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Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1 % dioxygen by mass.
By BYJU'S Exam Prep
Updated on: September 25th, 2023
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.
Table of content
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.
