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Bell Work: % Comp Review **Turn in late Folder Checks**

Bell Work: % Comp Review **Turn in late Folder Checks**. A 100 gram sample of carbon dioxide is 27.3% carbon. How many grams of C are in the sample? How many grams of O are in the sample? What’s the molar mass of CO 2 ?. Empirical and Molecular Formulas. Section 11.4.

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Bell Work: % Comp Review **Turn in late Folder Checks**

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  1. Bell Work: % Comp Review**Turn in late Folder Checks** A 100 gram sample of carbon dioxide is 27.3% carbon. • How many grams of C are in the sample? • How many grams of O are in the sample? • What’s the molar mass of CO2?

  2. Empirical and Molecular Formulas Section 11.4

  3. Empirical formulas are like reduced fractions. • can be reduced to . . . . . . MgO • Mg2O2 can be reduced to . . . . . . CH2O • C6H12O6 can be reduced to . . . . . .

  4. Types of Formulas The formulas for compounds can be expressed as an empirical formula and as a molecular(true) formula. Empirical Molecular (true) Name CH C2H2 acetylene CH C6H6 benzene CO2 CO2carbon dioxide CH2O C5H10O5 ribose Timberlake LecturePLUS

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

  6. 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. Timberlake LecturePLUS

  7. Learning Check EF-1 A. What is the empirical formula for C4H8? 1) C2H4 2) CH2 3) CH B. What is the empirical formula for C8H14? 1) C4H7 2) C6H12 3) C8H14 C. Which are possible molecular formulas for CH2O? 1) CH2O 2) C2H4O2 3) C3H6O3 Timberlake LecturePLUS

  8. Solution EF-1 A. What is the empirical formula for C4H8? 2) CH2 B. What is the empirical formula for C8H14? 1) C4H7 C. What is a molecular formula for CH2O? 1) CH2O 2) C2H4O2 3) C3H6O3 Timberlake LecturePLUS

  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. • HF5

  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. Empirical Formula • Example: glucose • Molecular formula = C6H12O6 • Empirical formula = CH2O • What is their common element ratio? • Ratio of one C to 2 H to 1 O • What is the scaling factor? • 6

  12. Finding Empirical Formulas from %Composition • 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.

  13. Empirical Formula • Next, find the amount of mol for each element. • 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.

  14. Empirical Formula • 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: SO3

  15. Example Problem 2 • Pg. 333: 46 N: 2.63 / 2.63 = 1 mol O: 3.95 / 2.63 = 1.5 mol Have to double moles to get whole # subscripts. N2O3

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

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

  18. Molecular Formula • 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

  19. Finding a Molecular FormulaExample 3: 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

  20. Molecular Formula 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: C2H3O2

  21. Molecular Formula 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 C2H3O2

  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: C4H6O4

  23. Learning Check EF-3 A compound has a formula mass of 176.0 and an empirical formula of C3H4O3. What is the molecular formula? 1) C3H4O3 2) C6H8O6 3) C9H12O9 Timberlake LecturePLUS

  24. Solution EF-3 A compound has a formula mass of 176.0 and an empirical formula of C3H4O3. What is the molecular formula? 2)C6H8O6 C3H4O3 = 88.0 g/EF 176.0 g = 2.00 88.0 Timberlake LecturePLUS

  25. 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. Timberlake LecturePLUS

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

  27. 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 Timberlake LecturePLUS

  28. 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?_____ Timberlake LecturePLUS

  29. 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?_____ Timberlake LecturePLUS

  30. Practice Problems • Pg 335: 52

  31. A Handy Flowchart

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