygen = 16 cc"),(therefore" ","Volume of CO"_(2)" formed "=56-16="40 cc"),("and ","Volume of oxygen used"=66-16="50 cc"):}`
Theoretical values : Let `C_(x)H_(y)` be the formula of the gaseous hydrocarbon. The oxidation equaiton will be
`{:(C_(x)H_(y),+,(x+y//4)O_(2),rarr,xCO_(2)+y//2H_(2)O" (negligible volume)"),("1 vol.",,x+y//4"vol.",,"x vol."),("1 cc",,x+y//4" cc",,"x cc"),("20 cc",,20(x+y//4)" cc",,"20 x cc"):}`
Equating experimental and theoretical values of carbon dioxide formed and oxygen used, we get
`{:(,20x=40,or,x=2,),(and,20(x+y//4)=50,or,20(2+y//4)=50,"("because x = 2")"),(,,or,y=2,):}`
Hence formula of hydrocarbon is `C_(2)H_(2)` (Acetylene).