Question #67dbe

1 Answer
Oct 5, 2015

Empirical Formula is #C_4H_5# and the molecular formula is #C_8H_10#.

Explanation:

#"Number of moles of organic compound" = 0.2612/106#
#=> n=2.4642xx10^-3mol#
#"Number of moles of "CO_2 = 0.8661/44#
#=> n= 0.0197mol#
#"Number of moles of " H_2O = 0.2250/18#
#=> n= 0.0125mol#
The equation for the reaction is:
#C_aH_b + a*b*O_2 -> a*CO_2 + b/2*H_2O#

Mole ratio of #CO_2:"organic compund"= 0.0197/(2.4642xx10^-3#
#= 8:1#
This means #8mol# of #CO_2# is produced for burning #1mol# of the organic compound. The value of #a# is therefore 8.
Mole ratio of #H_2O: "organic compound" = 0.0125/(2.4642xx10^-3#
#~~5:1#
This shows #5mol# of #H_2O# is produced. So, #b/2 = 5#, therefore #b# is equal to 10.
Hence, the organic compound has molecular formula #C8H10#. It's empirical formula would hence be #C_4H_5#. The name of this compound is Xylene, or dimethyl benzene.

**n.b. this is not an orthodox method. If this question is asked in an exam, you may not be awarded any marks for using this method.