When hydrocarbons are burned in limited amount of air both CO and CO_(2) are fomed. When 54 gm of a particular hydrocarbon was burned in air 56 gm of CO, 88 gm of CO_(2) and 54 gm of H_(2)O were formed. What is emperical formula of compound ?

When hydrocarbons are burned in limited amount of air both CO and CO_(

1.	When hydrocarbons are burned in limited amount of air both CO and CO_(2) are fomed. When 54 gm of a particular hydrocarbon was burned in air 56 gm of CO, 88 gm of CO_(2) and 54 gm of H_(2)O were formed. What is emperical formula of compound ?
Answer» `C_(2)H_(2)` `C_(2)H_(2)` `C_(4)H_(6)` `C_(2)H_(3)` Solution :`{:(Sol.(i),C_(x)H_(y),+,(x + (y)/(2)) O_(2),rarr,xCO,+,(y)/(2)H_(2)O,),("mole",a_(0),,(x + (y)/(4)) a_(0),,,,,),(,0,,"0",,xa_(0),,(y)/(2) a_(0),),(,CO,+,(1)/(2)O_(2),rarr,,CO_(2),,),(,xa_(0) - z,,(z)/(2),,,z,,):}` `n_(CO) = xa_(0) - z = (56)/(28) = 2` .........(i) `N_(CO_(2)) = z = (88)/(44) = 2` .......(ii) `xa_(0) = z + 2` `xa_(0) = 4` `n_(H_(2)O) = (y)/(2)a_(0) = (54)/(18) = 3` `ya_(0) = 6` .....(iii) Equation (ii)/(iii) `1(x)/(y) = (2)/(3)` Empirical formula `= C_(2)H_(3)` (ii) `C_(2)H_(3) + (11)/(2) O_(2) rarr 2CO_(2) + (3)/(2) H_(2)O` `n_(O_(2))` required `= (11)/(2)` `W_(O_(2)) = (11)/(2) XX 32 = 176 gm`