1. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 1 Introductory Chemistry: A Foundation FIFTH EDITION by Steven S. Zumdahl University of Illinois

2. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 2 Chemical Composition Chapter 8

3. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 3 These seashells from Sanibel Island, Florida, contain the mineral calcium carbonate. Source: Elaine Rebman/Photo Researchers, Inc.

4. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 4 The Enzo Ferrari is built of composite materials. Source: AP Photo

5. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 5 8.1 Counting by Weighing Aim-To understand the concept of average mass and explore how counting can be done by weighing.

6. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 6 The Jelly Bean Analogy If you worked in a candy store and someone came in and asked for 1000 jellybeans, no more and no less, how would you give the customer what they wanted?

7. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 7 JellyBean Analogy Suppose each jellybean has a mass of exactly 5 grams? You could simply weigh out 5000 grams of jellybeans and you would have 1000 of them. Counting out 1000 would take a long time.

8. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 8 Jellybean Analogy In reality, jellybeans do not all weigh exactly the same. You could weigh out 10 separate beans and find their average mass. Now you could simply multiply the average mass of the beans by the desired number and find how much 1000 beans would weigh.

9. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 9 8.2 Atomic Masses:Counting Atoms by Weighing Aim-To understand atomic mass and its experimental determination.

10. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 10 Atomic Masses Balanced equation tells us the relative numbers of molecules of reactants and products C + O 2  CO 2 1 atom of C reacts with 1 molecule of O 2 to make 1 molecule of CO 2 If I want to know how many O 2 molecules I will need or how many CO 2 molecules I can make, I will need to know how many C atoms are in the sample of carbon I am starting with

11. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 11 Atomic Masses Dalton used the percentages of elements in compounds and the chemical formulas to deduce the relative masses of atoms Unit is the amu. –Atomic Mass Unit –1 amu = 1.66 x 10 -24 g We define the masses of atoms in terms of atomic mass units –1 Carbon atom = 12.01 amu, –1 Oxygen atom = 16.00 amu

12. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 12 Atomic Mass Atomic mass is defined as the weighted average mass of all the isotopes of a particular element. –Includes a correction to allow for the relative abundance of each element. Atomic mass is found by looking on your periodic table, (number to the upper right corner, in red)

13. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 13

14. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 14 What is the mass of a molecule of Oxygen? Oxygen is diatomic, so the mass of a molecule (O 2 ) would apparently be more than the mass of one single atom of oxygen. The mass of O 2 would be simply the mass of 2 atoms of oxygen, added together, or 15.9994 amu + 15.9994 amu = 31.9988 amu

15. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 15 Formula Mass Formula mass is the mass of 1 formula unit of an ionic compound. Formula mass is found by adding the masses of all of the atoms that make up 1 formula unit of the compound. Example: CaCO 3 F.M. = 1-Ca 40.078 amu +1-C 12.0107 amu +3-O (3x15.9994) amu) 100.087 amu

16. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 16 Molecular Mass Molecular mass is the mass of 1 molecule of a covalently bonded compound. Calculated in the same manner as the formula mass mentioned earlier.

17. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 17 Atomic Masses Atomic/formula/molecular masses allow us to convert weights into numbers of particles If a sample of carbon weighs 3.00 x 10 20 amu, how many atoms of carbon are in the sample? Since our equation tells us that 1 C atom reacts with 1 O 2 molecule, if I have 2.50 x 10 19 C atoms, I will need 2.50 x 10 19 molecules of O 2

18. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 18 Example #1 Determine the mass of 1 Al atom 1 atom of Al = 26.98 amu Use the relationship as a conversion factor Calculate the Mass (in amu) of 75 atoms of Al

19. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 19 8.3 The Mole Aims-To understand the mole concept and Avogadro’s number. To learn to convert among moles, mass, and number of atoms in a given sample.

20. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 20 Chemical Packages - Moles We use a package for atoms and molecules called a mole A mole is the number of particles equal to the number of Carbon atoms in 12 g of C-12 One mole = 6.022 x 10 23 units One mole of particles= 1 molar mass (atomic/formula mass expressed in grams) 1 mole of C atoms weighs 12.01 g and has 6.02 x 10 23 atoms The number of particles in 1 mole is called Avogadro’s Number

21. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 21 Molar Mass The molar mass is the mass in grams of one mole of a compound The relative weights of atoms/molecules can be calculated from atomic masses water = H 2 O = 2(1.008 amu) + 16.00 amu = 18.02 amu 1 mole of H 2 O will weigh 18.02 g, therefore the molar mass of H 2 O is 18.02 g 1 mole of H 2 O will contain 16.00 g of oxygen and 2.02 g of hydrogen

22. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 22 A 1-mol sample of graphite (a form of carbon) weighs 12.01 g.

23. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 23 Figure 8.1: All these samples of pure elements contain the same number (a mole) of atoms: 6.022 x 10 23 atoms.

24. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 24 Example 8.6 from Text Calcium carbonate, CaCO 3, is the principal mineral found in limestone, marble, chalk, and pearls. Calculate the molar mass of calcium carbonate.

25. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 25 Example 8.6 cont. Calcium carbonate is composed of 1 ion of Ca 2+ and one ion of CO 3 2-. One mol of CaCO 3 contains one mol of Ca 2+ ions and one mol of CO 3 2- ions. We calculate the molar mass by summing the masses of the components.

26. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 26 Example 8.6 cont. Mass of one mol Ca 2+ = 40.08 grams Mass of one mol CO 3 2- (contains 1 mol C and 3 mol O) –1 mol C = 12.01 grams –3 mol O = 3 x 16.00 grams –One mol CO 3 2- = 60.01 grams

27. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 27 Example 8.6 cont. 40.08 grams + 60.01 grams = 100.09 grams

28. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 28 Figure 8.2: One-mole samples of iron (nails), iodine crystals, liquid mercury, and powdered sulfur. Source: Glenn Izett/U.S. Geological Survey

29. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 29

30. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 30 Example #2 How many atoms of boron are contained in 156.00 amu’s of boron?

31. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 31 Example #3 Calculate the equivalent number of moles of 26.2 grams of gold.

32. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 32 Example #4 How many atoms are there in 2.71 X 10 -4 mol of platinum?

33. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 33 A silicon chip of the type used in electronic equipment. Source: G.K. & Vikki Hart/The Image Bank/Getty Images

34. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 34 Example 8.4 from text A silicon chip used in an integrated circuit of a microcomputer has a mass of 5.68 milligrams. How many silicon (Si) atoms are present in this chip? The average atomic mass of silicon is 28.09 amu.

35. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 35 Example 8.4 cont. Convert mg to grams Convert grams to mol Si Convert mol Si to atoms of Si

36. 36 Figure 8.3: Various numbers of methane molecules showing their constituent atoms (cont’d ).

37. 37 Figure 8.3: Various numbers of methane molecules showing their constituent atoms (cont’d ).

38. Black walnuts with and without their green hulls.

39. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 39 8.5 Percent Composition Aim-To learn to find the mass percent of an element in a given compound

40. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 40 Percent Composition Percentage of each element in a compound –By mass Can be determined from Êthe formula of the compound or Ëthe experimental mass analysis of the compound The percentages may not always total to 100% due to rounding

41. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 41 ¬Determine the mass of each element in 1 mole of the compound 2 moles C = 2(12.01 g) = 24.02 g 6 moles H = 6(1.008 g) = 6.048 g 1 mol O = 1(16.00 g) = 16.00 g ­Determine the molar mass of the compound by adding the masses of the elements 1 mole C 2 H 5 OH = 46.07 g Determine the Percent Composition from the Formula C 2 H 5 OH Example #4

42. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 42 ®Divide the mass of each element by the molar mass of the compound and multiply by 100% Determine the Percent Composition from the Formula C 2 H 5 OH Example #4

43. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 43 8.6 Formulas of Compounds Aim-To understand the meaning of empirical formulas of compounds.

44. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 44 Empirical Formulas The simplest, whole-number ratio of atoms in a molecule is called the Empirical Formula –can be determined from percent composition or combining masses The Molecular Formula is a multiple of the Empirical Formula % A mass A (g) moles A 100g MM A % B mass B (g) moles B 100g MM B moles A moles B

45. Figure 8.4: The glucose molecul e.

46. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 46 ¬Find the greatest common factor (GCF) of the subscripts factors of 20 = (10 x 2), (5 x 4) factors of 12 = (6 x 2), (4 x 3) GCF = 4 ­Divide each subscript by the GCF to get the empirical formula C 20 H 12 = (C 5 H 3 ) 4 Empirical Formula = C 5 H 3 Determine the Empirical Formula of Benzopyrene, C 20 H 12 Example #5

47. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 47 8.7 Calculation of Empirical Formulas Aim-To learn to calculate empirical formulas

48. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 48 Steps for Determining the Empirical Formula of a Compound 1. Obtain the mass of each element present (in grams) 2. Determine the number of moles of each type of atom present 3. Divide the number of moles of each element by the smallest number of moles to convert the smallest number to 1. 4. If the numbers from 3 were not whole numbers, multiply the numbers by the smallest integer that will convert all of them to whole numbers. This set of numbers represents the subscripts.

49. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 49 ¬Convert the percentages to grams by assuming you have 100 g of the compound –Step can be skipped if given masses Determine the Empirical Formula of Acetic Anhydride if its Percent Composition is 47% Carbon, 47% Oxygen and 6.0% Hydrogen Example #6

50. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 50 ­Convert the grams to moles Determine the Empirical Formula of Acetic Anhydride if its Percent Composition is 47% Carbon, 47% Oxygen and 6.0% Hydrogen Example #6

51. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 51 ®Divide each by the smallest number of moles Determine the Empirical Formula of Acetic Anhydride if its Percent Composition is 47% Carbon, 47% Oxygen and 6.0% Hydrogen Example #6

52. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 52 ¯If any of the ratios is not a whole number, multiply all the ratios by a factor to make it a whole number –If ratio is ?.5 then multiply by 2; if ?.33 or ?.67 then multiply by 3; if ?.25 or ?.75 then multiply by 4 Multiply all the Ratios by 3 Because C is 1.3 Determine the Empirical Formula of Acetic Anhydride if its Percent Composition is 47% Carbon, 47% Oxygen and 6.0% Hydrogen Example #6

53. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 53 °Use the ratios as the subscripts in the empirical formula C4H6O3C4H6O3 Determine the Empirical Formula of Acetic Anhydride if its Percent Composition is 47% Carbon, 47% Oxygen and 6.0% Hydrogen Example #6

54. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 54 8.8 Calculation of Molecular Formulas Aim-To learn to calculate the molecular formula of a compound, given its empirical formula and molar mass

55. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 55 Molecular Formulas The molecular formula is a multiple of the empirical formula To determine the molecular formula you need to know the empirical formula and the molar mass of the compound

56. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 56 ¬Determine the empirical formula May need to calculate it as previous C5H3C5H3 ­Determine the molar mass of the empirical formula 5 C = 60.05 g, 3 H = 3.024 g C 5 H 3 = 63.07 g Determine the Molecular Formula of Benzopyrene if it has a molar mass of 252 g and an empirical formula of C 5 H 3 Example #7

57. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 57 ®Divide the given molar mass of the compound by the molar mass of the empirical formula –Round to the nearest whole number Determine the Molecular Formula of Benzopyrene if it has a molar mass of 252 g and an empirical formula of C 5 H 3 Example #7

58. Copyright©2004 by Houghton Mifflin Company. All rights reserved. 58 ¯Multiply the empirical formula by the calculated factor to give the molecular formula (C 5 H 3 ) 4 = C 20 H 12 Determine the Molecular Formula of Benzopyrene if it has a molar mass of 252 g and an empirical formula of C 5 H 3 Example #7

59. Figure 8.5: The structure of P 4 O 10 as a "ball- and- stick" model.