1. Moles Text reference for moles: PP 83 to 87 and PP 237 to 249

2. 2 The Mole A number of atoms, ions, or molecules that is large enough to see and handle. A mole = number of things – Just like a dozen = 12 things – One mole = 6.022 x 10 23 things Avogadro’s constant = 6.022 x 10 23 – Symbol for Avogadro’s constant is N A.

3. 3 The Mole How do we know when we have a mole? – count it out – weigh it out Molar mass - mass in grams numerically equal to the atomic weight of the element in grams. H has an atomic weight of 1.00794 g – 1.00794 g of H atoms = 6.022 x 10 23 H atoms Mg has an atomic weight of 24.3050 g – 24.3050 g of Mg atoms = 6.022 x 10 23 Mg atoms

4. 4 The Mole Example 2-1: Calculate the mass of a single Mg atom in grams to 3 significant figures.

5. 5 The Mole Example 2-1: Calculate the mass of a single Mg atom in grams to 3 significant figures.

6. 6 The Mole Example 2-1: Calculate the mass of a single Mg atom in grams to 3 significant figures.

7. 7 The Mole Example 2-1: Calculate the mass of a single Mg atom, in grams, to 3 significant figures.

8. 8 The Mole Example 2-2: Calculate the number of atoms in one-millionth of a gram of Mg to 3 significant figures.

9. 9 The Mole Example 2-2: Calculate the number of atoms in one-millionth of a gram of Mg to 3 significant figures.

10. 10 The Mole Example 2-2: Calculate the number of atoms in one-millionth of a gram of Mg to 3 significant figures.

11. 11 The Mole Example 2-2: Calculate the number of atoms in one-millionth of a gram of Mg to 3 significant figures.

12. 12 The Mole Example 2-3. How many atoms are contained in 1.67 moles of Mg?

13. 13 The Mole Example 2-3. How many atoms are contained in 1.67 moles of Mg?

14. 14 The Mole Example 2-3. How many atoms are contained in 1.67 moles of Mg?

15. 15 The Mole Example 2-3. How many atoms are contained in 1.67 moles of Mg?

16. 16 The Mole Example 2-3. How many atoms are contained in 1.67 moles of Mg?

17. 17 The Mole Example 2-4: How many moles of Mg atoms are present in 73.4 g of Mg? You do it!

18. 18 The Mole Example 2-4: How many moles of Mg atoms are present in 73.4 g of Mg?

19. 19 The Mole Example 2-4: How many moles of Mg atoms are present in 73.4 g of Mg?

20. 20 The Mole Example 2-4: How many moles of Mg atoms are present in 73.4 g of Mg? IT IS IMPERATIVE THAT YOU KNOW HOW TO DO THESE PROBLEMS

21. 21 Formula Weights, Molecular Weights, and Moles How do we calculate the molar mass of a compound? – add atomic weights of each atom The molar mass of propane, C 3 H 8, is: 3C X 12.01g = 36.03g 8H X 1.01g = 8.08g 44.11 g/mole

22. 22 Formula Weights, Molecular Weights, and Moles The molar mass of calcium nitrate, Ca(NO 3 ) 2, is: You do it!

23. 23 Formula Weights, Molecular Weights, and Moles One Mole ofContains – Cl 2 or70.90g6.022 x 10 23 Cl 2 molecules 2(6.022 x 10 23 ) Cl atoms You do it! – C 3 H 8 You do it!

24. 24 Formula Weights, Molecular Weights, and Moles One Mole ofContains – Cl 2 or70.90g6.022 x 10 23 Cl 2 molecules 2(6.022 x 10 23 ) Cl atoms – C 3 H 8 or 44.11 g6.022 x 10 23 C 3 H 8 molecules 3 (6.022 x 10 23 ) C atoms 8 (6.022 x 10 23 ) H atoms

25. 25 Formula Weights, Molecular Weights, and Moles Example 2-5: Calculate the number of C 3 H 8 molecules in 74.6 g of propane.

26. 26 Formula Weights, Molecular Weights, and Moles Example 2-5: Calculate the number of C 3 H 8 molecules in 74.6 g of propane.

27. 27 Formula Weights, Molecular Weights, and Moles Example 2-5: Calculate the number of C 3 H 8 molecules in 74.6 g of propane.

28. 28 Formula Weights, Molecular Weights, and Moles Example 2-5: Calculate the number of C 3 H 8 molecules in 74.6 g of propane.

29. 29 Formula Weights, Molecular Weights, and Moles Example 2-6. What is the mass of 10.0 billion propane molecules? You do it!

30. 30 Formula Weights, Molecular Weights, and Moles Example 2-6. What is the mass of 10.0 billion propane molecules?

31. 31 Formula Weights, Molecular Weights, and Moles Example 2-7. How many (a) moles, (b) molecules, and (c) oxygen atoms are contained in 60.0 g of ozone, O 3 ? The layer of ozone in the stratosphere is very beneficial to life on earth. You do it!

32. 32 Formula Weights, Molecular Weights, and Moles Example 2-7. How many (a) moles, (b) molecules, and (c) oxygen atoms are contained in 60.0 g of ozone, O 3 ? The layer of ozone in the stratosphere is very beneficial to life on earth. (a)

33. 33 Formula Weights, Molecular Weights, and Moles Example 2-7. How many (a) moles, (b) molecules, and (c) oxygen atoms are contained in 60.0 g of ozone, O 3 ? The layer of ozone in the stratosphere is very beneficial to life on earth. (b)

34. 34 Formula Weights, Molecular Weights, and Moles Example 2-7. How many (a) moles, (b) molecules, and (c) oxygen atoms are contained in 60.0 g of ozone, O 3 ? The layer of ozone in the stratosphere is very beneficial to life on earth. (c)

35. 35 Formula Weights, Molecular Weights, and Moles Example 2-8. Calculate the number of O atoms in 26.5 g of Li 2 CO 3. You do it!

36. 36 Formula Weights, Molecular Weights, and Moles Example 2-8. Calculate the number of O atoms in 26.5 g of Li 2 CO 3.

37. 37 Percent Composition and Formulas of Compounds % composition = mass of an individual element in a compound divided by the total mass of the compound x 100% Determine the percent composition of C in C 3 H 8.

38. 38 Percent Composition and Formulas of Compounds What is the percent composition of H in C 3 H 8 ? You do it!

39. 39 Percent Composition and Formulas of Compounds What is the percent composition of H in C 3 H 8 ?

40. 40 Percent Composition and Formulas of Compounds Example 2-10: Calculate the percent composition of Fe 2 (SO 4 ) 3 to 3 significant figures. You do it!

41. 41 Percent Composition and Formulas of Compounds Example 2-10: Calculate the percent composition of Fe 2 (SO 4 ) 3 to 3 sig. fig.

42. 42 Derivation of Formulas from Elemental Composition Example 2-11: A compound contains 24.74% K, 34.76% Mn, and 40.50% O by mass. What is its empirical formula? Make the simplifying assumption that we have 100.0 g of compound. In 100.0 g of compound there are: – 24.74 g of K – 34.76 g of Mn – 40.50 g of O

43. 43 Derivation of Formulas from Elemental Composition

44. 44 Derivation of Formulas from Elemental Composition

45. 45 Derivation of Formulas from Elemental Composition

46. 46 Derivation of Formulas from Elemental Composition

47. 47 Derivation of Formulas from Elemental Composition

48. 48 Derivation of Formulas from Elemental Composition Example 2-12: A sample of a compound contains 6.541g of Co and 2.368g of O. What is the empirical formula for this compound? You do it!

49. 49 Derivation of Formulas from Elemental Composition Example 2-12: A sample of a compound contains 6.541g of Co and 2.368g of O. What is the empirical formula for this compound?

50. 50 Derivation of Formulas from Elemental Composition Example 2-12: A sample of a compound contains 6.541g of Co and 2.368g of O. What is the empirical formula for this compound?

51. Determination of Molecular Formulas The formulas determined by the previous method produce empirical or simplest formulas. In the case of covalent compounds the empirical formula may or may not represent a real molecule. For example: CH 2 is an empirical formula by no molecule with this formula exists. C 2 H 4 is a real molecule with empirical formula CH 2 !

52. Determination of Molecular Formulas In order to determine a molecular formula from the empirical formula you need some additional information, usually the molar mass of the molecule. Since a molecular formula is a multiple of its empirical formula, you can divide the molar mass of the molecular formula by the molar mass of the empirical to find the multiplier. For example:

53. Determination of Molecular Formulas Suppose the empirical formula is CH 2 and we were told that the molar mass of a molecule of this substance is 70 g/mole. The molar mass of CH 2 is 14 g/mole So 70/14 = 5 so the molecular formula is 5 X the empirical or C 5 H 10

54. Determination of Molecular Formulas Example 2-13: A compound is found to contain 85.63% C and 14.37% H by mass. In another experiment its molar mass is found to be 56.1 g/mol. What is its molecular formula? » You do it

55. 55 Determination of Molecular Formulas Example 2-13: A compound is found to contain 85.63% C and 14.37% H by mass. In another experiment its molar mass is found to be 56.1 g/mol. What is its molecular formula? – short cut method

56. 56 Determination of Molecular Formulas