A compound contains 40% C, 6.67% H and 53.33% O. Determine its empirical formula. If its molar mass is 180 g mol⁻¹, determine its molecular formula.

📘 Complete Step-by-Step Solution

📌 Given Data

C = 40%

H = 6.67%

O = 53.33%

Molar Mass = 180 g mol⁻¹

🔵 Step 1: Assume 100 g of Compound

When percentage composition is given, we assume the sample mass to be 100 g.

Mass of C = 40 g
Mass of H = 6.67 g
Mass of O = 53.33 g

Total = 40 + 6.67 + 53.33 = 100 g

🟢 Step 2: Calculate Number of Moles of Each Element

Moles = Mass ÷ Atomic Mass

Carbon

C: 40 ÷ 12

= 3.333 mol

Hydrogen

H: 6.67 ÷ 1

= 6.67 mol

Oxygen

O: 53.33 ÷ 16

= 3.333 mol

🟠 Step 3: Divide by the Smallest Number of Moles

The smallest value is approximately 3.333.

C: 3.333 ÷ 3.333 = 1

H: 6.67 ÷ 3.333 ≈ 2

O: 3.333 ÷ 3.333 = 1

🟣 Step 4: Write the Empirical Formula

C : H : O

= 1 : 2 : 1

Empirical Formula = CH₂O

✅ Empirical Formula = CH₂O

🎬 Molecular Composition Animation

C H H O

C : H : O → 1 : 2 : 1 → CH₂O

🔴 Step 5: Calculate Empirical Formula Mass

CH₂O

= 12 + (2 × 1) + 16

= 12 + 2 + 16

= 30 g mol⁻¹

🟢 Step 6: Calculate n

Use the relation:

n = Molecular Mass ÷ Empirical Formula Mass

n = 180 ÷ 30

n = 6

The molecular formula is 6 times the empirical formula.

🟣 Step 7: Calculate Molecular Formula

Molecular Formula = (Empirical Formula) × n

= (CH₂O) × 6

= C₆H₁₂O₆

✅ Molecular Formula = C₆H₁₂O₆

✅ Final Answer

Empirical Formula = CH₂O

Empirical Formula Mass = 30 g mol⁻¹

n = 180 ÷ 30 = 6

Molecular Formula = C₆H₁₂O₆

🎯 Quick Revision

Percentage → Mass → Moles
Moles → Divide by Smallest
Ratio → Empirical Formula
n = Molecular Mass ÷ Empirical Formula Mass
Molecular Formula = Empirical Formula × n