1. Percent Composition 10.3

2. Percent Composition

3. I think you already know how to do this? What did Bob score on his test? How do you figure it out? 87 / 100 Bob

4. What did Bob score on his test? How do you figure it out? Part X 100 = 87 Whole 87 / 100 Bob

5. What did Sue score on her paper? Sue 174 / 205

6. What did Sue score on her paper? Part x 100 = 83 Whole.8341 x 100 = 83 Sue 171 / 205

7. Percent Composition = the relative amount of elements in a compound or substance. The percent by mass of an element in a compound is the number of grams of the element divided by the mass in grams of the compound, multiplied by 100%. % mass of element = mass of element x 100% mass of compound

8. Sample Problem 10.9 on pg. 306 When 13.60 gram sample of a compound containing only oxygen and magnesium is decomposed, 5.40 g of oxygen is obtained. What is the percent composition of this compound? Knowns Unknowns Mass of compound=% Mg = Mass of oxygen= % O = Mass of magnezium= Analyze CalculateEvaluate

9. % Composition from Chemical Formula What is the percent composition of propane? C 3 H 8 Knowns Unknowns

10. Calculate the percent composition of H 2 S?

11. Calculate the percent composition of Mg(OH) 2 ?

13. 10.2 HW 25. 22.4 L 26. 567 g CaCO 3 27. 11.0 mol C 2 H 6 O 28. 33.6 L Cl 2 29. 39.9 g/mol 30. Gas A: 28.0 g, nitrogen 31. The balloons have the same number of molecules. Each has 1 mole of gas or Avogadro’s number of particles. They will have different masses.

15. Percent Composition as a Conversion Factor If you know the % composition you can use it to determine how much of a specific element you have.

16. Percent Composition as a Conversion Factor If you know the % composition you can use it to determine how much of a specific element you have. Propane is 82% C and 18% H. If you have 300 g of C 3 H 8, how much C and H do you have?

17. Percent Composition as a Conversion Factor Propane is 82% C and 18% H. If you have 300 g of C 3 H 8, how much C and H do you have? 300 g C 3 H 8 x 82 g C = 100 g C 3 H 8 300 g C 3 H 8 x 18 g H = 100 g C 3 H 8 Analyze

18. Percent Composition as a Conversion Factor Propane is 82% C and 18% H. If you have 300 g of C 3 H 8, how much C and H do you have? 300 g C 3 H 8 x 82 g C = 246 g C 100 g C 3 H 8 300 g C 3 H 8 x 18 g H = 54 g H 100 g C 3 H 8 Calculate

19. Percent Composition as a Conversion Factor Propane is 82% C and 18% H. If you have 300 g of C 3 H 8, how much C and H do you have? 300 g C 3 H 8 x 82 g C = 246 g C 100 g C 3 H 8 300 g C 3 H 8 x 18 g H = 54 g H 100 g C 3 H 8 Evaluate 246 + 54 _______ 300

20. Empirical Formula = the empirical formula of a compound shows the smallest whole number ratio of the elements in the compound. C 2 H 2 Acetylene Empirical Formula C 8 H 8 Styrene

21. Empirical Formula and % Composition A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the empirical formula?

22. Empirical Formula and % Composition A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the empirical formula?

23. Empirical Formula and % Composition A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the empirical formula? Know Unknown % N = 25.9 % N N ? O ? % O = 74.1 % O Because percent means parts per 100, you can assume the 100.0g of the compound contains 25.9g of N and 74.1g of O. Use these values and convert to moles.

24. Empirical Formula and % Composition A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the empirical formula? Know Unknown % N = 25.9 % N N ? O ? % O = 74.1 % O 25.9 g N X 1 mol N = 1.85 mol N 14.0 g N 74.1 g O X 1mol O = 4.63 mol O 16.0 g O

25. Empirical Formula and % Composition A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the empirical formula? Know Unknown % N = 25.9 % N N ? O ? % O = 74.1 % O 25.9 g N X 1 mol N = 1.85 mol N 14.0 g N 74.1 g O X 1mol O = 4.63 mol O 16.0 g O Remember …. The amount of N to O is a ratio between the two

26. Empirical Formula and % Composition A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the empirical formula? Know Unknown % N = 25.9 % N N ? O ? % O = 74.1 % O 25.9 g N X 1 mol N = 1.85 mol N 14.0 g N 74.1 g O X 1mol O = 4.63 mol O 16.0 g O 1.85 mol N = 1 mol N : 4.63 mol O = 2.5 mol O 1.85

27. Empirical Formula and % Composition A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the empirical formula? Know Unknown % N = 25.9 % N N ? O ? % O = 74.1 % O 25.9 g N X 1 mol N = 1.85 mol N 14.0 g N 74.1 g O X 1mol O = 4.63 mol O 16.0 g O 1.85 mol N = 1 mol N ; 4.63 mol O = 2.5 mol O 1.85 1 mol N X 2 = 2 mol N 2.5 mol O X 2 = 5 mol O N 2 O 5 is the empirical formula