What is the percent composition of a compound with the empirical formula #MgCl_2#?

1 Answer
Apr 21, 2016

Mg#Cl_2# ----> Mg + #Cl_2#

One mole of Mg#Cl_2# , gives one mole of Magnesium , and one mole of Chlorine gas. In terms of mass , 95 g of Mg#Cl_2# on decomposition gives 24 g of Magnesium, 71 g of Chlorine.

Mass percentage of Mg = (mass of Mg / mass of Mg#Cl_2# ) x 100

mass percentage of Mg = ( 24 g / 95 g) x 100 = 25.26 %

Mass percentage of Cl = (mass of Cl / mass of Mg#Cl_2# ) x 100

mass percentage of Cl = ( 71 g / 95 g) x 100 = 74.74 %