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

By Ritesh|Updated : November 13th, 2022

The empirical formula of an oxide of iron which has 69.9% iron and 30.1 % dioxygen by mass is Fe2O3. The simplest whole-number ratio of the atoms of the various elements present in one compound molecule can be found in the empirical formula of the compound.

Step 1: Given, that iron oxide has a mass percentage of 69.9% iron and 30.1% dioxygen.

Thus, 69.9 g of iron and 30.1 g of dioxygen make up 100 g of iron oxide.

The number of moles of iron present in 100 g of iron oxide is 69.9/55.8 = 1.25

The number of moles of dioxygen present in 100 g of iron oxide is 30.1/32 = 0.94

Step 2:

The amount of oxygen atoms to iron atoms in a single formula unit of iron oxide is equal to (2 x 0.94)/ 1.25 = 1.5: 1 = 3: 2

Hence, the formula of iron oxide is Fe2O3.

Uses of Iron oxide

  • Iron is produced using iron(III) oxide as a feedstock.
  • A pigment is made from it. Illustration: Pigment Red 101 and Pigment Brown 6.
  • In cosmetics, it is employed.
  • In dental composites, it is utilised.
  • It is a crucial component in calamine lotion.
  • On metallic jewelry, it is used to apply the finishing polish.
  • Magnetic discs and magnetic tapes both use it.

Summary:

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

Fe2O3 is the empirical formula of an oxide of iron which has 69.9% iron and 30.1 % dioxygen by mass. In chemistry, the empirical formula of a chemical compound is the simplest whole-number ratio of atoms present in a compound.

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