Empirical & Molecular Formula Calculator

Calculate empirical formulas from percent composition or mass data, and determine molecular formulas from empirical formulas and molar mass.

Percent Composition → Empirical Formula

Enter the percent composition of each element. Total should equal 100%.

Example Problems:
C: 40.0%, H: 6.7%, O: 53.3% (CH₂O) C: 92.3%, H: 7.7% (CH) Na: 32.4%, S: 22.6%, O: 45% (Na₂SO₄) C: 85.7%, H: 14.3% (CH₂) Fe: 69.9%, O: 30.1% (Fe₂O₃) Ca: 40%, C: 12%, O: 48% (CaCO₃)
Mass Data → Empirical Formula

Enter the mass (in grams) of each element in the compound.

Example Problems:
12.01g C, 2.02g H, 16.00g O 2.30g Na, 3.55g Cl 11.16g Fe, 4.80g O 1.50g C, 0.25g H, 1.75g N
Empirical → Molecular Formula

Convert an empirical formula to its molecular formula using the compound's molar mass.

Enter the empirical formula
Enter the experimentally determined molar mass
Example Problems:
CH₂O → ? (MM: 180 g/mol) CH → ? (MM: 78 g/mol) NO₂ → ? (MM: 92 g/mol) CH₂ → ? (MM: 84 g/mol) C₂H₅ → ? (MM: 58 g/mol)
Combustion Analysis

Determine empirical formula from combustion products (CO₂ and H₂O).

Example Problems:
Glucose: 1.00g → 2.20g CO₂, 0.90g H₂O Hydrocarbon: 0.50g → 1.47g CO₂, 0.30g H₂O Amine: 1.00g → 1.47g CO₂, 0.60g H₂O, 0.47g N₂
Results


Enter data and click Calculate to see results

Understanding Empirical & Molecular Formulas
Empirical Formula

The simplest whole-number ratio of atoms in a compound.

Example: Glucose (C₆H₁₂O₆) has empirical formula CH₂O

Molecular Formula

The actual number of atoms of each element in a molecule.

Example: Glucose molecular formula is C₆H₁₂O₆

Key Concepts
  • Percent Composition: % of each element by mass
  • Mole Ratio: Convert mass/percent to moles, then divide by smallest
  • Combustion Analysis: Burn compound, analyze CO₂ and H₂O products
  • Multiplier (n): Molecular Mass / Empirical Mass

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