3.5 Determining Empirical and Molecular Formulas

Introduction

Chemists often need to know the exact composition of a compound — not just the percentage of each element, but the ratio in which the atoms are present. This is where empirical and molecular formulas come in.

• The empirical formula gives the simplest whole-number ratio of elements in a compound.
• The molecular formula gives the actual number of atoms of each element in a molecule.

Understanding how to determine these formulas is essential in:

• Drug design — to determine exact compositions of chemical compounds used in medications
• Materials science — to understand the structure of new materials
• Analytical chemistry — to identify unknown compounds
• Biochemistry — to understand the composition of biological molecules

Definition

Empirical Formula: The empirical formula of a compound tells you the simplest ratio of elements in the compound.

Molecular Formula: The molecular formula tells you the actual ratio of elements in the compound.

In simple words:

• Empirical formula = simplest whole-number ratio of atoms.
• Molecular formula = actual number of atoms in a molecule.

Example: For hydrogen peroxide:

• Empirical formula = HO (1:1 ratio)
• Molecular formula = H₂O₂ (actual numbers)

Main Idea

• The empirical formula is found from the percentage composition or mass data.
• The molecular formula is found by comparing the molar mass of the compound to the empirical formula mass.
• The molecular formula is always a whole-number multiple of the empirical formula.

Molecular formula = (Empirical formula) × n

Where n is a whole number.

Core Concept

Key point: The molecular formula is always a multiple of the empirical formula.

n = Molar mass of compoundMolar mass of empirical formula

Detailed Explanation

Determining the Empirical Formula

Determining the empirical formula of a compound involves the following steps:

Step 1: Find the number of moles of each element from its mass or percentage composition.

Moles = Mass or percentageAtomic mass

Step 2: Divide each element's molar amount by the smallest molar amount to get the simplest whole-number ratio.

Step 3: Write the empirical formula using the simplest ratio as subscripts.

Determining the Molecular Formula

Determining the molecular formula of a compound involves the following steps:

Step 1: Find the molar mass of the empirical formula.

Step 2: Divide the compound's molar mass by the empirical formula's molar mass. This will give you a whole number (n).

n = Molar mass of compoundMolar mass of empirical formula

Step 3: Multiply all the subscripts in the empirical formula by n to get the molecular formula.

Intermediate Level

Steps for empirical formula:

• Moles = mass or % / atomic mass.
• Divide by smallest moles.
• Write simplest ratio.

Steps for molecular formula:

• Find empirical formula mass.
• n = Molar mass of compound / Empirical formula mass.
• Multiply subscripts by n.

Example: 40% C, 6.7% H, 53.3% O:

• Moles: C = 3.33, H = 6.7, O = 3.33
• Ratio: 1:2:1 → CH₂O
• Empirical formula mass = 30
• n = 18030 = 6
• Molecular formula = C₆H₁₂O₆

Formula / Mathematical Form

Empirical Formula Steps:

Moles of each element = Mass or percentageAtomic mass

Simplest ratio = Moles of each elementSmallest number of moles

Molecular Formula Steps:

n = Molar mass of compoundMolar mass of empirical formula

Molecular formula = (Empirical formula) × n

Solved Example

Example 3.6

A compound has the following composition:

Carbon = 40%, Hydrogen = 6.7%, Oxygen = 53.3%

The molar mass of the compound is 180 g/mol. Determine its empirical and molecular formulas.

Solution:

Step 1: Find moles of each element.

C = 4012 = 3.33

H = 6.71 = 6.7

O = 53.316 = 3.33

Step 2: Divide by the smallest number (3.33).

C : H : O

3.33 : 6.7 : 3.33

3.333.33 : 6.73.33 : 3.333.33

1 : 2 : 1

Empirical formula = CH₂O

Step 3: Find the molar mass of the empirical formula.

Molar mass of CH₂O = 12 + (2 × 1) + 16 = 30\ g/mol

Step 4: Find n.

n = Molar mass of compoundMolar mass of empirical formula

n = 18030 = 6

Step 5: Multiply subscripts by n.

Molecular formula = (CH₂O) × 6 = C₆H₁₂O₆

• Empirical formula = CH₂O
• Molecular formula = C₆H₁₂O₆

Concept Assessment Exercise 3.7

1. Determine the empirical and molecular formula of a compound from the following data: 26.4% Carbon, 3.3% Hydrogen, 70.3% Oxygen, Molar Mass: 91.0 g/mol

Solution:

Step 1: Find moles.

C = 26.412 = 2.20

H = 3.31 = 3.30

O = 70.316 = 4.39

Step 2: Divide by smallest (2.20).

C : H : O

2.202.20 : 3.302.20 : 4.392.20

1 : 1.5 : 2

Multiply by 2 to get whole numbers:

2 : 3 : 4

Empirical formula = C₂H₃O₄

Step 3: Molar mass of empirical formula.

= (2 × 12) + (3 × 1) + (4 × 16)

= 24 + 3 + 64 = 91\ g/mol

Step 4: Find n.

n = 9191 = 1

Molecular formula = C₂H₃O₄

• Empirical formula = C₂H₃O₄
• Molecular formula = C₂H₃O₄

2. Determine the empirical and molecular formula of a compound composed of 18.24 g Carbon, 0.51 g Hydrogen, and 16.91 g Fluorine, which has a molar mass 562.0 g/mol.

Solution:

Step 1: Find moles.

C = 18.2412 = 1.52

H = 0.511 = 0.51

F = 16.9119 = 0.89

Step 2: Divide by smallest (0.51).

C : H : F

1.520.51 : 0.510.51 : 0.890.51

2.98 : 1 : 1.75

Multiply by 4 to get whole numbers:

12 : 4 : 7

Empirical formula = C₁₂H₄F₇

Step 3: Molar mass of empirical formula.

= (12 × 12) + (4 × 1) + (7 × 19)

= 144 + 4 + 133 = 281\ g/mol

Step 4: Find n.

n = 562281 = 2

Step 5: Multiply subscripts by n.

Molecular formula = (C₁₂H₄F₇) × 2 = C₂₄H₈F₁₄

• Empirical formula = C₁₂H₄F₇
• Molecular formula = C₂₄H₈F₁₄

Summary Table

Important Points

• - Empirical formula = simplest whole-number ratio of elements.
• - Molecular formula = actual number of atoms of each element.
• - Empirical formula steps:
• 1. Convert mass or % to moles.
• 2. Divide by smallest moles.
• 3. Write simplest ratio.
• - Molecular formula steps:
• 1. Find empirical formula.
• 2. Find empirical formula mass.
• 3. n = Molar mass of compound / Empirical formula mass.
• 4. Multiply subscripts by n.
• - The molecular formula is always a whole-number multiple of the empirical formula.
• - Percentage composition is used to find the empirical formula.
• - Molar mass is used to find the molecular formula.

Remember

• - Empirical formula = simplest ratio.
• - Molecular formula = actual ratio.
• - Empirical formula steps: moles → divide → ratio.
• - Molecular formula steps: empirical formula → n → multiply.
• - n = Molar mass / Empirical formula mass.
• - Example: CH₂O (empirical) → C₆H₁₂O₆ (molecular).
• - Percentage composition → empirical formula.
• - Molar mass → molecular formula.

Common Mistakes

Quick Revision

• Empirical Formula:
• The empirical formula of a compound tells you the simplest ratio of elements in the compound.

• Molecular Formula:
• The molecular formula tells you the actual ratio of elements in the compound.

Determining the Empirical Formula:

• Find the number of moles of each element from its mass or percentage composition.
• Divide each element's molar amount by the smallest or least molar amount to get the simplest whole-number ratio of the compound.
• Write the empirical formula using the simplest ratio as subscripts.

Determining the Molecular Formula:

• Find the molar mass of the empirical formula.
• Divide the compound's molar mass by the empirical formula's molar mass. This will give you a whole number (n).
• Multiply all the subscripts in the empirical formula by n to get the molecular formula.

• Example:
• 40% C, 6.7% H, 53.3% O; molar mass = 180 g/mol:

• Moles: C = 3.33, H = 6.7, O = 3.33
• Ratio: 1:2:1 → CH₂O
• Empirical formula mass = 30
• n = 18030 = 6
• Molecular formula = C₆H₁₂O₆

60-Second Revision

• Empirical formula = simplest ratio of elements.
• Molecular formula = actual ratio of elements.
• Empirical steps: moles → divide by smallest → ratio.
• Molecular steps: empirical formula → n → multiply.
• n = Molar mass / Empirical formula mass.
• Example: CH₂O (empirical) → C₆H₁₂O₆ (molecular).
• Percentage composition → empirical formula.
• Molar mass → molecular formula.
• ⚠️ Warning: always divide by the smallest number of moles to get the simplest ratio.