Empirical and Molecular Formulas

Overview

The empirical formula shows the simplest whole-number ratio of atoms in a compound. The molecular formula shows the actual number of atoms in one molecule and is a whole-number multiple of the empirical formula.

Key Definitions

Empirical Formula

• Simplest whole-number ratio of atoms
• Example: CH₂O (glucose's empirical formula)

Molecular Formula

• Actual number of atoms in one molecule
• Example: C₆H₁₂O₆ (glucose's molecular formula)

Relationship

$$\text{Molecular Formula} = (\text{Empirical Formula})_n$$

Where $n = \frac{\text{Molar mass}}{\text{Empirical formula mass}}$

Finding Empirical Formula from Percent Composition

Step-by-Step Method

1. Assume 100 g sample (so percentages = grams)
2. Convert grams to moles for each element
3. Divide all by the smallest number of moles
4. Round to whole numbers (or multiply if needed)

Example 1: Basic Calculation

A compound is 40.0% C, 6.7% H, and 53.3% O.

Step 1: Assume 100 g

• C: 40.0 g, H: 6.7 g, O: 53.3 g

Step 2: Convert to moles

$$\text{C: } \frac{40.0}{12.01} = 3.33 \text{ mol}$$ $$\text{H: } \frac{6.7}{1.008} = 6.65 \text{ mol}$$ $$\text{O: } \frac{53.3}{16.00} = 3.33 \text{ mol}$$

Step 3: Divide by smallest (3.33)

$$\text{C: } \frac{3.33}{3.33} = 1, \quad \text{H: } \frac{6.65}{3.33} = 2, \quad \text{O: } \frac{3.33}{3.33} = 1$$

Empirical formula: CH₂O

Example 2: Non-Whole Numbers

A compound is 43.6% P and 56.4% O.

$$\text{P: } \frac{43.6}{30.97} = 1.408 \text{ mol}$$ $$\text{O: } \frac{56.4}{16.00} = 3.525 \text{ mol}$$

Divide by smallest:

$$\text{P: } \frac{1.408}{1.408} = 1, \quad \text{O: } \frac{3.525}{1.408} = 2.50$$

Multiply by 2:

$$\text{P: } 1 \times 2 = 2, \quad \text{O: } 2.50 \times 2 = 5$$

Empirical formula: P₂O₅

Common Multipliers

Decimal	Multiply by
0.25	4
0.33	3
0.50	2
0.67	3
0.75	4

Finding Molecular Formula

Given: Empirical formula and molar mass

$$n = \frac{\text{Molar mass}}{\text{Empirical formula mass}}$$ $$\text{Molecular formula} = n \times \text{Empirical formula}$$

Example

Empirical formula: CH₂O (mass = 30.03 g/mol) Molar mass: 180.18 g/mol

$$n = \frac{180.18}{30.03} = 6$$ $$\text{Molecular formula: } (\text{CH}_2\text{O})_6 = \text{C}_6\text{H}_{12}\text{O}_6$$

Combustion Analysis

Used to determine formulas of organic compounds (C, H, O).

Process

1. Burn sample completely in excess O₂
2. Collect and weigh CO₂ (contains all C)
3. Collect and weigh H₂O (contains all H)
4. Calculate O by difference (if present)

Formulas

$$\text{mol C} = \text{mol CO}_2 = \frac{\text{mass CO}_2}{44.01}$$ $$\text{mol H} = 2 \times \text{mol H}_2\text{O} = 2 \times \frac{\text{mass H}_2\text{O}}{18.02}$$ $$\text{mass O} = \text{sample mass} - \text{mass C} - \text{mass H}$$

Example: Combustion Analysis

0.255 g of a compound containing C, H, and O produces 0.561 g CO₂ and 0.306 g H₂O.

$$\text{mol C} = \frac{0.561}{44.01} = 0.01275 \text{ mol}$$ $$\text{mass C} = 0.01275 \times 12.01 = 0.153 \text{ g}$$ $$\text{mol H} = 2 \times \frac{0.306}{18.02} = 0.0340 \text{ mol}$$ $$\text{mass H} = 0.0340 \times 1.008 = 0.0343 \text{ g}$$ $$\text{mass O} = 0.255 - 0.153 - 0.0343 = 0.0677 \text{ g}$$ $$\text{mol O} = \frac{0.0677}{16.00} = 0.00423 \text{ mol}$$

Ratio (divide by 0.00423):

• C: $0.01275 / 0.00423 \approx 3$
• H: $0.0340 / 0.00423 \approx 8$
• O: $0.00423 / 0.00423 = 1$

Empirical formula: C₃H₈O

Hydrates

Compounds that include water molecules in their crystal structure.

General Form

$$\text{Salt} \cdot n\text{H}_2\text{O}$$

Example: Finding x in CuSO₄ · xH₂O

5.00 g hydrate heated, 3.19 g anhydrous salt remains

$$\text{mass H}_2\text{O} = 5.00 - 3.19 = 1.81 \text{ g}$$ $$\text{mol CuSO}_4 = \frac{3.19}{159.61} = 0.0200 \text{ mol}$$ $$\text{mol H}_2\text{O} = \frac{1.81}{18.02} = 0.100 \text{ mol}$$ $$x = \frac{0.100}{0.0200} = 5$$

Formula: CuSO₄ · 5H₂O