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

Apr 21, 2016

Mg$C {l}_{2}$ ----> Mg + $C {l}_{2}$

One mole of Mg$C {l}_{2}$ , gives one mole of Magnesium , and one mole of Chlorine gas. In terms of mass , 95 g of Mg$C {l}_{2}$ on decomposition gives 24 g of Magnesium, 71 g of Chlorine.

Mass percentage of Mg = (mass of Mg / mass of Mg$C {l}_{2}$ ) x 100

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

Mass percentage of Cl = (mass of Cl / mass of Mg$C {l}_{2}$ ) x 100

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