Empirical and Molecular Formulas

The empirical and molecular formulas are vital in representing the composition of chemical compounds. They help in understanding the relative and actual number of atoms of each element in a compound. The distinction between them is crucial for chemical calculations and formula derivation in stoichiometry. Empirical and Molecular Formula

Empirical Formula

The empirical formula of a compound shows the simplest whole number ratio of atoms of each element in the compound. It does not necessarily represent the actual number of atoms, just the ratio.

Example: The empirical formula of glucose (C₆H₁₂O₆) is CH₂O.

Molecular Formula

The molecular formula represents the actual number of atoms of each element in a molecule of the compound. It is a whole-number multiple of the empirical formula.

Example: The molecular formula of glucose is C₆H₁₂O₆, and its empirical formula is CH₂O. Here, the molecular formula = (Empirical Formula) × 6.

Formula Relationship

The relationship between molecular and empirical formula is:

Molecular Formula = n × Empirical Formula
where n is an integer and calculated as:
n = (Molar Mass of compound) / (Empirical Formula Mass)

Comparison Table

Property	Empirical Formula	Molecular Formula
Represents	Simplest whole number ratio	Actual number of atoms
Uniqueness	May represent multiple compounds	Unique for a compound
Determination	From elemental analysis	From molar mass and empirical formula

Example Problem

Given: A compound contains 40% Carbon, 6.7% Hydrogen, and 53.3% Oxygen. Its molar mass is 180 g/mol. Find its molecular formula.

1. Assume 100 g sample → C = 40 g, H = 6.7 g, O = 53.3 g
2. Moles: C = 40 / 12 = 3.33, H = 6.7 / 1 = 6.7, O = 53.3 / 16 = 3.33
3. Divide by smallest: C = 1, H = 2, O = 1 → Empirical formula = CH₂O
4. Empirical mass = 12 + 2×1 + 16 = 30 g/mol
5. n = 180 / 30 = 6 → Molecular formula = C₆H₁₂O₆

Quiz

1. The empirical formula of C₄H₈ is:
   a) C₄H₈
   b) CH₂
   c) C₂H₄
   d) CH
   Answer: b)
   Explanation: Divide subscripts by 4 → CH₂ is the simplest whole number ratio.
2. If a compound has empirical formula CH and molar mass 78, what is its molecular formula?
   a) C₂H₂
   b) C₃H₃
   c) C₆H₆
   d) C₄H₄
   Answer: c)
   Explanation: CH = 13 g/mol. 78 / 13 = 6 → C₆H₆.
3. Which of the following cannot be an empirical formula?
   a) CH
   b) CH₂O
   c) C₂H₆
   d) NH₂
   Answer: c)
   Explanation: C₂H₆ can be simplified to CH₃.
4. Which technique helps determine empirical formula?
   a) X-ray crystallography
   b) Combustion analysis
   c) Spectroscopy
   d) Colorimetry
   Answer: b)
   Explanation: Combustion analysis provides elemental composition percentages.

Next Up: Stoichiometry and Calculations → Understand mole concepts and chemical equations.