1. Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA Introductory Chemistry, 2 nd Edition Nivaldo Tro Chapter 6 Chemical Composition 2006, Prentice Hall

2. Tro's Introductory Chemistry, Chapter 6 Why is Knowledge of Composition Important? everything in nature is either chemically or physically combined with other substances to know the amount of a material in a sample, you need to know what fraction of the sample it is Some Applications: the amount of sodium in sodium chloride for diet the amount of iron in iron ore for steel production the amount of hydrogen in water for hydrogen fuel the amount of chlorine in freon to estimate ozone depletion

3. Tro's Introductory Chemistry, Chapter 6 3 Counting Nails by the Pound I want to buy a certain number of nails for a project, but the hardware store sells nails by the pound! How do I know how many nails I am buying when I buy a pound of nails? Analogy How many atoms in a given mass of an element?

4. Tro's Introductory Chemistry, Chapter 6 4 Counting Nails by the Pound A hardware store customer buys 2.60 pounds of nails. A dozen of the nails has a mass of 0.150 pounds. How many nails did the customer buy? Solution map:

5. Tro's Introductory Chemistry, Chapter 6 5 Counting Nails by the Pound The customer bought 2.60 lbs of nails and received 208 nails. He counted the nails by weighing them!

6. Tro's Introductory Chemistry, Chapter 6 6 Counting Nails by the Pound What if he bought a different size nail? Would the mass of a dozen be 0.150 lbs? Would there be 208 nails in 2.60 lbs? How would this effect the conversion factors?

7. Tro's Introductory Chemistry, Chapter 6 7 Counting Atoms by Moles If we can find the mass of a particular number of atoms, we can use this information to convert the mass of a element sample to the number of atoms in the sample. The number of atoms we will use is 6.022 x 10 23 and we call this a mole 1 mole = 6.022 x 10 23 things  Like 1 dozen = 12 things

8. Tro's Introductory Chemistry, Chapter 6 8 Chemical Packages - Moles mole = number of particles equal to the number of atoms in 12 g of C-12 1 atom of C-12 weighs exactly 12 amu 1 mole of C-12 weighs exactly 12 g The number of particles in 1 mole is called Avogadro’s Number = 6.0221421 x 10 23 1 mole of C atoms weighs 12.01 g and has 6.022 x 10 23 atoms  the average mass of a C atom is 12.01 amu

9. Example 6.1 Converting Between Moles and Number of Atoms

10. Tro's Introductory Chemistry, Chapter 6 10 Example: A silver ring contains 1.1 x 10 22 silver atoms. How many moles of silver are in the ring?

11. Tro's Introductory Chemistry, Chapter 6 11 Example: A silver ring contains 1.1 x 10 22 silver atoms. How many moles of silver are in the ring? Write down the given quantity and its units. Given:1.1 x 10 22 Ag atoms

12. Tro's Introductory Chemistry, Chapter 6 12 Write down the quantity to find and/or its units. Find: ? moles Information Given:1.1 x 10 22 Ag atoms Example: A silver ring contains 1.1 x 10 22 silver atoms. How many moles of silver are in the ring?

13. Tro's Introductory Chemistry, Chapter 6 13 Collect Needed Conversion Factors: 1 mole Ag atoms = 6.022 x 10 23 Ag atoms Information Given:1.1 x 10 22 Ag atoms Find:? moles Example: A silver ring contains 1.1 x 10 22 silver atoms. How many moles of silver are in the ring?

14. Tro's Introductory Chemistry, Chapter 6 14 Write a Solution Map for converting the units : Information Given:1.1 x 10 22 Ag atoms Find:? moles Conv. Fact.: 1 mole = 6.022 x 10 23 Example: A silver ring contains 1.1 x 10 22 silver atoms. How many moles of silver are in the ring? atoms Ag moles Ag

15. Tro's Introductory Chemistry, Chapter 6 15 Apply the Solution Map: = 1.8266 x 10 22 moles Ag = 1.8 x 10 22 moles Ag Sig. Figs. & Round: Information Given:1.1 x 10 22 Ag atoms Find:? moles Conv. Fact.: 1 mole = 6.022 x 10 23 Sol’n Map: atoms  mole Example: A silver ring contains 1.1 x 10 22 silver atoms. How many moles of silver are in the ring?

16. Tro's Introductory Chemistry, Chapter 6 16 Check the Solution: 1.1 x 10 22 Ag atoms = 1.8 x 10 -2 moles Ag The units of the answer, moles, are correct. The magnitude of the answer makes sense since 1.1 x 10 22 is less than 1 mole. Information Given:1.1 x 10 22 Ag atoms Find:? moles Conv. Fact.: 1 mole = 6.022 x 10 23 Sol’n Map: atoms  mole Example: A silver ring contains 1.1 x 10 22 silver atoms. How many moles of silver are in the ring?

17. Tro's Introductory Chemistry, Chapter 6 17 Relationship Between Moles and Mass The mass of one mole of atoms is called the molar mass The molar mass of an element, in grams, is numerically equal to the element’s atomic mass, in amu

18. Tro's Introductory Chemistry, Chapter 6 18 Mole and Mass Relationships 1 mole Sulfur 32.06 g 1 mole Carbon 12.01 g

19. Example 6.2 Converting Between Grams and Moles of Atoms

20. Tro's Introductory Chemistry, Chapter 6 20 Example: Calculate the number of moles of sulfur in 57.8 g of sulfur

21. Tro's Introductory Chemistry, Chapter 6 21 Example: Calculate the number of moles of sulfur in 57.8 g of sulfur Write down the given quantity and its units. Given:57.8 g S

22. Tro's Introductory Chemistry, Chapter 6 22 Write down the quantity to find and/or its units. Find: ? moles S Information Given:57.8 g S Example: Calculate the number of moles of sulfur in 57.8 g of sulfur

23. Tro's Introductory Chemistry, Chapter 6 23 Collect Needed Conversion Factors: 1 mole S atoms = 32.06 g Information Given:57.8 g S Find:? moles S Example: Calculate the number of moles of sulfur in 57.8 g of sulfur

24. Tro's Introductory Chemistry, Chapter 6 24 Write a Solution Map for converting the units : Information Given:57.8 g S Find:? moles S Conv. Fact.: 1 mole S = 32.06 g g S moles S Example: Calculate the number of moles of sulfur in 57.8 g of sulfur

25. Tro's Introductory Chemistry, Chapter 6 25 Apply the Solution Map: = 1.80287 moles S = 1.80 moles S Sig. Figs. & Round: Information Given:57.8 g S Find:? moles S Conv. Fact.: 1 mole S = 32.06 g Sol’n Map:g  moles Example: Calculate the number of moles of sulfur in 57.8 g of sulfur

26. Tro's Introductory Chemistry, Chapter 6 26 Check the Solution: 57.8 g sulfur = 1.80 moles sulfur The units of the answer, moles, are correct. The magnitude of the answer makes sense since 57.8 g is more than 1 mole. Information Given:57.8 g S Find:? moles S Conv. Fact.: 1 mole S = 32.06 g Sol’n Map:g  moles Example: Calculate the number of moles of sulfur in 57.8 g of sulfur

27. Example 6.3 Converting Between Grams and Number of Atoms

28. Tro's Introductory Chemistry, Chapter 6 28 Example: How many aluminum atoms are in an aluminum can with a mass of 16.2 g?

29. Tro's Introductory Chemistry, Chapter 6 29 Example: How many aluminum atoms are in an aluminum can with a mass of 16.2 g? Write down the given quantity and its units. Given:16.2 g Al

30. Tro's Introductory Chemistry, Chapter 6 30 Write down the quantity to find and/or its units. Find: ? atoms Al Information Given:16.2 g Al Example: How many aluminum atoms are in an aluminum can with a mass of 16.2 g?

31. Tro's Introductory Chemistry, Chapter 6 31 Collect Needed Conversion Factors: 1 mole Al atoms = 26.98 g Al 1 mole = 6.022 x 10 23 Information Given:16.2 g Al Find:? atoms Al Example: How many aluminum atoms are in an aluminum can with a mass of 16.2 g?

32. Tro's Introductory Chemistry, Chapter 6 32 Write a Solution Map for converting the units : g Almol Alatoms Al Information Given:16.2 g Al Find:? atoms Al C F: 1 mol Al = 26.98 g 1 mol = 6.022 x 10 23 Example: How many aluminum atoms are in an aluminum can with a mass of 16.2 g?

33. Tro's Introductory Chemistry, Chapter 6 33 Apply the Solution Map: = 3.6159 x 10 23 atoms Al = 3.62 x 10 23 atoms Al Sig. Figs. & Round: Information Given:16.2 g Al Find:? atoms Al CF: 1 mol Al = 26.98 g 1 mol = 6.022 x 10 23 SM: g  mol  atoms Example: How many aluminum atoms are in an aluminum can with a mass of 16.2 g?

34. Tro's Introductory Chemistry, Chapter 6 34 Check the Solution: 16.2 g Al = 3.62 x 10 23 atoms Al The units of the answer, atoms, are correct. The magnitude of the answer makes sense since 16.2 g is less than the mass of 1 mole, 26.98 g. Information Given:16.2 g Al Find:? atoms Al CF: 1 mol Al = 26.98 g 1 mol = 6.022 x 1023 SM: g  mol  atoms Example: How many aluminum atoms are in an aluminum can with a mass of 16.2 g?

35. Tro's Introductory Chemistry, Chapter 6 35 Molar Mass of Compounds the relative weights of molecules can be calculated from atomic weights Formula Mass = 1 molecule of H 2 O = 2(1.01 amu H) + 16.00 amu O = 18.02 amu since 1 mole of H 2 O contains 2 moles of H and 1 mole of O Molar Mass = 1 mole H 2 O = 2(1.01 g H) + 16.00 g O = 18.02 g

36. Example 6.4 Converting Between Grams and Moles of Compound

37. Tro's Introductory Chemistry, Chapter 6 37 Example: Calculate the mass (in grams) of 1.75 mol of water

38. Tro's Introductory Chemistry, Chapter 6 38 Example: Calculate the mass (in grams) of 1.75 mol of water Write down the given quantity and its units. Given:1.75 mol H 2 O

39. Tro's Introductory Chemistry, Chapter 6 39 Write down the quantity to find and/or its units. Find: ? g H 2 O Information Given:1.75 mol H 2 O Example: Calculate the mass (in grams) of 1.75 mol of water

40. Tro's Introductory Chemistry, Chapter 6 40 Collect Needed Conversion Factors: Molar Mass H 2 O = 2(atomic mass H) + 1(atomic mass O) = 2(1.01) + 1(16.00) = 18.02 g/mol 1 mol H 2 O = 18.02 g H 2 O Information Given:1.75 mol H 2 O Find:? g H 2 O Example: Calculate the mass (in grams) of 1.75 mol of water

41. Tro's Introductory Chemistry, Chapter 6 41 Write a Solution Map for converting the units : Information Given:1.75 mol H 2 O Find:? g H 2 O C F: 1 mole H 2 O = 18.02 g H 2 O mol H 2 O g H 2 O Example: Calculate the mass (in grams) of 1.75 mol of water

42. Tro's Introductory Chemistry, Chapter 6 42 Apply the Solution Map: = 31.535 g H 2 O = 31.5 g H 2 O Sig. Figs. & Round: Information Given:1.75 mol H 2 O Find:? g H 2 O C F: 1 mole H 2 O = 18.02 g H 2 O Sol’n Map:mol  g Example: Calculate the mass (in grams) of 1.75 mol of water

43. Tro's Introductory Chemistry, Chapter 6 43 Check the Solution: 1.75 mol H 2 O = 31.5 g H 2 O The units of the answer, g, are correct. The magnitude of the answer makes sense since 31.5 g is more than 1 mole. Information Given:1.75 mol H 2 O Find:? g H 2 O C F: 1 mole H 2 O = 18.02 g H 2 O Sol’n Map:mol  g Example: Calculate the mass (in grams) of 1.75 mol of water

44. Example 6.5 Converting Between Grams and Number of Molecules

45. Tro's Introductory Chemistry, Chapter 6 45 Example: Find the mass of 4.78 x 10 24 NO 2 molecules?

46. Tro's Introductory Chemistry, Chapter 6 46 Example: Find the mass of 4.78 x 10 24 NO 2 molecules Write down the given quantity and its units. Given:4.78 x 10 24 NO 2 molecules

47. Tro's Introductory Chemistry, Chapter 6 47 Write down the quantity to find and/or its units. Find: ? g NO 2 Information Given:4.78 x 10 24 molec NO 2 Example: Find the mass of 4.78 x 10 24 NO 2 molecules

48. Tro's Introductory Chemistry, Chapter 6 48 Collect Needed Conversion Factors: Molar Mass NO 2 = 1(atomic mass N) + 2(atomic mass O) = 1(14.01) + 2(16.00) = 46.01 g/mol 1 mole NO 2 molec = 46.01 g NO 2 1 mole = 6.022 x 10 23 Information Given:4.78 x 10 24 molec NO 2 Find:? g NO 2 Example: Find the mass of 4.78 x 10 24 NO 2 molecules

49. Tro's Introductory Chemistry, Chapter 6 49 Write a Solution Map for converting the units : molec NO 2 mol NO 2 g NO 2 Information Given:4.78 x10 24 NO 2 molec Find:? g NO 2 C F: 1 mol NO 2 = 46.01 g NO 2 1 mol = 6.022 x 10 23 Example: Find the mass of 4.78 x 10 24 NO 2 molecules

50. Tro's Introductory Chemistry, Chapter 6 50 Apply the Solution Map: = 365.21 g NO 2 = 365 g NO 2 Sig. Figs. & Round: Information Given:4.78 x 10 24 molec NO 2 Find:? g NO 2 CF: 1 mol NO 2 = 46.01 g 1 mol = 6.022 x 10 23 SM: molec  mol  g Example: Find the mass of 4.78 x 10 24 NO 2 molecules

51. Tro's Introductory Chemistry, Chapter 6 51 Check the Solution: 4.78 x 10 24 molecules NO 2 = 365 g NO 2 The units of the answer, g, are correct. The magnitude of the answer makes sense since 4.78 x 10 24 molecules is more than 1 mole. Information Given:4.78 x 10 24 molec NO 2 Find:? g NO 2 CF: 1 mol NO 2 = 46.01 g 1 mol = 6.022 x 10 23 SM: molec  mol  g Example: Find the mass of 4.78 x 10 24 NO 2 molecules

52. Tro's Introductory Chemistry, Chapter 6 52 Chemical Formulas as Conversion Factors 1 spider  8 legs 1 chair  4 legs 1 H 2 O molecule  2 H atoms  1 O atom

53. Tro's Introductory Chemistry, Chapter 6 53 Mole Relationships in Chemical Formulas since we count atoms and molecules in mole units, we can find the number of moles of a constituent element if we know the number of moles of the compound Moles of CompoundMoles of Constituents 1 mol NaCl1 mole Na, 1 mole Cl 1 mol H 2 O2 mol H, 1 mole O 1 mol CaCO 3 1 mol Ca, 1 mol C, 3 mol O 1 mol C 6 H 12 O 6 6 mol C, 12 mol H, 6 mol O

54. Example 6.7 Converting Between Grams of a Compound and Grams of a Constituent Element

55. Tro's Introductory Chemistry, Chapter 6 55 Example: Carvone, (C 10 H 14 O), is the main component in spearmint oil. It has a pleasant odor and mint flavor. It is often added to chewing gum, liquers, soaps and perfumes. Find the mass of carbon in 55.4 g of carvone.

56. Tro's Introductory Chemistry, Chapter 6 56 Example: Find the mass of carbon in 55.4 g of carvone, (C 10 H 14 O). Write down the given quantity and its units. Given:55.4 g C 10 H 14 O

57. Tro's Introductory Chemistry, Chapter 6 57 Write down the quantity to find and/or its units. Find: ? g C Information Given:55.4 g C 10 H 14 O Example: Find the mass of carbon in 55.4 g of carvone, (C 10 H 14 O).

58. Tro's Introductory Chemistry, Chapter 6 58 Collect Needed Conversion Factors: Molar Mass C 10 H 14 O = 10(atomic mass C) + 14(atomic mass H) + 1(atomic mass O) = 10(12.01) + 14(1.01) + (16.00) = 150.2 g/mol 1 mole C 10 H 14 O = 150.2 g C 10 H 14 O 1 mole C 10 H 14 O  10 mol C 1 mole C = 12.01 g C Information Given:55.4 g C 10 H 14 O Find: g C Example: Find the mass of carbon in 55.4 g of carvone, (C 10 H 14 O).

59. Tro's Introductory Chemistry, Chapter 6 59 Write a Solution Map for converting the units : g C 10 H 14 O Information Given:55.4 g C 10 H 14 O Find: g C CF: 1 mol C 10 H 14 O = 150.2 g 1 mol C 10 H 14 O  10 mol C 1 mol C = 12.01 g Example: Find the mass of carbon in 55.4 g of carvone, (C 10 H 14 O). mol C 10 H 14 O mol C gCgC

60. Tro's Introductory Chemistry, Chapter 6 60 Apply the Solution Map: = 44.2979 g C = 44.3 g C Sig. Figs. & Round: Information Given:55.4 g C 10 H 14 O Find: g C CF: 1 mol C 10 H 14 O = 150.2 g 1 mol C 10 H 14 O  10 mol C 1 mol C = 12.01 g SM: g C 10 H 14 O  mol C 10 H 14 O  mol C  g C Example: Find the mass of carbon in 55.4 g of carvone, (C 10 H 14 O).

61. Tro's Introductory Chemistry, Chapter 6 61 Check the Solution: 55.4 g C 10 H 14 O = 44.3 g C The units of the answer, g C, are correct. The magnitude of the answer makes sense since the amount of C is less than the amount of C 10 H 14 O. Information Given:55.4 g C 10 H 14 O Find: g C CF: 1 mol C 10 H 14 O = 150.2 g 1 mol C 10 H 14 O  10 mol C 1 mol C = 12.01 g SM: g C 10 H 14 O  mol C 10 H 14 O  mol C  g C Example: Find the mass of carbon in 55.4 g of carvone, (C 10 H 14 O).

62. Tro's Introductory Chemistry, Chapter 6 62 Percent Composition Percentage of each element in a compound By mass Can be determined from 1.the formula of the compound 2.the experimental mass analysis of the compound The percentages may not always total to 100% due to rounding

63. Tro's Introductory Chemistry, Chapter 6 63 Mass Percent as a Conversion Factor the mass percent tells you the mass of a constituent element in 100 g of the compound the fact that NaCl is 39% Na by mass means that 100 g of NaCl contains 39 g Na this can be used as a conversion factor 100 g NaCl  39 g Na

64. Tro's Introductory Chemistry, Chapter 6 64 Example - Percent Composition from the Formula C 2 H 5 OH 1.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 2.Determine the molar mass of the compound by adding the masses of the elements 1 mole C 2 H 5 OH = 46.07 g

65. Tro's Introductory Chemistry, Chapter 6 65 Sample - Percent Composition from the Formula C 2 H 5 OH 3.Divide the mass of each element by the molar mass of the compound and multiply by 100%

66. Tro's Introductory Chemistry, Chapter 6 66 Example – Percent Composition of Carvone if a 30.0 g sample contains 24.0 g of C, 3.2 g O and the rest H? 1.Determine the masses of all the elements in the sample C = 24.0 g, O = 3.2 g H = 30.0 g – (24.0 g + 3.2 g) = 2.8 g

67. 67 Example – Percent Composition of Carvone if a 30.0 g sample contains 24.0 g of C, 3.2 g O and the rest H? 2.Divide the mass of each element by the total mass of the sample then multiply by 100% to give its percentage

68. Tro's Introductory Chemistry, Chapter 6 68 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

69. Tro's Introductory Chemistry, Chapter 6 69 Empirical Formulas Hydrogen Peroxide Molecular Formula = H 2 O 2 Empirical Formula = HO Benzene Molecular Formula = C 6 H 6 Empirical Formula = CH Glucose Molecular Formula = C 6 H 12 O 6 Empirical Formula = CH 2 O

70. Tro's Introductory Chemistry, Chapter 6 70 Finding an Empirical Formula 1)convert the percentages to grams a)skip if already grams 2)convert grams to moles a)use molar mass of each element 3)write a pseudoformula using moles as subscripts 4)divide all by smallest number of moles 5)multiply all mole ratios by number to make all whole numbers a)if ratio ?.5, multiply all by 2; if ratio ?.33 or ?.67, multiply all by 3, etc. b)skip if already whole numbers

71. Example 6.11 Finding an Empirical Formula from Experimental Data

72. Tro's Introductory Chemistry, Chapter 6 72 Example: A laboratory analysis of aspirin determined the following mass percent composition. Find the empirical formula. C = 60.00% H = 4.48% O = 35.53%

73. Tro's Introductory Chemistry, Chapter 6 73 Example: Find the empirical formula of aspirin with the given mass percent composition. Write down the given quantity and its units. Given:C = 60.00% H = 4.48% O = 35.53% Therefore in 100 g of aspirin there are 60.00 g C, 4.48 g H and 35.53 g O

74. Tro's Introductory Chemistry, Chapter 6 74 Write down the quantity to find and/or its units. Find: empirical formula, C x H y O z Information Given:60.00 g C, 4.48 g H, 35.53 g O Example: Find the empirical formula of aspirin with the given mass percent composition.

75. Tro's Introductory Chemistry, Chapter 6 75 Collect Needed Conversion Factors: 1 mole C = 12.01 g C 1 mole H = 1.01 g H 1 mole O = 16.00 g O Information Given:60.00 g C, 4.48 g H, 35.53 g O Find: empirical formula, C x H y O z Example: Find the empirical formula of aspirin with the given mass percent composition.

76. Tro's Introductory Chemistry, Chapter 6 76 Write a Solution Map: g C, H, O mol C, H, O mol ratio empirical formula Information Given:60.00 g C, 4.48 g H, 35.53 g O Find: empirical formula, C x H y O z CF: 1 mol C = 12.01 g; 1 mol H = 1.01 g; 1 mol O = 16.00 g Example: Find the empirical formula of aspirin with the given mass percent composition.

77. 77 Apply the Solution Map: calculate the moles of each element Information Given:60.00 g C, 4.48 g H, 35.53 g O Find: empirical formula, C x H y O z CF: 1 mol C = 12.01 g; 1 mol H = 1.01 g; 1 mol O = 16.00 g SM: g C,H,O  mol C,H,O  mol ratio  empirical formula Example: Find the empirical formula of aspirin with the given mass percent composition.

78. Tro's Introductory Chemistry, Chapter 6 78 Apply the Solution Map: write a pseudoformula Information Given:4.996 mol C, 4.44 mol H, 2.221 mol O Find: empirical formula, C x H y O z CF: 1 mol C = 12.01 g; 1 mol H = 1.01 g; 1 mol O = 16.00 g SM: g C,H,O  mol C,H,O  mol ratio  empirical formula Example: Find the empirical formula of aspirin with the given mass percent composition. C 4.996 H 4.44 O 2.221

79. Tro's Introductory Chemistry, Chapter 6 79 Apply the Solution Map: find the mole ratio by dividing by the smallest number of moles Information Given:C 4.996 H 4.44 O 2.221 Find: empirical formula, C x H y O z CF: 1 mol C = 12.01 g; 1 mol H = 1.01 g; 1 mol O = 16.00 g SM: g C,H,O  mol C,H,O  mol ratio  empirical formula Example: Find the empirical formula of aspirin with the given mass percent composition.

80. Tro's Introductory Chemistry, Chapter 6 80 Apply the Solution Map: multiply subscripts by factor to give whole number Information Given:C 2.25 H 2 O 1 Find: empirical formula, C x H y O z CF: 1 mol C = 12.01 g; 1 mol H = 1.01 g; 1 mol O = 16.00 g SM: g C,H,O  mol C,H,O  mol ratio  empirical formula Example: Find the empirical formula of aspirin with the given mass percent composition. { } x 4 C9H8O4C9H8O4

81. Tro's Introductory Chemistry, Chapter 6 81 All these molecules have the same Empirical Formula. How are the molecules different? NameMolecular Formula Empirical Formula glyceraldehydeC3H6O3C3H6O3 CH 2 O erythroseC4H8O4C4H8O4 CH 2 O arabinoseC 5 H 10 O 5 CH 2 O glucoseC 6 H 12 O 6 CH 2 O

82. Tro's Introductory Chemistry, Chapter 6 82 All these molecules have the same Empirical Formula. How are the molecules different? NameMolecular Formula Empirical Formula Molar Mass, g glyceraldehydeC3H6O3C3H6O3 CH 2 O90 erythroseC4H8O3C4H8O3 CH 2 O120 arabinoseC 5 H 10 O 5 CH 2 O150 glucoseC 6 H 12 O 6 CH 2 O180

83. Tro's Introductory Chemistry, Chapter 6 83 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 Molar Mass real formula Molar Mass empirical formula = factor used to multiply subscripts

84. Tro's Introductory Chemistry, Chapter 6 84 Example – Determine the Molecular Formula of Cadinene if it has a molar mass of 204 g and an empirical formula of C 5 H 8 1.Determine the empirical formula May need to calculate it as previous C5H8C5H8 2.Determine the molar mass of the empirical formula 5 C = 60.05 g, 8 H = 8.064 g C 5 H 8 = 68.11 g

85. Tro's Introductory Chemistry, Chapter 6 85 3.Divide the given molar mass of the compound by the molar mass of the empirical formula Round to the nearest whole number Example – Determine the Molecular Formula of Cadinene if it has a molar mass of 204 g and an empirical formula of C 5 H 8

86. Tro's Introductory Chemistry, Chapter 6 86 4.Multiply the empirical formula by the factor above to give the molecular formula (C 5 H 8 ) 3 = C 15 H 24 Example – Determine the Molecular Formula of Cadinene if it has a molar mass of 204 g and an empirical formula of C 5 H 8