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Percentage Composition

Law of Multiple Proportions : If two elements combine to form more than one compound, the amounts of one element that combine with a fixed amount of the other element are in the ratio of small whole numbers.

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Percentage Composition

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  1. Law of Multiple Proportions: If two elements combine to form more than one compound, the amounts of one element that combine with a fixed amount of the other element are in the ratio of small whole numbers.

  2. Law of Multiple Proportions: If two elements combine to form more than one compound, the amounts of one element that combine with a fixed amount of the other element are in the ratio of small whole numbers. Example: Nitrogen and oxygen combine to form several different oxides, of which three are NO, N2O, and NO2. From experiment it is known that 16 g of oxygen will combine with 14 g, 28 g, and 7 g of nitrogen, respectively, for the three compounds. These masses are in the ratio 14: 28: 7 or 2 : 4 : 1.

  3. Percentage Composition Determining a chemical formula of an unknown compound begins with finding the elements present.

  4. Percentage Composition Determining a chemical formula of an unknown compound begins with finding the elements present. The next step is to determine the percentage composition – which allows the empirical formula to be determined.

  5. Percentage Composition Determining a chemical formula of an unknown compound begins with finding the elements present. The next step is to determine the percentage composition – which allows the empirical formula to be determined. Percentage composition: The percent by mass of each element present in a compound.

  6. Example: For H2O2 the molar mass is 34.0147 g/mol. There are 2.01588 g of H and 31.9988 g of O. % H % O

  7. Example: For H2O2 the molar mass is 34.0147 g/mol. There are 2.01588 g of H and 31.9988 g of O. % H % O The same percentage results would be obtained if we started with the empirical formula HO.

  8. Example: Calculating % composition from mass data.

  9. Example: Calculating % composition from mass data. A sample of a liquid with a mass of 8.657 g was decomposed into its elements to give 5.217 g of carbon, 0.9620 g of hydrogen, and 2.478 g of oxygen. What is the % by mass of each element in the liquid?

  10. Example: Calculating % composition from mass data. A sample of a liquid with a mass of 8.657 g was decomposed into its elements to give 5.217 g of carbon, 0.9620 g of hydrogen, and 2.478 g of oxygen. What is the % by mass of each element in the liquid? Answer: Do a percentage calculation.

  11. Example: Calculating % composition from mass data. A sample of a liquid with a mass of 8.657 g was decomposed into its elements to give 5.217 g of carbon, 0.9620 g of hydrogen, and 2.478 g of oxygen. What is the % by mass of each element in the liquid? Answer: Do a percentage calculation.

  12. Example: Calculating % composition from mass data. A sample of a liquid with a mass of 8.657 g was decomposed into its elements to give 5.217 g of carbon, 0.9620 g of hydrogen, and 2.478 g of oxygen. What is the % by mass of each element in the liquid? Answer: Do a percentage calculation.

  13. Empirical Formulas To determine the empirical formula from percentage composition data – find the number of moles of each element present.

  14. Empirical Formulas To determine the empirical formula from percentage composition data – find the number of moles of each element present. Example 1: Ascorbic acid (vitamin C) is composed of 40.92% C, 4.58% H and 54.50% O. Determine the empirical formula.

  15. Since we have % data, it is convenient to start with 100 g of the substance. Therefore, in 100 g of ascorbic acid, there are 40.92 g of C, 4.58 g of H, and 54.50 g of O. Now calculate the number of moles:

  16. Since we have % data, it is convenient to start with 100 g of the substance. Therefore, in 100 g of ascorbic acid, there are 40.92 g of C, 4.58 g of H, and 54.50 g of O. Now calculate the number of moles: moles of carbon =

  17. Since we have % data, it is convenient to start with 100 g of the substance. Therefore, in 100 g of ascorbic acid, there are 40.92 g of C, 4.58 g of H, and 54.50 g of O. Now calculate the number of moles: moles of carbon = moles of hydrogen =

  18. Since we have % data, it is convenient to start with 100 g of the substance. Therefore, in 100 g of ascorbic acid, there are 40.92 g of C, 4.58 g of H, and 54.50 g of O. Now calculate the number of moles: moles of carbon = moles of hydrogen = moles of oxygen =

  19. Hence, the empirical formula is obtained as follows: C3.406H4.54O3.406

  20. Hence, the empirical formula is obtained as follows: C3.406H4.54O3.406 Now divide by the smallest number,

  21. Hence, the empirical formula is obtained as follows: C3.406H4.54O3.406 Now divide by the smallest number, that is, C1.000H1.33O1.000 = C1H4/3O1

  22. Hence, the empirical formula is obtained as follows: C3.406H4.54O3.406 Now divide by the smallest number, that is, C1.000H1.33O1.000 = C1H4/3O1 Now eliminate the fraction (multiply by 3) to obtain the empirical formula as C3H4O3.

  23. Example: A major air pollutant in coal burning countries is a colorless pungent smelling gas containing sulfur and oxygen. Chemical analysis of a 1.078 g sample of this gas showed that it contained 0.540 g of sulfur and 0.538 g of oxygen. What is the empirical formula of this gas?

  24. Example: A major air pollutant in coal burning countries is a colorless pungent smelling gas containing sulfur and oxygen. Chemical analysis of a 1.078 g sample of this gas showed that it contained 0.540 g of sulfur and 0.538 g of oxygen. What is the empirical formula of this gas? It is not necessary in this problem to calculate the % composition – just calculate the number of moles of each element present.

  25. moles of sulfur = moles of oxygen =

  26. moles of sulfur = moles of oxygen = Hence, S0.0168O0.0336

  27. moles of sulfur = moles of oxygen = Hence, S0.0168O0.0336 Now divide by the smallest number to obtain, SO2 which is the empirical formula.

  28. Molecular Formulas A molecular formula gives the actual composition of a molecule. Example: Calculating a molecular formula from an empirical formula and molar mass information.

  29. Molecular Formulas A molecular formula gives the actual composition of a molecule. Example: Calculating a molecular formula from an empirical formula and molar mass information. An oxide of nitrogen gave the following analysis: 1.52 g of nitrogen combined with 3.47 g of oxygen. The molar mass of the compound was found to be 92.0 g/mol. Determine the molecular formula.

  30. Approach 1: Determine the empirical formula and then determine the molecular formula.

  31. Approach 1: Determine the empirical formula and then determine the molecular formula. moles of oxygen = moles of nitrogen =

  32. Approach 1: Determine the empirical formula and then determine the molecular formula. moles of oxygen = moles of nitrogen = Empirical formula is N0.109O0.217NO2

  33. Approach 1: Determine the empirical formula and then determine the molecular formula. moles of oxygen = moles of nitrogen = Empirical formula is N0.109O0.217NO2 The mass of the empirical formula unit is 14.0067 + 2 x 15.9994 = 46.0055 g

  34. Approach 1: Determine the empirical formula and then determine the molecular formula. moles of oxygen = moles of nitrogen = Empirical formula is N0.109O0.217NO2 The mass of the empirical formula unit is 14.0067 + 2 x 15.9994 = 46.0055 g Since the mass of 1 mol is 92.0 g, the number of empirical formula units in the compound is

  35. equal to

  36. equal to Hence, the molecular formula is (NO2)2, that is N2O4 (dinitrogentetraoxide).

  37. equal to Hence, the molecular formula is (NO2)2, that is N2O4 (dinitrogentetraoxide). Example: A compound of carbon, hydrogen, and oxygen has the following % composition (by mass): C 42.10 %, H 6.479 %, O 51.421 % The molar mass of the compound is 342.3 g/mol. What is the molecular formula?

  38. equal to Hence, the molecular formula is (NO2)2, that is N2O4 (dinitrogentetraoxide). Example: A compound of carbon, hydrogen, and oxygen has the following % composition (by mass): C 42.10 %, H 6.479 %, O 51.421 % The molar mass of the compound is 342.3 g/mol. What is the molecular formula? Approach 2: Work directly with the given molar mass and find the moles of each element present.

  39. moles of C = moles of H = moles of O = Hence, the molecular formula is C12H22O11

  40. Exercise: A compound of carbon, hydrogen, and oxygen has the following composition: C 40.00% and H 6.714 %. The molar mass of the compound is 180.16 g/mol. What is the molecular formula? Answer is C6H12O6

  41. Types of chemical reactions

  42. Types of chemical reactions Combination Reactions: Two or more substances combine to produce one substance.

  43. Types of chemical reactions Combination Reactions: Two or more substances combine to produce one substance. Examples: 2 H2 + O2 2 H2O 2 CO + O2 2 CO2 2 Na + Cl2 2 NaCl

  44. Decomposition Reactions: In decomposition reactions, one substance undergoes a reaction to produce two or more substances.

  45. Decomposition Reactions: In decomposition reactions, one substance undergoes a reaction to produce two or more substances. Examples: 2 H2O2 2 H2O + O2 2 NaHCO3 Na2CO3 + H2O + CO2 CuSO4.5H2O(s)CuSO4(s) + 5H2O(g)

  46. Displacement Reactions (single replacement): In displacement reactions, an atom of one element or an ion in a compound is replaced by an atom of another element or another ion.

  47. Displacement Reactions (single replacement): In displacement reactions, an atom of one element or an ion in a compound is replaced by an atom of another element or another ion. Examples: Zn + CuSO4(aq)Cu + ZnSO4(aq) Fe + SnCl2(aq)Sn + FeCl2(aq)

  48. Displacement Reactions (single replacement): In displacement reactions, an atom of one element or an ion in a compound is replaced by an atom of another element or another ion. Examples: Zn + CuSO4(aq)Cu + ZnSO4(aq) Fe + SnCl2(aq)Sn + FeCl2(aq) In both cases the metal in elemental form displaces another metal in combined form.

  49. Activity series of metals

  50. In examples like: Zn + CuSO4(aq)Cu + ZnSO4(aq) Fe + SnCl2(aq)Sn + FeCl2(aq)

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