Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass.



Q.3:- Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass.

 

 

Answer:-

% of iron by mass = 69.9 % [Given]
% of oxygen by mass = 30.1 % [Given]



Atomic mass of iron = 55.85 amu
Atomic mass of oxygen = 16.00 amu
Relative moles of iron in iron oxide = %mass of iron by mass/Atomic mass of iron = 69.9/55.85 = 1.25
Relative moles of oxygen in iron oxide = %mass of oxygen by mass/Atomic mass of oxygen = 30.01/16=1.88
Simplest molar ratio = 1.25/1.25 : 1.88/1.25
 ⇒ 1 : 1.5 = 2 : 3

∴ The empirical formula of the iron oxide is Fe2O3.