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.

20 cc of a hydrocarbon mixed with 66 cc of oxygen were exploded in a e

1.	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.
Answer» Solution :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).