1. Empirical and Molecular formulas

2. Empirical – lowest whole number ratio of elements in a compound Molecular – some multiple of the empirical formula Examples: CH 4 C 6 H 12 O 6 Na 2 SO 4 C 3 H 6 Can an empirical formula also be the molecular formula? YES C 12 H 22 O 11 E – 1:4 M- 6:12:6 reduced to 1:2:1 E – 2:1M -3:6 reduced to 1:2

3. Steps for determining empirical formulas. 1)Assume a 100 g sample when given percents. This makes the 10.3 % Z = 10.3 g Z 2)Change grams into moles for each element. 3)Divide the all the moles by smallest number of moles to get the lowest whole number ratio. 4)Write the empirical formula.

4. A compound was found to contain 36.11 % calcium and 63.89 % chlorine by mass. What is its empirical formula? What assumption did you make? 100 g sample 36.11 % Ca = 36.11 g Ca x 63.89 % Cl = 63.89 g Cl x = 0.9009 mol Ca = 1.802 mol Cl 0.9009 = 1 mol Ca = 2 mol Cl Therefore the empirical formula is CaCl 2 Step 1Step 2 Step 3 Step 4

5. GOOFY MATH(.33 = 1/3.5 = ½.67 = 2/3) A compound was found to contain 68.42 % chromium and the rest oxygen by mass. What is its empirical formula? What assumption did you make? 100 g sample 68.42 g Cr 31.58 g O = 1.316 mol Cr = 1.974 mol O 1.316 = 1 mol Cr = 1.5 mol O X 2 = 2 Cr = 3 O Cr 2 O 3 Way to much to round off so you have to get rid of the fraction.

6. Ethene is a compound containing only carbon and hydrogen. It contains 85.63 % carbon and the rest hydrogen by mass. It also has a molar mass of 28.05 g/mol. What are empirical and molecular formulas of this compound? What assumption did you make? 85.63 g C x 14.37 g H x 100 g sample = 7.130 mol C = 14.26 mol H 7.130 = 1 mol C = 2 mol H Empirical formula CH 2 Remember molecular is some multiple of the empirical. So take the molar mass of the compound and divide it by the molar mass of the empirical formula. 28.05/14.03 = 2 So take 2 x CH 2 = C 2 H 4