Empirical and Molecular Formula of a Compound

Ask a Question, Get an Answer!

Type your question here

Hundreds of tutors are online and ready to help you right now!

Empirical formula is the formula of a compound, which shows the simplest whole number ratio between the atoms of the elements in the compound. It does not indicate the actual number of atoms of the elements present but the simplest whole number ratio.

Example

How can we differentiate a molecular formula from an empirical formula?

If the subscripts in the formula have a common divisor, it is usually a molecular formula. Generally the empirical formula is multiplied by this common divisor to get the molecular formula.

Example:

Empirical formula of acetic acid is CH₂O.

Molecular formula is CH₃ COOH = C₂H₄O₂

C₁H₂ O₁ x 2 = C₂H₄O₂ [Molecular formula]

Numericals based on empirical formula

Example:1

An oxide of iron contains 72.41% of iron. Calculate the empirical formula for the oxide of iron [Fe = 56; O=16].

Solution

Therefore simple ratio = Fe₃O₄.

Empirical formula = Fe₃O₄.

Steps for calculation

• Calculate the percentage by weight of each element.

• Find out relative number of atoms by dividing percentage of weight by atomic weight.

• Choose the simplest ratio and the smallest, divide all the ratios by it.

• If whole numbers are not obtained, then multiply it by a smallest integer to make it whole.

Example: 2

The percentage composition of a compound is 71.8% antimony (Sb) and 28.2% sulphur. What is the empirical formula of this compound?

Solution:

In 100g of the compound, masses of elements are as follows:

In 100 grams of compound, 71.8 g are antimony and 28.2% sulphur.

Empirical formula is Sb₂S₃.

Determination of molecular formula from empirical formula

Molecular formula is the chemical formula, which represents the actual numbers of atoms of each element present in a compound.

Example: 1

Calculate the molecular formula of a compound with vapor density of 30 having 40% carbon; 6.67% of hydrogen and the rest is oxygen.

Empirical formula = C₁H₂O₁

Empirical formula weight = 12 x 1 + 2 x 1 + 1 x 16

= 12 + 2 + 16

= 30 g

Molecular weight = 2 x vapor density

= 2 x 30

= 60

Molecular weight = n x empirical weight

60 = n x 30

Molecular formula = n x empirical formula

= 2 x CH₂O = C₂H₄O₂

Steps

• Calculate empirical formula.

• Use vapor density if given.

• If molecular weight given, calculate 'n' using this formula.

• Molecular formula = n x empirical formula

Example: 2

A compound has molecular formula C₅H_10. What is its empirical formula?

Solution:

Ratio of C atoms to H atoms is 5 : 10 = 1:2.

Empirical formula is C₁H_2.