Define empirical formula and molecular formula. Explain the method of determining the empirical formula of a compound from its percentage composition.

📘 Complete Explanation

🔵 1. Empirical Formula

The empirical formula represents the simplest whole-number ratio of atoms of different elements present in a compound.

Example: C₆H₁₂O₆ → CH₂O

For glucose, the ratio 6 : 12 : 6 is simplified to 1 : 2 : 1. Therefore, its empirical formula is CH₂O.

🟠 2. Molecular Formula

The molecular formula shows the actual number of atoms of each element present in one molecule of a compound.

Glucose:

C₆H₁₂O₆

The molecular formula may be an integral multiple of the empirical formula.

Molecular Formula = n × Empirical Formula

🔗 Difference Between Them

Empirical Formula	Molecular Formula
Simplest ratio	Actual number of atoms
Example: CH₂O	Example: C₆H₁₂O₆
May not represent one molecule	Represents one molecule

🧮 Method to Determine Empirical Formula from Percentage Composition

The empirical formula can be determined from percentage composition using the following steps:

Percentage → Mass → Moles → Simplest Ratio → Formula

Step 1: Assume 100 g of the Compound

When percentages are given, assume that the total mass of the compound is 100 g.

Thus, the percentage of each element becomes its mass in grams.

40% C = 40 g C

6.67% H = 6.67 g H

53.33% O = 53.33 g O

Step 2: Convert Mass into Moles

Use:

Number of moles = Mass ÷ Atomic Mass

Element	Mass	Atomic Mass	Moles
C	40 g	12	3.33
H	6.67 g	1	6.67
O	53.33 g	16	3.33

Step 3: Divide by the Smallest Number

The smallest number of moles is approximately 3.33. Divide all mole values by 3.33.

C = 3.33 ÷ 3.33 = 1

H = 6.67 ÷ 3.33 ≈ 2

O = 3.33 ÷ 3.33 = 1

C : H : O

1 : 2 : 1

🎬 Percentage Composition → Empirical Formula

C H H O

Moles: 3.33 : 6.67 : 3.33

↓
Divide by 3.33

↓
1 : 2 : 1

↓
CH₂O

The simplest whole-number ratio gives the empirical formula.

📝 Complete Example

A compound contains 40% carbon, 6.67% hydrogen and 53.33% oxygen. Find its empirical formula.

Assume 100 g

C = 40 g, H = 6.67 g, O = 53.33 g

Moles = 3.33 : 6.67 : 3.33

Simplest Ratio = 1 : 2 : 1

Empirical Formula = CH₂O

✅ Final Answer

Empirical Formula
Simplest whole-number ratio of atoms

Molecular Formula
Actual number of atoms in one molecule

Method: % → Mass → Moles → Divide by Smallest → Simplest Ratio

🎯 Quick Revision

1️⃣ Assume 100 g
2️⃣ Write masses of elements
3️⃣ Convert masses into moles
4️⃣ Divide by the smallest mole value
5️⃣ Obtain the simplest whole-number ratio
6️⃣ Write the empirical formula

Example: C₆H₁₂O₆ → CH₂O