On reduction with hydrogen 3.6 gm of an oxide left 3.2 gm of tje metal. If he atomic weight of the metal is 64, the simplest formula of the oxide would be:
Given,
3.6 g of the oxide contains 3.2 g of the metal
Atomic mass of metal = 64
Metal : 3.2 g
Mole ratio = 3.2 / 64 = 1/20
Oxygen : 3.6 - 3.2 = 0.4 g
Mole ratio = 0.4 / 16 = 1/40
Hence, simplest formula is M2O
3.6 g of the oxide contains 3.2 g of the metal
Atomic mass of metal = 64
Metal : 3.2 g
Mole ratio = 3.2 / 64 = 1/20
Oxygen : 3.6 - 3.2 = 0.4 g
Mole ratio = 0.4 / 16 = 1/40
Hence, simplest formula is M2O