Empirical and Molecular Formulas - Chemistry

1. Fundamental Concepts

Empirical Formula: The simplest whole-number ratio of atoms of each element in a compound (e.g., CH₂ for ethene, C₂H₄).
Molecular Formula: The actual number of atoms of each element in a molecule of the compound (a multiple of the empirical formula, e.g., C₂H₄ for ethene).

2. Key Concepts

The empirical formula may equal the molecular formula (e.g., H₂O, CO₂) if the molecular formula is already in the simplest ratio.
The molar mass of the molecular formula = n × (molar mass of the empirical formula), where n is a positive integer (n = 1, 2, 3...).
Empirical formulas are determined from percent composition or mass data; molecular formulas require additional molar mass information.

3. Examples

Easy

Question: A compound has an empirical formula of CH and a molar mass of 78 g/mol. The molecular formula is _______ (molar mass of CH = 13 g/mol).

Calculate n: n = molecular molar mass / empirical formula mass = 78 g/mol / 13 g/mol = 6
Multiply empirical formula subscripts by n: C₁×₆H₁×₆ = C₆H₆

Answer: C₆H₆

Medium

Question: A sample of a compound contains 4.0 g of hydrogen and 32.0 g of oxygen. Its empirical formula is _______; if its molar mass is 36 g/mol, its molecular formula is _______.
(Empirical Formula):

Convert mass to moles:
- H: 4.0 g / 1.0 g/mol = 4.0 mol
- O: 32.0 g / 16.00 g/mol = 2.0 mol
Find mole ratio (divide by smallest mole value):
- H: 4.0 / 2.0 = 2; O: 2.0 / 2.0 = 1
Empirical formula: H₂O

(Molecular Formula):
Empirical formula mass (H₂O): (2×1) + 16 = 18 g/mol
Calculate n: 36 g/mol / 18 g/mol = 2 (round to whole number)
Molecular formula: (H₂O)₂ = H₄O₂

Answer: H₂O; H₄O₂

Hard

Question: A compound is 40.0% C, 6.7% H, and 53.3% O by mass, with a molar mass of 180 g/mol. Its empirical formula is _______ and molecular formula is _______.
(Empirical Formula):

Assume 100 g sample (mass = % value):
- C: 40.0 g; H: 6.7 g; O: 53.3 g
Convert to moles:
- C: 40.0 g / 12.01 g/mol ≈ 3.33 mol
- H: 6.7 g / 1.008 g/mol ≈ 6.65 mol
- O: 53.3 g / 16.00 g/mol ≈ 3.33 mol
Find mole ratio (divide by smallest mole value, 3.33):
- C: 3.33 / 3.33 = 1; H: 6.65 / 3.33 ≈ 2; O: 3.33 / 3.33 = 1
Empirical formula: CH₂O

(Molecular Formula):
Empirical formula mass (CH₂O): 12.01 + (2×1.008) + 16.00 ≈ 30.03 g/mol
Calculate n: 180 g/mol / 30.03 g/mol ≈ 6
Molecular formula: (CH₂O)₆ = C₆H₁₂O₆

Answer: CH₂O; C₆H₁₂O₆

4. Problem-Solving Techniques

Step 1: Find Empirical Formula

Assume a 100 g sample (convert percent composition to grams).
Convert grams to moles (using molar masses of elements).
Divide each mole value by the smallest mole value to get mole ratios.
Round ratios to whole numbers (multiply by a factor if needed) to get the empirical formula.

Step 2: Find Molecular Formula

1. Calculate the empirical formula’s molar mass.
2. Divide the compound’s molar mass by the empirical formula mass to find n.
3. Multiply each subscript in the empirical formula by n to get the molecular formula.