Answer: Given
Mass % of iron = 69.9 %
Mass % of oxygen = 30.1 %
Element |
Atomic mass |
Mass % |
Mass % / atomic mass |
Fe |
55.85 |
69.9 |
69.9/55.85 = 1.25 |
O |
16.00 |
30.1 |
30.1/16.00 = 1.88 |
Fe : O = 1.25 : 1.88
Converting in simple ratio we get
Fe : O = 2 : 3
Hence, the empirical formula of the iron oxide is Fe2O3.
% 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/Atomic mass of iron = 69.9/55.85 = 1.25 Relative moles of oxygen in iron oxide = %mass of oxygen /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.