1. Percent Composition and Empirical Formulas

2. Terms: u The law of definite proportions describes that elements in a given compound are always present in the same proportions by mass. u The Percentage composition of a compound refers to the relative mass of each element in the compound. u The term molecular mass describes the mass of a molecule. u The term formula mass describes the mass of the smallest repeating unit of an ionic compound.

3. Finding Percentage composition from a chemical formula: 1. Find the mass of the compound, then find the mass of the individual parts (atoms). 2. Divide the atoms by the compound and multiply by 100% For example: H 2 O H 2 = 2 x 1.01 = 2.02 +O = +16 Total molecular mass = 18.02 Percentage by mass of H in water = 2/18.02 =.11 x 100% = 11% Percentage by mass of O in water = 16/18.02 =.888 x 100% = 89%

4. Percentage Composition from mass 1.Take the mass of the atom and divide by the total mass of the compound, then multiply by 100% Example: Mass of compound is 48.72g, it contains 32.69g of zinc and 16.03g of sulfur %Zn = (32.69g/ 48.72g) x 100% = 67.10% %S = (16.03g/ 48.72g) x 100% = 32.90%

5. Practice: 1.Calculate the mass percentage of nitrogen in each compound: a)N 2 O b)Sr(NO 3 ) 2 c)NH 4 NO 3 d)HNO 3 e)NH 4 f)KNO 3

6. Empirical Formulas The empirical formula is simply the simplest whole number ratio of the elements in a compound. The molecular formula is the actual number of atoms that makes up the molecule For example, the empirical formula of ethene (C 2 H 4 ) is CH 2 the empirical formula of butene (C 4 H 8 ) is CH 2

7. Comparing NameMolecularEmpiricalLow ratio Hydrogen peroxideH2O2H2O2 HO1:1 GlucoseC 6 H 12 O 6 CH 2 O1:2:1 Benzene C6H6C6H6 CH1:1 Ethene C2H4C2H4 CH 2 1:2 Butene C4H8C4H8 CH 2 1:2 Aniline C6H7NC6H7NC6H7NC6H7N 6:7:1 WaterH2OH2OH2OH2O2:1

8. Finding Empirical formula: Calculate empirical formula of a compound that is 85.6% carbon and 14.4% hydrogen. First, assume that you have 100g of the substance, so you really have 85.6g of C and 14.4g of H. Next, convert it to moles using molar mass. Finally, find the lowest whole ration between the two.

9. 1.85.6g C and 14.4g H 2.mol C = 85.6g x mol = 7.13mol C 12.01g mol H = 14.4g x mol = 14.3 mol H 1.01g 3.Lowest mole ratio: C = 7.13 /7.13 = 1 H = 14.3/ 7.13 = 2.01, so 2 ANSWER: CH 2

10. Practice: Find the empirical formula for the following: 1.17.5% hydrogen and 82.4% nitrogen 2.46.3% lithium and 53.7% oxygen 3.15.9% boron and 84.1% fluorine 4.52.51% chlorine and 47.48% sulfur

11. Finding Molecular Formula The empirical formula is CH 2 O, and its molar mass is 150g/mol. What is molecular formula? Find empirical molar mass of CH 2 O: 12.01 + 2.02 + 15.99 g/mol =30.02 g/mol Divide real molar mass by empirical molar mass: 150/30.02 = 5 Multiply subscripts of empirical formula through by ratio (5): C 5 H 10 O 5

12. Do for Practice: Chem 11A Pg. 376 # 10,11 Pg. 377#12,13 Pg. 378-379 #21,25,39,40 Do for Practice: Chem 11 Pg. 82 #1-3 Pg. 91 #13-15 Pg. 94 #2-4, 6,7 Pg. 97 #17-19