1. Bell Work: Empirical 1.__________ formulas show the actual ratio of elements found in a compound in nature. 2._________ formulas show the simplified ratio of elements in a compound. 3.Name three compounds that have identical molecular and empirical formulas. 4.Percent to ____, mass to ___, _____ by small, multiply ‘til _____.

2. Bell Work: Empirical vs. Molecular 1.Draw a Venn diagram to compare and contrast Empirical and Molecular formulas. 2.Draw a Venn diagram to compare and contrast moles and grams. 3.Draw a Venn diagram to compare and contrast formula units and molecules.

3. The molecular formula needs to be found by going one step further Molecular formula = (Empirical formula) x scaling factor To find the scaling factor 1) Determine the empirical mass (total mass of all elements in empirical formula) 2) Divide molecular mass by empirical mass Molecular Formula

4. Chemical analysis of succinic acid indicates it is composed of 40.68% C, 5.08% H, and 54.24 % O, and has a molar mass of 118.1 g/mol. Determine the empirical and molecular formulas for succinic acid. 1)Convert the percent for each element into moles (use the percent given as the amount in grams for each element in 100 g of the compound) 40.68 g C x (1 mol C/12.0 g C) = 3.39 mol C 5.08 g H x (1 mol H/1.0 g H) = 5.08 mol H 54.24 g O x (1 mol O/16.0 g O) = 3.39 mol O Finding a Molecular Formula Example 3:

5. 2)Next, divide each mol amount by the smallest mol amount. 3.39 mol C/ 3.39 = 1 mol C 5.08 mol H/ 3.39 = 1.5 mol H 3.39 mol O/ 3.39 = 1 mol O Ratio of C : H : O = 1 : 1.5 : 1 3) Write the Empirical Formula: You can’t have half-moles, so multiply everything by 2. Empirical Formula: C 2 H 3 O 2 Molecular Formula

6. 4) We need to find the empirical mass using the masses of each element. 2 mol C x (12.0 g C/1 mol C) = 24.0 g C. 3 mol H x (1.0 g H/1 mol H) = 3.0 g H. 2 mol O x (16.0 g O/1 mol O) = 32.0 g O. Empirical Mass: 59.0 g/ mol C 2 H 3 O 2

7. Molecular Formula 5) Now, divide the molar mass by the empirical mass to determine the scaling factor. 118.1 / 59.0 = 2.00 Multiply the subscripts of the empirical formula by 2 to find the molecular formula. Molecular Formula: C 4 H 6 O 4

8. Timberlake LecturePLUS8 Learning Check EF-3 A compound has a formula mass of 176.0 and an empirical formula of C 3 H 4 O 3. What is the molecular formula? 1) C 3 H 4 O 3 2) C 6 H 8 O 6 3) C 9 H 12 O 9

9. Timberlake LecturePLUS9 Solution EF-3 A compound has a formula mass of 176.0 and an empirical formula of C 3 H 4 O 3. What is the molecular formula? 2)C 6 H 8 O 6 C 3 H 4 O 3 = 88.0 g/EF 176.0 g = 2.00 88.0

10. Timberlake LecturePLUS10 Learning Check EF-5 Aspirin is 60.0% C, 4.5 % H and 35.5 O. Calculate its simplest formula. In 100 g of aspirin, there are 60.0 g C, 4.5 g H, and 35.5 g O.

11. Timberlake LecturePLUS11 Solution EF-5 60.0 g C x ___________= ______ mol C 4.5 g H x ___________ = _______mol H 35.5 g O x ___________ = _______mol O

12. Timberlake LecturePLUS12 Solution EF-5 60.0 g C x 1 mol C = 5.00 mol C 12.0 g C 4.5 g H x 1 mol H = 4.5 mol H 1.01 g H 35.5 g O x 1mol O= 2.22 mol O 16.0 g O

13. Timberlake LecturePLUS13 Divide by the smallest # of moles. 5.00 mol C = ________________ ______ mol O 4.5 mol H = ________________ ______ mol O 2.22 mol O = ________________ ______ mol O Are are the results whole numbers?_____

14. Timberlake LecturePLUS14 Divide by the smallest # of moles. 5.00 mol C = ___2.25__ 2.22 mol O 4.5 mol H = ___2.00__ 2.22 mol O 2.22 mol O = ___1.00__ 2.22 mol O Are are the results whole numbers?_____

15. Pg 335: 52 Practice Problems

16. A Handy Flowchart