Share with your friends
Call

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.

Talk Doctor Online in Bissoy App
Call

% 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.

Talk Doctor Online in Bissoy App