5.2 cc of a gaseous hydrocarbon was exploded with excess of oxygen and product cooled. A contraction of 7.8 cc was observed. A further contraction of 10.4 cc was noted on treatment with aqueous potash. Find the formula of the hydrocarbon and give I.U.P.A.C. name to it.

5.2 cc of a gaseous hydrocarbon was exploded with excess of oxygen and

1.	5.2 cc of a gaseous hydrocarbon was exploded with excess of oxygen and product cooled. A contraction of 7.8 cc was observed. A further contraction of 10.4 cc was noted on treatment with aqueous potash. Find the formula of the hydrocarbon and give I.U.P.A.C. name to it.
Answer» Solution :Experimental values : `"Volume of HYDROCARBON taken = 5.2 cc"` `"Contraction on explosion and cooling = 7.8 cc"` `"Contraction on introducing KOH"` `"i.e.Volume of "CO_(2)" formed = 10.4 cc"` Theoretical values : Combustion of hydrocarbon (say `C_(X)H_(y)`) can be represented as follows :- `{:(C_(x)H_(y),+,(x+y//4)O_(2),rarr,xCO_(2),+,""y//2H_(2)O),("1 vol.",,x+y//4"vol.",,"x vol.",,"(negligible volume)"),("1 cc",,x+y//4" cc",,"x cc",,),("5.2 cc",,5.2(x+y//4)" cc",,"5.2 x cc",,):}` Equating experimental values and the theoretical values of carbon dioxide formed, we have `5.2x=10.4"or"x=2` Theoretical value of contraction after explosion and coling for 1 cc. of the hydrocarbon `=(1+x+y//4)-x=1+y//4" cc"` Since for 5.2 cc of hydrocarbon, contraction on explosion and cooling = 7.8 cc `therefore"For 1 cc of hydrocarbon, contrction on explosion and colling "=(7.8)/(5.2)=(3)/(2)` Equating the two values, we have `1+(y)/(4)=(3)/(2)"or"(y)/(4)=(1)/(2)"or"y=2` Hence the formula of the hydrocarbon is `C_(2)H_(2)" (acetylene)"` I.U.P.A.C. name for this compound is Ethyne.