Question

determine the empirical formula of an oxide of iron which has 69.9 percent iron 30.1% oxygen by mass​

Answer 1

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.