Empirical Formula and Molecular Formula

A brief representation of the molecule of the substance in terms of symbols of various elements present in it is called chemical formula of the compound.

The relative number of different types of atoms present in a compound gives the empirical formula of the compound.

In other words we can say that the empirical formula gives the simplest whole number ratio of different types of atoms present in a compound e.g., formula of hydrogen peroxide is H₂O₂. The simplest whole number ratio of the atoms is 1:1 and the empirical formula is HO.

Molecular formula expresses the exact number of different atoms in a compound. In other words molecular formula of the compound gives the actual ratio of the atoms of various elements present in a compound.

The molecular formula of glucose is C₆H₁₂O₆. One molecule of glucose contains 6 atoms of carbon, 12 atoms of hydrogen and 6 atoms of oxygen. The empirical formula for glucose is CH₂O. Glucose has 2 moles of hydrogen for every mole of carbon and oxygen. The formulas for water and hydrogen peroxide are:

• Water Molecular Formula: H₂O
• Water Empirical Formula: H₂O
• Hydrogen Peroxide Molecular Formula: H₂O₂
• Hydrogen Peroxide Empirical Formula: HO

In the case of water, the molecular formula and empirical formula are the same. Molecular formula = n × empirical formula

Where n is a whole number.

Sometimes, the empirical formula and molecular formula both can be same.

Example:- The empirical formula of Boron Hydride is BH₃. Calculate molecular formula when measured mass of compound is 27.66.

Solution: The atomic mass is given by = B + 3(H) = 10.81 + 3(1) = 13.81u

But, the measured molecular mass for Boron is given as 27.66u

By using the expression, Molecular formula = n × empirical formula

n = molecular formula/empirical formula = 27.66/13.81 = 2

Putting value of n = 2 in the empirical formula we get molecular formula as

Molecular formula = 2(BH₃) = B₂H₆.

Determination of Empirical formula

Example: What is the empirical formula of a compound containing 50.05% sulphur and 49.95% oxygen by mass?

Solution: Convert the percent to grams assuming that 100 g of the compound is present.

S = 50.05 g 
O = 49.95 g

2) Convert the mass to mole:

S = 50.05 g / 32.066 g/mol = 1.5608 mol 
O = 49.95 g / 16.00 g/mol = 3.1212 mol

3) Divide by the lowest, thus obtaining the smallest whole-number ratio:

S = 1.5608 / 1.5608 = 1 
O = 3.1212 / 1.5608 = 2

Element	%composition	Atomic mass	Relative no. of atoms	Simplest ratio	Whole number
S	50.05	32.066	50.05/32.06= 1.5608	1.5608/1.5608=1	1
O	49.95	16	49.95/16.00=3.1212	3.1212/1.5608=1.99	2

The emperical formula is SO₂

Element	% composition	Atomic mass	Gram atoms	Atomic ratio	Simplest Whole no. ratio
C	34.6	12	34.6/12=2.88	2.88/2.88=1	3
H	3.85	1	3.85/1=3.85	3.85/2.88=1.337	4
O	61.55	16	61.55/16=3.85	3.85/2.88=1.337	4

 Thus the empirical formula of the compound is C₃H₄O₄

Example: What will be the molecular formula for the above compound if the molecular weight for this compound is 64.07 g/mol?

Solution: Calculate the empirical formula weight (SO₂):

                           32 + 16 + 16 = 64

Divide the molecule weight by the empirical formula weight:

                               64.07 / 64 = 1

Thus n = 1

By using the expression, Molecular formula = n × empirical formula

Hence Molecular formula = Empirical formula= SO₂

Example: The molecular mass of benzene is 78g and its percentage composition is 92.3% C and 7.69% H. Determine the molecular formula of benzene.

Solution: Percentage composition = 92.3% C, 7.69% H

Mole ratio = 92.3/22.01 C = 7.69 C;       7.69/1.008 H = 7.63 H

Dividing both mole ratios by 7.63 gives 1.008C: 1H

Thus the empirical formula is CH. The molecular formula will be (CH)n

Empirical formula mass = 12.01+1.008 = 13.018amu

n = Molecular mass/ Empirical formula mass = 78/13.018 = 5.99 = 6

Therefore, molecular formula is C₆H₆