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Percent Composition

Percent Composition. 10.3. Percent Composition. 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. What did Bob score on his test? How do you figure it out? Part X 100 = 87

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Percent Composition

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  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? KnownsUnknowns Mass of compound= % Mg = Mass of oxygen= % O = Mass of magnezium= Analyze Calculate Evaluate

  9. % Composition from Chemical Formula What is the percent composition of propane? C3H8 Knowns Unknowns

  10. Calculate the percent composition of H2S?

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

  12. 10.2 HW 25. 22.4 L 26. 567 g CaCO3 27. 11.0 mol C2H6O 28. 33.6 L Cl2 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.

  13. 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.

  14. 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 C3H8, how much C and H do you have?

  15. Percent Composition as a Conversion Factor Propane is 82% C and 18% H. If you have 300 g of C3H8, how much C and H do you have? 300 g C3H8 x 82 g C = 100 g C3H8 300 g C3H8 x 18 g H = 100 g C3H8 Analyze

  16. Percent Composition as a Conversion Factor Propane is 82% C and 18% H. If you have 300 g of C3H8, how much C and H do you have? 300 g C3H8 x 82 g C = 246 g C 100 g C3H8 300 g C3H8 x 18 g H = 54 g H 100 g C3H8 Calculate

  17. Percent Composition as a Conversion Factor Propane is 82% C and 18% H. If you have 300 g of C3H8, how much C and H do you have? 300 g C3H8 x 82 g C = 246 g C 100 g C3H8 300 g C3H8 x 18 g H = 54 g H 100 g C3H8 246 + 54 _______ 300 Evaluate

  18. Empirical Formula = the empirical formula of a compound shows the smallest whole number ratio of the elements in the compound. C2H2 Acetylene Empirical Formula C8H8 Styrene

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

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

  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? 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.

  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? 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

  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 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

  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 1.85 mol N = 1 mol N :4.63 mol O = 2.5 mol O 1.85 1.85

  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 1.85 mol N = 1 mol N ; 4.63 mol O = 2.5 mol O 1.85 1.85 1 mol N X 2 = 2 mol N 2.5 mol O X 2 = 5 mol O N2O5 is the empirical formula

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