1. Bell Work: % Comp Review **Turn in late Folder Checks** A 100 gram sample of carbon dioxide is 27.3% carbon. 1.How many grams of C are in the sample? 2.How many grams of O are in the sample? 3.What’s the molar mass of CO 2 ?

2. Empirical and Molecular Formulas Section 11.4

3. Mg 2 O 2 can be reduced to...... MgO C 6 H 12 O 6 can be reduced to...... CH 2 O

4. Timberlake LecturePLUS4 Types of Formulas The formulas for compounds can be expressed as an empirical formula and as a molecular(true) formula. Empirical Molecular (true)Name CHC 2 H 2 acetylene CHC 6 H 6 benzene CO 2 CO 2 carbon dioxide CH 2 OC 5 H 10 O 5 ribose

5. Timberlake LecturePLUS5 Empirical Formulas Write your own one-sentence definition for each of the following: Empirical formula Molecular formula

6. Timberlake LecturePLUS6 An empirical formula represents the simplest whole number ratio of the atoms in a compound. The molecular formula is the true or actual ratio of the atoms in a compound.

7. Timberlake LecturePLUS7 Learning Check EF-1 A. What is the empirical formula for C 4 H 8 ? 1) C 2 H 4 2) CH 2 3) CH B. What is the empirical formula for C 8 H 14 ? 1) C 4 H 7 2) C 6 H 12 3) C 8 H 14 C. Which are possible molecular formulas for CH 2 O? 1) CH 2 O2) C 2 H 4 O 2 3) C 3 H 6 O 3

8. Timberlake LecturePLUS8 Solution EF-1 A. What is the empirical formula for C 4 H 8 ? 2) CH 2 B. What is the empirical formula for C 8 H 14 ? 1) C 4 H 7 C. What is a molecular formula for CH 2 O? 1) CH 2 O2) C 2 H 4 O 2 3) C 3 H 6 O 3

9. Empirical Formulas What if you don’t know the chemical formula? We can use the percent composition to find the empirical formula. An empirical formula represents the simplest whole number ratio of the atoms in a compound. HF 5

10. Empirical Formula The empirical formula may or may not be the same as the actual molecular formula. Molecular Formula: the actual number of atoms of each element in one molecule or formula unit of a substance.

11. Example: glucose Molecular formula = C 6 H 12 O 6 Empirical formula = CH 2 O What is their common element ratio?  Ratio of one C to 2 H to 1 O What is the scaling factor? 66 Empirical Formula

12. If you have the percent composition given to you, you can determine the empirical formula. You need to assume a few things.  The total mass of the compound is 100.0g.  The percent composition of the element is equal to the mass in grams of the element. Example 1: An oxide of sulfur has a percent composition of 40.05% S and 59.95% O. In 100 g of the compound, 40.05 g are S and 59.95 g are O. Next, find the amount of mol for each element. Finding Empirical Formulas from %Composition

13. OK, Now what? These numbers will help us determine the subscripts for the empirical formula. You cannot use the exact numbers that we just found because they are not whole numbers. Empirical Formula Next, find the amount of mol for each element.

14. Divide all numbers by the smallest number. So, S has a subscript of one:  1.249 mol S / 1.249 = 1 mol S Then, divide the mol O by the same number to find its subscript in the empirical formula.  3.747 mol O/ 1.249 = 3 mol O  Then write your empirical formula using your subscripts: SO 3 Empirical Formula

15. Pg. 333: 46 Example Problem 2

16. 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 _____.

17. 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.

18. 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

19. 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:

20. 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

21. 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

22. 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

23. Timberlake LecturePLUS23 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

24. Timberlake LecturePLUS24 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

25. Timberlake LecturePLUS25 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.

26. Timberlake LecturePLUS26 Solution EF-5 60.0 g C x ___________= ______ mol C 4.5 g H x ___________ = _______mol H 35.5 g O x ___________ = _______mol O

27. Timberlake LecturePLUS27 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

28. Timberlake LecturePLUS28 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?_____

29. Timberlake LecturePLUS29 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?_____

30. Pg 335: 52 Practice Problems

31. A Handy Flowchart