Empirical & Molecular Formula Calculator

Empirical and Molecular Formula Guide

Empirical vs Molecular Formula

The empirical formula shows the simplest whole-number ratio of atoms in a compound. The molecular formula shows the actual number of atoms. Benzene: empirical formula CH (1:1 ratio), molecular formula C₆H₆ (actual molecules have 6 of each). Glucose: empirical formula CH₂O (1:2:1 ratio), molecular formula C₆H₁₂O₆ (actual). The molecular formula is always a whole-number multiple of the empirical formula: molecular formula = (empirical formula) × n.

Calculating Empirical Formula from % Composition

Step 1: assume 100g of compound, so % = grams. Step 2: divide each mass by the element's atomic mass to get moles. Step 3: divide all mole values by the smallest mole value to get the ratio. Step 4: if ratio isn't whole numbers, multiply all by the appropriate factor (×2, ×3, etc.). Example: 40% C, 6.7% H, 53.3% O → moles: C = 40/12 = 3.33, H = 6.7/1 = 6.7, O = 53.3/16 = 3.33 → ratio: C:H:O = 1:2.01:1 → empirical formula CH₂O.

Finding Molecular Formula from Empirical Formula

Molar mass of empirical formula = sum of atomic masses in empirical formula. Whole-number multiplier n = molar mass of compound ÷ empirical formula mass. Molecular formula = empirical formula × n. Example: empirical formula CH₂O, molecular mass = 180 g/mol. Empirical formula mass = 12 + 2 + 16 = 30 g/mol. n = 180 ÷ 30 = 6. Molecular formula = C₆H₁₂O₆ (glucose). This method requires the molecular molar mass from another source (mass spectrometry in practice).

Combustion Analysis

Organic compounds are commonly analysed by combustion: burn a known mass, then measure CO₂ and H₂O produced. Mass C = mass CO₂ × (12/44). Mass H = mass H₂O × (2/18). Mass O = original mass − mass C − mass H. This gives the same data as % composition, allowing empirical formula determination. Combustion analysis is standard in pharmaceutical and organic chemistry for confirming compound identity after synthesis.