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

2. 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:

3. Percent Composition What is the percent of C & H in C 2 H 6 ? – 79.89% carbon & 20.11% hydrogen CARBON HYDROGEN

4. Percent Composition What is the percent of each element in sodium sulfite, Na 2 SO 3 ? – 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 Na 2 SO 3 =

5. 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 (C 8 N 4 O 2 H 10 ) is C 4 N 2 OH 5 – C 8 N 4 O 2 H 10 is a completely different chemical than C 4 N 2 OH 5 ! What is the empirical formula of glucose (C 6 H 12 O 6 )? CH 2 O Back = Carbon Blue = Nitrogen Red = Oxygen Gray = Hydrogen

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

7. Empirical Formula from Mole Ratios To calculate the empirical formula from mole ratios, follow these steps: 1)Divide each mole quantity by the smallest mole quantity 2)Assign ratios as subscripts 3)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.

8. 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 = CH 3 LOWEST

9. Empirical Formula from Grams To calculate the empirical formula from grams, follow these steps: 1)Convert the grams to moles (divide grams by molar mass) 2)Divide each mole quantity by the smallest mole quantity 3)Assign ratios as subscripts 4)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

10. 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 = 1 = 3 empirical formula = CaCO 3 Empirical Formula from Grams LOWEST 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.

11. Empirical Formula from Percents To calculate the empirical formula from percent composition, follow these steps: 1)Change % sign to grams. You assume 100 g of the compound 2)Convert the grams to moles (divide grams by molar mass) 3)Divide each mole quantity by the smallest mole quantity 4)Assign ratios as subscripts 5)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

12. Empirical Formula from Percents Calculate the empirical formula of a compound containing 18.8% nickel and 81.2% iodine. nickeliodine 18.8 g 81.2 g 58.69 g/mol 126.905 g/mol 0.3200.6400.320 = 0.320 mol = 0.640 mol = 1 = 2 empirical formula = NiI 2 LOWEST

13. 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.

14. Determining Molecular Formulas 1)Determine the empirical formula 2)Find the empirical formula mass (you are given the molecular formula mass) 3)Divide the molecular formula mass by the empirical formula mass. This is the scalar (multiplier). 4)Multiply the subscripts in the empirical formula by the multiplier.

15. Calculating Molecular Formula A compound has an empirical formula of Sc 2 O 3. The molecular mass is 414 g/mol. What is the molecular formula? – Empirical Formula = Sc 2 O 3 – 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 Sc 2 O 3 – Find the multiplier… 414 g / 137.909 g = 3 (the multiplier is 3) – Molecular Formula = Sc 6 O 9

16. 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?

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