(but not the kind that lives underground)

1. The Mole (but not the kind that lives underground)

2. Where Are We Going? Count large objects by ones (cars). Count smaller objects by dozens (eggs). Count tiny objects by hundreds (ream of paper).

3. Where Are We Going? Smaller objects need larger numbers to be sizeable. Atoms and molecules are so small that we must have a HUGE number of them to work with them.

4. Avogadro’s Number 2 is a pair. 12 is a dozen. 144 is a gross. 602,200,000,000,000,000,000,000 is Avogadro’s number. It can be written 6.022 × 1023.

5. The Mole A mole is the amount of matter in Avogadro’s number of particles (i.e., 6.022 × 1023 particles). Use four significant digits for calculations.

6. Why That Number? That many particles weighs (in grams) the exact weight (in u) of one particle. 1 atom of H weighs 1.008 u. 1 mole of H weighs 1.008 g.

7. Molar Mass The terms “gram-atomic mass,” “gram-molecular mass,” and “gram-formula mass” are sometimes used. We will use “molar mass” (MM) for all these terms to avoid confusion.

8. Molar Mass Molar mass is the weight of 1 mole of the substance. If the substance is more than one element, add up the molar mass of the elements.

9. Example 1 1 mole of H2SO4 has: 2 moles of H at 1 g each 1 mole of S at 32 g each 4 moles of O at 16 g each (2)(1) + (1)(32) + (4)(16) = 98 So, H2SO4 has 98 g/mole.

10. Question What is the MM of KMnO4? • 158 g/mole • 115 g/mole • 207 g/mole • 131 g/mole • 147 g/mole

11. 1 mol molar mass molar mass 1 mol × × NA units1 mol 1 mol NA units × × Mass(grams) Moles No. of particles

12. 0.500 mole K atoms

13. 4.00 g Cu

14. 33.3 mg Au

15. 70 g Al2(SO4)3

16. Types of Formulas • Empirical formula • Molecular formula • Structural formula

17. Empirical Formula Shows only the ratio of atoms Ex: CH2 for ethene (really C2H4)

18. Molecular Formula Shows the actual number of atoms of each element in the substance Ex: C2H4 for ethene

19. H H C = C H H Structural Formula Not only has the exact number of atoms, but also shows the structure Ex:

20. Percent Composition Is always percent by weight, unless stated to be by volume Deals with mass, not number of particles Equals × 100% part whole

21. 60.00 g H2O 53.28 g O 6.72 g H

22. Example 2 A laboratory analysis of a 30.00 g sample of Al2(SO4)3 showed that it contained 4.731 g of aluminum, 8.436 g of sulfur, and 16.833 g of oxygen. What is the percent composition of this compound?

23. 4.731 g Al 30.00 g Al2(SO4)3 8.436 g S 30.00 g Al2(SO4)3 16.833 g O 30.00 g Al2(SO4)3 Example 2 Al: S: O: × 100% = 15.77% Al × 100% = 28.12% S × 100% = 56.11% O

24. Formula-to-Percent Problems • Find the number of grams of each element in 1 mole of the substance. • Find the molar mass of substances.

25. Formula-to-Percent Problems • Divide the number of grams of each element by the molar mass of substances. • Multiply by 100 to get percent.

26. Example 3 Find the percent composition of this compound: Al2(SO4)3. 2 mol of Al (26.98 g/mol) = 53.96 g 3 mol of S (32.07 g/mol) = 96.21 g 12 mol of O (16.00 g/mol) = 192.0 g Total 342.2 g

27. 53.96 g Al 342.2 g Al2(SO4)3 96.21 g S 342.2 g Al2(SO4)3 192.0 g O 342.2 g Al2(SO4)3 Example 3 Al: S: O: × 100% = 15.77% Al × 100% = 28.12% S × 100% = 56.11% O

28. Example 4 How many grams of oxygen would a 65.00 g sample of Al2(SO4)3 contain? 65.00 g Al2(SO4)3 16.833 g O 30.00 g Al2(SO4)3 = 36.47 g O

29. Percent-to-Formula Problems • Assume there is 100 g of the substance. • Divide the number of grams of each element by its molar mass to find the number of moles.

30. Percent-to-Formula Problems • Divide the number of moles of each element by the smallest number of moles. This gives the empirical formula.

31. 100.00 g unknown 75.00 g C 25.00 g H

32. Example 5 An unknown gas is 72.55% O and 27.45% C by mass. What is the empirical formula?

33. Example 5 Percent Composition: 72.55% O 27.45% C

34. Example 5 Mass Composition: In a 100 g sample of the gas, there will be 72.55 g of O and 27.45 g of C.

35. 1 mole O 16.00 g O 72.55 g O 1 mol C 12.01 g C 27.45 g C Example 5 Mole Composition: = 4.534 mole O = 2.286 mol C

36. 4.534 2.286 2.286 2.286 : Example 5 Mole Ratio: mol O : mol C = 4.534 : 2.286 reduced to lowest terms is mol O : mol C = = 1.983 : 1.000

37. Example 5 Empirical Formula: For every 2 mol of oxygen, there is 1 mol of carbon. The empirical formula must be CO2.

38. Example 6 A 5.000 g sample of an unknown compound contains 1.844 g of N and 3.156 g of O. Find the empirical formula.

39. Example 6 Mass Composition: 1.844 g N 3.156 g O

40. 1 mol N 14.01 g N 1.844 g N 1 mol O 16.00 g O 3.156 g O Example 6 Mole Composition of Sample: = 0.1316 mol N = 0.1973 mol O

41. 0.1316 0.1973 0.1316 0.1316 : Example 6 Mole Ratio: mol N : mol O = 0.1316 : 0.1973 reduced to lowest terms is mol N : mol O = = 1.000 : 1.499 The ratio should be 1 : 1.5 (N : O).

42. 2 3 2 2 : 1 : 1.5 = 2 2 3 2 1 2 3 1 : Example 6 Empirical Formula: = 2:3 The empirical formula of the compound is N2O3.

43. Example 7 Find the molecular formula of caffeine, which has a molar mass of 194.20 g/mol. Percent composition: 5.170% H 16.49% O 28.86% N 49.48% C

44. Example 7 Mass composition: In a 100 g sample of the compound, there will be 5.170 g H 16.49 g O 28.86 g N 49.48 g C

45. 5.170 g H 1 mol H 1.008 g H 1 mol O 16.00 g O 16.49 g O Example 7 Mole Composition of Sample: = 5.129 mol H = 1.031 mol O

46. 28.86 g N 1 mol N 14.01 g N 1 mol C 12.01 g C 49.48 g C Example 7 Mole Composition of Sample: = 2.060 mol N = 4.120 mol C

47. 5.129 mol 1.031 mol H: = 4.974 1.031 mol 1.031 mol O: = 1.000 2.060 mol 1.031 mol N: = 1.998 4.120 mol 1.031 mol C: = 3.996 Example 7 Mole Ratio:

48. 194.20 g/mol 97.10 g/mol Example 7 Molecular Formula: empirical formula = C4H5N2O molar mass of empirical formula = 97.10 g/mol molar mass of caffeine = 194.20 g/mol = 2.000

49. Example 7 Molecular Formula: C4H5N2O 2 CH N O 8 10 4 2