Empirical and molecular formulae
Mole Concept and Stoichiometry · Chemistry
Study notes
A compound has 40% C, 6.67% H, 53.33% O by mass; molar mass 180 g/mol. Find both formulae. Step 1: Assume 100 g: 40 g C, 6.67 g H, 53.33 g O. Step 2: Moles: C = 40/12 = 3.33; H = 6.67/1 = 6.67; O = 53.33/16 = 3.33. Step 3: Divide by smallest (3.33): C₁H₂O₁ → empirical CH₂O. Step 4: M_empirical = 30; n = 180/30 = 6 → molecular C₆H₁₂O₆ (glucose!).