Determining the Empirical Formula for a Compound The empirical formula for a compound is the simplest __________ number __________ of the atoms in the compound. Examples: H2O is the empirical formula for water. (can’t be reduced) Glucose, C6H12O6, has the empirical formula ____________. Step 1: Change % to grams. Step 2: Convert grams to moles using the mole chart. Step 3: Divide each of these answers by the smallest number. Step 4: Your answers will be the subscripts for the empirical formula. whole ratio C1H2O1 Helpful Rhyme: % to mass, mass to mole, divide by small, times ’til whole.

Assume 100g since % , just change to % sign to g. Practice Problems: Calculate the empirical formula if a compound is found to be composed of 7.8% Carbon and 92.2% Cl Assume 100g since % , just change to % sign to g. Find moles of each element 7.8g C 1 mole = 92.2g Cl 1 mole = (% to mass) (mass to moles) 0.65 moles C 12.0 g C 2.63 moles Cl 35.0 g Cl

Divide each mole found by lowest mole. (Round to whole #’s) 0.65 mole C = 2.63 mole Cl = 0.65 0.65 4. Final answer, putting numbers in as subscripts. 1 4 (÷ by small) C1Cl4 CCl4

Assume 100g since % , just change to % sign to g. Practice Problems: Calculate the empirical formula if a compound is found to be composed of 42.9% C and 57.1% O Assume 100g since % , just change to % sign to g. Find moles of each element 42.9g C 1 mole = 57.1g O 1 mole = (% to mass) (mass to moles) 3.57 moles C 12.0 g C 3.57 moles O 16.0 g O

Divide each mole found by lowest mole. (Round to whole #’s) 3.575 mole C = 3.569 mole O = 3.569 3.569 4. Final answer, putting numbers in as subscripts. 1 1 (÷ by small) C1O1 CO

Determining the Molecular Formula for a Compound The molecular formula for a compound is either the same as the empirical formula ratio or it is a “_________ _________ of this ratio. It represents the true # of atoms in the molecule. Examples: 1) H2O is the empirical & molecular formula for water. 2) CH2O is the empirical formula for sugar, ethanoic acid, and methanol. The molecular formula for glucose is C6H12O6, (___times the empirical ratio!) Step 1: Determine the empirical formula for the compound. (See the previous steps in the notes.) Step 2: Determine the “whole # multiple” by dividing the molecular formula mass (given in the problem) by the empirical formula molar mass. Multiply each of the empirical ratios by this whole number. whole # multiple 6

H5.9O5.9 H1O1 HO 5.9 5.9 H2O2 Practice Problems: 1) An unknown compound is composed of 5.9% hydrogen and 94.1% oxygen. The molecular formula mass is 34 g. Determine the molecular formula for the compound. H = 5.9% = 5.9 g (mass to moles) 5.9 g H ÷ 1.0 = 5.9 moles H O = 94.1% = 94.1 g 94.1 g O ÷ 16.0 = 5.9 moles O (÷ by small) H5.9O5.9 H1O1 HO 5.9 5.9 H2O2 [H1O1 ]x 2 = Molar mass of H1O1 = 17 g (34 ÷ 17 = 2) Molecular formula mass divided by molar mass

C3.3H6.6O3.3 C1H2O1 CH2O 3.3 3.3 3.3 C4H8O4 Practice Problems: 2) An unknown compound is composed of 40% carbon, 6.6% hydrogen, and 53.4% oxygen. Determine the molecular formula for the compound if the mass of one mole of the compound is 120 g. C = 40% = 40 g (mass to moles) 40 g C ÷ 12.0 = 3.3 moles C H = 6.6% = 6.6 g 6.6 g H ÷ 1.0 = 6.6 moles H O = 53.4% = 53.4 g 53.4 g O ÷ 16.0 = 3.3 moles O (÷ by small) C3.3H6.6O3.3 C1H2O1 CH2O 3.3 3.3 3.3 (Compare) C4H8O4 [C1H2O1 ]x 4 = Our formula mass = 30 g (120 ÷ 30 = 4)