Unit 9: Covalent Bonding Chapters 8 & 9 Chemistry 1L Cypress Creek High School

Percent Composition Percent Composition is the percentage by mass of each element in a compound The percent composition is found by using the following formula:

Percent Composition What is the percent of C & H in C2H6? 79.89% carbon & 20.11% hydrogen HYDROGEN CARBON

Percent Composition What is the percent of each element in sodium sulfite, Na2SO3? 2 mole Na x 22.990 g/mol Na = 45.980 g Na 1 mole S x 32.066 g/mol S = 32.066 g S 3 moles O x 15.999 g/mol O = + 47.997 g O 126.043 g Sodium % = (45.980 g / 126.043 g) x 100 = 36.486 % Na Sulfur % = (32.066 g / 126.043 g) x 100 = 25.441 % S Oxygen % = (47.997 g / 126.043 g) x 100 = 38.080 % O Molar Mass of Na2SO3 =

Empirical Formula An Empirical Formula is the LOWEST whole number ratio of the elements in a compound. The subscript cannot be a fraction. Ex: The empirical formula for caffeine (C8N4O2H10) is C4N2OH5 C8N4O2H10 is a completely different chemical than C4N2OH5! What is the empirical formula of glucose (C6H12O6)? CH2O Back = Carbon Blue = Nitrogen Red = Oxygen Gray = Hydrogen

Empirical Formula Calculations There are 3 types of empirical calculations: Mole Ratio Grams Percent

Empirical Formula from Mole Ratios To calculate the empirical formula from mole ratios, follow these steps: Divide each mole quantity by the smallest mole quantity Assign ratios as subscripts If any of the ratios are not even (ex: 0.5, 1.25, 1.3), multiply them all by the lowest common multiple to achieve whole numbers then place these numbers as the subscripts.

Empirical Formula from Mole Ratios Find the empirical formula for a compound containing 2 mole carbon and 6 mole hydrogen. C = 2/2 = 1 H = 6/2 = 3 empirical formula = CH3 LOWEST

Empirical Formula from Grams To calculate the empirical formula from grams, follow these steps: Convert the grams to moles (divide grams by molar mass) Divide each mole quantity by the smallest mole quantity Assign ratios as subscripts If any of the ratios are not even (ex: 0.5, 1.25, 1.3), multiply them all by the lowest common multiple to achieve whole numbers then place these numbers as the subscripts

Empirical Formula from Grams Find the empirical formula of a compound that contains 13.5 grams of calcium, 4.05 grams of carbon, and 16.2 grams of oxygen. calcium carbon oxygen 13.5 g 4.05 g 16.2 g 40.08 g/mol 12.011 g/mol 15.999 g/mol 0.337 0.337 1.01 0.337 0.337 0.337 = 0.337 mol = 0.337 mol = 1.01 mol LOWEST = 1 = 1 = 3 empirical formula = CaCO3

Empirical Formula from Percents To calculate the empirical formula from percent composition, follow these steps: Change % sign to grams. You assume 100 g of the compound Convert the grams to moles (divide grams by molar mass) Divide each mole quantity by the smallest mole quantity Assign ratios as subscripts If any of the ratios are not even (ex: 0.5, 1.25, 1.3), multiply them all by the lowest common multiple to achieve whole numbers then place these numbers as the subscripts

Empirical Formula from Percents Calculate the empirical formula of a compound containing 18.8% nickel and 81.2% iodine. nickel iodine 18.8 g 81.2 g 58.69 g/mol 126.905 g/mol 0.320 0.640 0.320 0.320 = 0.320 mol = 0.640 mol LOWEST = 1 = 2 empirical formula = NiI2

Molecular Formulas For many compounds, the empirical formula is not the true formula. A molecular formula tells the exact number of atoms of each element in a molecule or formula unit of a compound. The actual formula The molecular formula for a compound is always a whole-number multiple of the empirical formula.

Determining Molecular Formulas Determine the empirical formula Find the empirical formula mass (you are given the molecular formula mass) Divide the molecular formula mass by the empirical formula mass. This is the scalar (multiplier). Multiply the subscripts in the empirical formula by the multiplier.

Calculating Molecular Formula A compound has an empirical formula of Sc2O3. The molecular mass is 414 g/mol. What is the molecular formula? Empirical Formula = Sc2O3 Empirical Formula Mass = 2 mol Sc x 44.956 g/mol Sc = 89.912 g Sc 3 mol O x 15.999 g/mol O = + 47.997 g O molar mass = 137.909 g Sc2O3 Find the multiplier… 414 g / 137.909 g = 3 (the multiplier is 3) Molecular Formula = Sc6O9

Practice Molecular Formula A molecule has 85.6% carbon and 14.5% hydrogen. If the molecule has a mass of 42.1 grams, what is the molecular formula?

Be Prepared for Unit 9 Test. THE END. Be Prepared for Unit 9 Test.