Question

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

Answer 1

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 %