1. 1

2. 2 The mass of a single atom is too small to measure on a balance. mass of hydrogen atom = 1.673 x 10 -24 g

3. 3 This is an infinitesimal mass 1.673 x 10 -24 g

4. 4 Chemists have chosen a unit for counting atoms. That unit is the Chemists require a unit for counting which can express large numbers of atoms using simple numbers. MOLE

5. 5 1 mole = 6.02 x 10 23 objects

6. 6 LARGE 6.02 x 10 23 is a very number

7. 7 6.02 x 10 23 is number Avogadro’s Number

8. 8 Amadeo (Amedeo) Avogadro

9. 9 If 10,000 people started to count Avogadro’s number and counted at the rate of 100 numbers per minute each minute of the day, it would take over 1 trillion years to count the total number. http://youtu.be/1R7NiIum2TI

10. 10 1 mole of any element contains 6.02 x 10 23 particles of that substance.

11. 11 The atomic weight in grams of any element 23 contains 1 mole of atoms.

12. 12 This is the same number of particles 6.02 x 10 23 as there are in exactly 12 grams of

13. 13ExamplesExamples

14. 14 Species Quantity Number of H atoms H 1 mole 6.02 x 10 23

15. 15 Species Quantity Number of H 2 molecules H2H2 1 mole 6.02 x 10 23

16. 16 Species Quantity Number of Na atoms Na 1 mole 6.02 x 10 23

17. 17 Species Quantity Number of Fe atoms Fe 1 mole 6.02 x 10 23

18. 18 Species Quantity Number of C 6 H 6 molecules C6H6C6H6 1 mole 6.02 x 10 23

19. 19 1 mol of atoms =6.02 x 10 23 atoms 6.02 x 10 23 molecules 6.02 x 10 23 ions 1 mol of molecules = 1 mol of ions =

20. 20 The mole weight of an element is its atomic weight in grams. It contains 6.02 x 10 23 atoms (Avogadro’s number) of the element.

21. 21 Element Atomic mass Mole weight Number of atoms H1.008 amu1.008 g6.02 x 10 23 Mg24.31 amu24.31 g6.02 x 10 23 Na22.99 amu22.99 g6.02 x 10 23

22. 22ProblemsProblems

23. 23 Convert To Moles!

24. 24 Weight (mass)Numbers Moles  Periodic TableAvogadro’s Number 6.02 × 10 23

25. 25 Atomic weight iron = 55.85 How many moles of iron does 25.0 g of iron represent? Conversion sequence: grams Fe → moles Fe Set up the calculation using a conversion factor between moles and grams.

26. 26 Atomic weight iron = 55.85 Conversion sequence is: grams Fe → moles Fe → atoms Fe How many iron atoms are contained in 40.0 grams of iron?

27. 27 Mole weight Na = 22.99 g Conversion sequence: atoms Na → moles Na → grams Na What is the mass of 3.01 x 10 23 atoms of sodium (Na)?

28. 28 Conversion sequence: moles O 2 → molecules O 2 → atoms O How many oxygen atoms are present in 2.00 moles of oxygen molecules?

29. 29 Mole Weight of Compounds

30. 30 The mole weight of a compound can be determined by adding the mole weights of all of the atoms in its formula.

31. 31 2 C = 2(12.01 g) = 24.02 g 6 H = 6(1.01 g) = 6.06 g 1 O = 1(16.00 g) = 16.00 g 46.08 g Calculate the mole weight of C 2 H 6 O.

32. 32 1 Li = 1(6.94 g) = 6.94 g 1 Cl = 1(35.45 g) = 35.45 g 4 O = 4(16.00 g) = 64.00 g 106.39 g Calculate the mole weight of lithium perchlorate. LiClO 4

33. 33 Calculate the mole weight of ammonium phosphate. 3 N = 3(14.01 g) = 42.03 g 12 H = 12(1.01 g) = 12.12 g 1 P = 1(30.97 g) = 30.97 g 4 O = 4(16.00 g) = 64.00 g 149.12 g (NH 4 ) 3 PO 4

34. 34 Water in a hydrate is known as water of hydration or water of crystallization. Solids that contain water as part of their crystalline structure are known as hydrates.

35. 35 Formulas of hydrates are written by first writing the formula for the anhydrous compound and then adding a dot followed by the number of water molecules present.anhydrous 6H 2 OCoCl 2

36. 36 Calculate the mole weight of NaC 2 H 3 O 2 · 3 H 2 O 1 Na = 1(22.99 g) = 22.99 g 2 C = 2(12.01 g) = 24.02 g 9 H = 9(1.01 g) = 9.09 g 5 O = 5(16.00 g) = 80.00 g 136.10 g Sodium acetate trihydrate

37. 37 Avogadro’s Number of Particles 6.02 x 10 23 Particles Mole Weight 1 MOLE

38. 38 1 MOLE Ca Avogadro’s Number of Ca atoms 6.02 x 10 23 Ca atoms 40.078 g Ca

39. 39 1 MOLE H 2 O Avogadro’s Number of H 2 O molecules 6.02 x 10 23 H 2 O molecules 18.02 g H 2 O

40. 40 In dealing with diatomic elements (H 2, O 2, N 2, F 2, Cl 2, Br 2, and I 2 ), distinguish between one mole of atoms and one mole of molecules.

41. 41 Calculate the mole weight of 1 mole of H atoms. 1 H = 1(1.01 g) = 1.01 g Calculate the mole weight of 1 mole of H 2 molecules. 2 H = 2(1.01 g) = 2.02 g

42. 42ProblemsProblems

43. 43 How many moles of benzene, C 6 H 6, are present in 390.0 grams of benzene? Conversion sequence: grams C 6 H 6 → moles C 6 H 6 The mole weight of C 6 H 6 is 78.12 g.

44. 44 How many grams of (NH 4 ) 3 PO 4 are contained in 2.52 moles of (NH 4 ) 3 PO 4 ? Conversion sequence: moles (NH 4 ) 3 PO 4 → grams (NH 4 ) 3 PO 4 The mole weight of (NH 4 ) 3 PO 4 is 149.12 g.

45. 45 56.04 g of N 2 contains how many N 2 molecules? The mole weight of N 2 is 28.02 g. Conversion sequence: g N 2 → moles N 2 → molecules N 2 Use the conversion factors

46. 46 56.04 g of N 2 contains how many N atoms? The mole weight of N 2 is 28.02 g. Conversion sequence: g N 2 → moles N 2 → molecules N 2 → atoms N Use the conversion factors

47. 47 Percent Composition of Compounds

48. 48 Percent composition of a compound is the mass percent of each element in the compound. H2OH2O 11.19% H by mass88.79% O by mass

49. 49 Percent Composition From Formula

50. 50 If the formula of a compound is known, a two-step process is needed to calculate the percent composition. Step 1 Calculate the mole weight of the formula. Step 2 Divide the total mass of each element in the formula by the mole weight and multiply by 100.

51. 51

52. 52 Step 1 Calculate the mole weight of H 2 S. 2 H = 2 (1.01 g) = 2.02 g 1 S = 1 (32.07 g) = 32.07 g 34.09 g Calculate the percent composition of hydrosulfuric acid H 2 S (aq).

53. 53 Calculate the percent composition of hydrosulfuric acid H 2 S. Step 2 Divide the mass of each element by the mole weight and multiply by 100. H 5.93% S 94.07%

54. 54 Percent Composition From Experimental Data

55. 55 Percent composition can be calculated from experimental data without knowing the composition of the compound. Step 1 Calculate the mass of the compound formed. Step 2 Divide the mass of each element by the total mass of the compound and multiply by 100.

56. 56 Step 1 Calculate the total mass of the compound 1.52 g N 3.47 g O 4.99 g A compound containing nitrogen and oxygen is found to contain 1.52 g of nitrogen and 3.47 g of oxygen. Determine its percent composition. = total mass of product

57. 57 Step 2 Divide the mass of each element by the total mass of the compound formed. N 30.5% O 69.5% A compound containing nitrogen and oxygen is found to contain 1.52 g of nitrogen and 3.47 g of oxygen. Determine its percent composition.

58. 58 Empirical Formula versus Molecular Formula

59. 59 The empirical formula or simplest formula gives the smallest whole- number ratio of the atoms present in a compound. The empirical formula gives the relative number of atoms of each element present in the compound.

60. 60 The molecular formula is the true formula of a compound. The molecular formula represents the total number of atoms of each element present in one molecule of a compound.

61. 61ExamplesExamples

62. 62 C2H4C2H4 Molecular Formula CH 2 Empirical Formula C:H 1:2 Smallest Whole Number Ratio

63. 63 C6H6C6H6 Molecular Formula CH Empirical Formula C:H 1:1 Smallest Whole Number Ratio

64. 64 H2O2H2O2 Molecular Formula HOEmpirical Formula H:O 1:1 Smallest Whole Number Ratio

65. 65

66. 66 Two compounds can have identical empirical formulas and different molecular formulas.

67. 67

68. 68 Calculating Empirical Formulas

69. 69 The analysis of a compound shows that it contains 27% carbon and 73% oxygen. Calculate the empirical formula for this substance. C___O___

70. 70 Step 1 Assume a definite starting quantity (usually 100.0 g) of the compound, if not given, and express the mass of each element in grams. Step 2 Convert the grams of each element into moles of each element using each element’s mole weight.

71. 71 Step 3 Divide the moles of atoms of each element by the moles of atoms of the element that had the smallest value – If the numbers obtained are whole numbers, use them as subscripts and write the empirical formula. – If the numbers obtained are not whole numbers, go on to step 4.

72. 72 Step 4 Multiply the values obtained in step 3 by the smallest numbers that will convert them to whole numbers Use these whole numbers as the subscripts in the empirical formula. FeO 1.5 Fe 1 x 2 O 1.5 x 2 Fe 2 O 3

73. 73 The results of calculations may differ from a whole number. –If they differ ±0.1 round off to the next nearest whole number. 2.9 33 –Deviations greater than 0.1 unit from a whole number usually mean that the calculated ratios have to be multiplied by a whole number.

74. 74ProblemsProblems

75. 75 The analysis of a salt shows that it contains 56.58% potassium (K); 8.68% carbon (C); and 34.73% oxygen (O). Calculate the empirical formula for this substance. Step 1 Express each element in grams. Assume 100 grams of compound. K = 56.58 g C = 8.68 g O = 34.73 g

76. 76 The analysis of a salt shows that it contains 56.58% potassium (K); 8.68% carbon (C); and 34.73% oxygen (O). Calculate the empirical formula for this substance. Step 2 Convert the grams of each element to moles. C has the smallest number of moles

77. 77 The analysis of a salt shows that it contains 56.58% potassium (K); 8.68% carbon (C); and 34.73% oxygen (O). Calculate the empirical formula for this substance. Step 3 Divide each number of moles by the smallest value. The simplest ratio of K:C:O is 2:1:3 Empirical formula K 2 CO 3 C has the smallest number of moles

78. 78 The percent composition of a compound is 25.94% nitrogen (N), and 74.06% oxygen (O). Calculate the empirical formula for this substance. Step 1 Express each element in grams. Assume 100 grams of compound. N = 25.94 g O = 74.06 g

79. 79 Step 2 Convert the grams of each element to moles. The percent composition of a compound is 25.94% nitrogen (N), and 74.06% oxygen (O). Calculate the empirical formula for this substance.

80. 80 Step 3 Divide each number of moles by the smallest value. The percent composition of a compound is 25.94% nitrogen (N), and 74.06% oxygen (O). Calculate the empirical formula for this substance. This is not a ratio of whole numbers.

81. 81 Step 4 Multiply each of the values by 2. The percent composition of a compound is 25.94% nitrogen (N), and 74.06% oxygen (O). Calculate the empirical formula for this substance. Empirical formula N 2 O 5 N: (1.000)2 = 2.000 O: (2.500)2 = 5.000

82. 82 Calculating the Molecular Formula from the Empirical Formula

83. 83 The molecular formula can be calculated from the empirical formula if the mole weight is known. The molecular formula will be equal to the empirical formula or some multiple n of it. To determine the molecular formula evaluate n. n is the number of units of the empirical formula contained in the molecular formula.

84. 84 What is the molecular formula of a compound which has an empirical formula of CH 2 and a mole weight of 126.2 g? The molecular formula is (CH 2 ) 9 = C 9 H 18 Let n = the number of formula units of CH 2. Calculate the mass of each CH 2 unit 1 C = 1(12.01 g) = 12.01g 2 H = 2(1.01 g) = 2.02g 14.03g

85. 85 A compound made of 30.4% N and 69.6% O has a mole weight of 138 grams/mole. Find the molecular formula. Old Way:

86. 86 A compound made of 30.4% N and 69.6% O has a mole weight of 138 grams/mole. Find the molecular formula. New Way: N3O6N3O6

87. 87