Empirical vs. Molecular Formulas Empirical Formula The lowest whole number ratio (subscripts) of elements in a compound. Cannot be reduced!!! not empirical empirical Ex: C6H12O6 CH2O Molecular Formula Actual number of atoms in a chemical compound molecular EX: C12H24O12 Molecular Formulas can be reduced to Empirical Formulas molecular EX: C12H24O12 CH2O empirical Different molecular formulas can have similar empirical formulas molecular EX: N3O9 N12O36 empirical NO3 PRACTICE: 1. Identify each as empirical (can’t be reduced) or molecular (can be reduced) 2. If its molecular – write the empirical 1. 2. 3. 4. 5. C2H4 NO3 S9Cl12 C3Cl9 N4S9 molecular CH2 - empirical empirical molecular molecular empirical S3Cl4 - empirical CCl3 - empirical Finding Empirical Formula from Percent Composition • Ex: A compound was found to be 54.53% Carbon, 9.15% Hydrogen, and 36.32% Oxygen. Find its Empirical Formula. Steps: 1. Assume a 100g sample (change % g) 2. ÷ by the molar mass of that element to find moles (sig fig it!) 3. Identify the lowest # of moles and ÷ them all by that number 4. Round each to the nearest whole # (if .0 or .9) (if not - special) 5. The resulting whole #are the subscripts for that element in the empirical formula Calculating Empirical Formula 63.5% Silver 8.2% Nitrogen 28.3% Oxygen 63.5 g Ag 107.87 8.2 g N 14.01 28.3 g O 16.00 .589 mole Ag .589 .59 mole N .589 1.77 mole O .589 1 1 AgNO3 3 Calculating Empirical Formula (special) 60.00%C 60.00g C 12.01 4.48%H 4.48g H 1.01 4.996 mole C 2.221 4.44 mole H 2.221 2.249 2 35.53%O 35.53g O 16.00 2.221 mole N 2.221 1 x4 x4 x4 9 8 4 C9H8O4 Calculating Molecular Formula 1.Find the empirical formula 2.Calculate the molar mass of your empirical formula 3.Identify the molar mass of your molecular (GIVEN in the problem every time!) 4.Divide the molecular mass / empirical mass 5.Round to the nearest whole # 6.Multiply the whole # by the subscripts in the Empirical formula Practice If a compound has an empirical formula of NO3 and a molecular mass of 186g – what is the molecular formula? Empirical formula: NO3 Molecular mass (given) empirical mass 3 x NO3 = N3O9 molar mass: 62.01g 186g 62.01