1. The Mole Chapter 10 1

2. Objectives  Use the mole and molar mass to make conversions among moles, mass, and number of particles  Determine the percent composition of the components of a compound  Calculate empirical and molecular formulas for compounds  Determine the formulas for hydrates  Examine relationship between volume and amount of gas (Avogadro’s Law – Ch. 13.2) 2

3. Dimensional Analysis CCCConversion Factors: –W–W–W–What is 8a*5b/2a= - Real Example: 1 mile = 5280 ft –C–C–C–Convert 5.5 miles to feet. –5–5–5–5.5 miles x 5280 ft = = = = 1 mi 20 b 29040 ft 3

4. The Mole  Mole (mol) measures the number of particles of a substance (atom, molecule, formula unit)  Using “mole” is just a shorthand way for saying 6.02 x 10 23 particles  6.02 x 10 23 particles = Avogadro’s number 4

5. How much mass is in one atom of carbon-12 ? Exactly 12 amu (by definition) 5

6. Molar Mass of Atoms  Definition: 12.0000g of Carbon-12 is exactly 1 mol (6.02x10 23 ) of Carbon-12 atoms. 12.0000g of Carbon-12 is exactly 1 mol (6.02x10 23 ) of Carbon-12 atoms.  Therefore: 6

7. Finding Molar Mass of Other Elements  All other elements are determined with respect to Carbon-12 by mass spectroscopy.  For example: the He-4 has 4/12 or 1/3 the mass of Carbon-12 7

8. Finding Molar Mass of Other Elements  Another element: Fe-56 has 55.935 amu –So it is 55.935/12 or 4.67 x the mass of C-12 –If we put all the iron isotopes together, the ‘average’ Fe atom is 55.85 amu –Each amu has a mass of 1.66 x 10 g, so: –Each amu has a mass of 1.66 x 10 -24 g, so: 8

9. Finding the Molar Mass of an Element 9

10. Mass of Atoms  If you have two 1 g samples of different substances, what do they have in common? 10

11. Mass of Atoms 11

12.  The atomic mass of ONE iron atom is 55.85 AMU (atomic mass units).  The molar mass of (a mole of) iron is 55.85 grams. 12

13. The number of grams in a mole is called the molar mass. 13

14. Why does a mole of iron weigh more than a mole of carbon? For the same reason that a dozen bowling balls weighs more than a dozen golf balls! For the same reason that a dozen bowling balls weighs more than a dozen golf balls! Bowling balls are heavier than golf balls. 14

15. Mass of Iron on Balance  55.85 g of iron (iron’s molar mass) on scale is:  1 mole of iron (1 mol Fe = 55.85 g Fe)  6.02 x 10 23 atoms of iron 15

16. A conversion you already know.  How many eggs are in 14 dozen eggs? 14 dozen eggs x 12 eggs = 168 eggs 1 dozen eggs 1 dozen eggs  How many dozens of eggs would you need to buy if you were going to feed 504 people 1 egg each? 504 eggs x 1 dozen eggs = 42 dozen eggs 12 eggs 12 eggs 16

17. Moles and Number of Particles Examples  If have 2 mole of iron how many atoms of iron do you have? 2 moles Fe x 6.02x10 atoms = 12.04x10 atoms 2 moles Fe x 6.02x10 23 atoms = 12.04x10 23 atoms 1 mole 1 mole  If have 12.04 x 10 23 atoms of iron how many moles of iron have you? 12.04x10 atoms x 1 mole____ = 2.0 moles 12.04x10 23 atoms x 1 mole____ = 2.0 moles 6.02x10 atoms 6.02x10 23 atoms 17

18. Mass and Moles Examples WWWWhat is the mass in grams of 2.00 moles of Cu? 2.00 moles Cu x 63.5 grams = 127 grams Cu 1 mole Cu HHHHow moles is 190. g of Cu? 190. g Cu x 1 mole Cu = 3.00 mol Cu 63.5 g 18

19. Mass ↔ Particles Examples 19

20. Mass ↔ Particles Examples 20

21. Flowchart Atoms or Molecules Moles Mass (grams) Divide by 6.02 X 10 23 Multiply by 6.02 X 10 23 Multiply by molar mass from periodic table Divide by molar mass from periodic table 21

22. Mass, Moles, and Number of Particles Practice   How many moles are in 25.5 g of Ag?   How many grams are 42.60 mol of silicon (Si)?   How many atoms are in 15.0 mol of Xe?   How many moles are in 7.23 x 10 24 atoms of Xe?   How many atoms are in 6.50 g of B?   How many grams are in 5.53 x 10 22 atoms of Mg? 22

23. Practice  How many moles are in 25.5 g of Ag? 23

24. Practice  How many grams are 42.60 mol of silicon (Si)? 24

25. Practice  How many atoms are in 15.0 mol of Xe? 25

26. Practice  How many moles are in 7.23 x 10 24 atoms of Xe? 26

27. Practice  How many atoms are in 6.50 g of B? 27

28. Practice  How many grams are in 5.53 x 10 22 atoms of Mg? 28

29. One Mole of Four Elements One mole each of helium, sulfur, copper, and mercury. How many atoms of helium are present? Of sulfur? Of copper? Of mercury? 29

30. One-Mole Quantities of Some Elements & Compounds 30

31. We can also calculate the molar mass of compounds like carbon dioxide. molar massmolar mass The formula for carbon dioxide is CO 2. The formula for carbon dioxide is CO 2. That means there is one carbon atom and two oxygen atoms in every molecule of CO 2. That means there is one carbon atom and two oxygen atoms in every molecule of CO 2. 31

32. We can also calculate the molar mass of compounds like carbon dioxide.  Carbon weighs 12.01 grams/mole  Oxygen weighs 16.00 grams/mole. Therefore CO 2 has a molar mass of: Therefore CO 2 has a molar mass of: C + O + O = CO 2 C + O + O = CO 2 12.01 + 16.00 + 16.00 = 44.01 g 12.01 + 16.00 + 16.00 = 44.01 g 32

33. Finding Molar Mass of CaCl Finding Molar Mass of CaCl 2 (assume 1 mole compound) Atom# mol Atoms in 1 mol of compound Molar Mass (g/mol) Total (g) Ca Cl CaCl 2 33

34. Let’s find out the molar mass of glucose (C 6 H 12 O 6 ). Use the molar masses from the periodic table:  Carbon – 12.0 grams/mole  Hydrogen – 1.0 gram/mole  Oxygen – 16.0 grams/mole 34

35. How many grams are in a mole of glucose (C 6 H 12 O 6 )?  Carbon: 6 x 12.0 g/mol = 72.0 g  Hydrogen: 12 x 1.0 g/mol =12.0 g  Oxygen: 6 x 16.0 g/mol = 96.0 g 180.0 g/mol 180.0 g/mol (For consistency, round all molar masses (elements & compounds) to 0.1 g.) 35

36. Moles to Mass of Compound  How many grams of KCl are in 2.30 mol of KCl?  First, find molar mass. K: 1 K x 39.1 g/mol = 39.1 g K: 1 K x 39.1 g/mol = 39.1 g Cl: 1 Cl x 35.5 g/mol = 35.5 g Cl: 1 Cl x 35.5 g/mol = 35.5 g 74.6 g/mol 74.6 g/mol 36

37. Moles to Mass of Compound HHHHow many grams of KCl are in 2.30 mol of KCl? 2222.30 mol KCl x 74.6 g KCl = mol KCl =171.6 g KCl = 172 g KCl (proper Sig Figs) # of moles Molar mass 37

38. Mass to Moles in a Compound  How many moles of KCl are present in 253.6 grams KCl ? First, determine the molar mass of KCl. First, determine the molar mass of KCl. (Same as before, which is 74.6 g/mol) (Same as before, which is 74.6 g/mol) Then, divide mass by molar mass to get moles. 253.6 g KCl x 1 mol =3.40 mol KCl 74.6 g KCl 74.6 g KCl  38

39. Mass to Moles in a Compound  Can determine the number of moles of each of the atoms/ions that make up the compound. Multiply by ion/compound conversion factor. Conversion factor: mole of specific atom 1 mol of compound 1 mol of compound Example: 2 mol Cl - 1 mol CaCl 2 1 mol CaCl 2 39

40. Mass to Moles in a Compound  How many moles of Cl - ions are there 5.5 mol of CaCl 2 ? 5.5 mol CaCl 2 x 2 mol Cl - = 11.0 mol Cl - 5.5 mol CaCl 2 x 2 mol Cl - = 11.0 mol Cl - 1 mol CaCl 2 1 mol CaCl 2 40

41. Mass, Moles & Particles  Use the flowchart from the elemental calculations and use them for compounds.  Moles are still central  Get to mass (g) by multiplying by molar mass  Get to # molecules (covalent compounds) or formula units (ionic compounds) by multiplying by Avogadro’s number. 41

42. Mass to Moles in a Compound HHHHow many moles of Na+ are there in 561.8 g of Na2CO3? FFFFirst, find Molar Mass of Na2CO3: Na: 2 x 23.0 g/mol = 46.0 g C: 1 x 12.0 g/mol = 12.0 g O O O O: 3 x 16.0 g/mol = 48.0 g Molar Mass = 106.0 g/mol 42

43. Mass to Moles in a Compound HHHHow many moles of Na+ are there in 561.8 g of Na2CO3? NNNNext, find the number of moles of Na2CO3. 561.8 g x 1 1 mol = = 5.300 mol Na2CO3 106.0 g UUUUse conversion of moles of ion/mole of compound. 5555.300 mol Na2CO3 x 2 mol Na+ = 10.6 mol Na+ 1 mol Na2CO3 43

44. Mass to Number of Particles  How many Na + ions are in 561.8 g of Na 2 CO 3 ?  Note: Looking for individual # of ions. Will be a very large number.  First, find # of moles of the ion or element you are interested in.  In this case it was 10.6 mol Na +. 44

45. Mass to Number of Particles FFFFinally, multiply # of mol by Avogadro’s Constant. 11110.6 mol Na+ x 6.02 x 1023 ions 1 1 mol = = 6.38 x 1024 Na+ ions 45

46. Mass to Number of Particles  How many formula units of Na 2 CO 3 are there in the 5.30 mol of Na 2 CO 3 ?  5.30 mol Na 2 CO 3 x 6.02 x 10 23 Frm Unts 1 mol of Na 2 CO 3 1 mol of Na 2 CO 3 = 31.9 x 10 23 formula Units = 31.9 x 10 23 formula Units 46

47. Practice Problems  Determine the # of formula units, the number of moles of each ion, and the number of each ion in: a)2.50 mol ZnCl 2 b)623.7 g of Fe 2 S 3  Determine the mass in grams of 2.11x10 24 formula units of Na 2 S. 47

48. Practice Problems  Determine the # of formula units in 2.50 mole of ZnCl 2. 48

49. Practice Problems  Determine the # of moles Zn 2+ ions in 2.50 mol of ZnCl 2. 49

50. Practice Problems  Determine the # of moles Cl - ions in 2.50 mol of ZnCl 2. 50

51. Practice Problems  Determine the # of Zn 2+ ions in 2.50 mol of ZnCl 2. 51

52. Practice Problems  Determine the # of Cl - ions in 2.50 mol of ZnCl 2. 52

53. Practice Problems DDDDetermine the # of formula units in in 623.7g of Fe2S3. 53

54. Practice Problems  Determine the # of formula units in in 623.7g of Fe 2 S 3. 54

55. Practice Problems  Determine the # of moles of Fe 3+ ions in 623.7g of Fe 2 S 3. 55

56. Practice Problems  Determine the # of moles of S 2- ions in 623.7g of Fe 2 S 3. 56

57. Practice Problems  Determine the # Fe 3+ ions ions in 623.7g of Fe 2 S 3. 57

58. Practice Problems  Determine the # S 2- ions in 623.7g of Fe 2 S 3. 58

59. Practice Problems  Determine the # Fe 3+ ions and S 2- ions in 623.7g of Fe 2 S 3. 59

60. Practice Problems DDDDetermine the mass in grams of 2.11x1024 formula units of Na2S. HHHHint: First find Molar Mass. 60

61. Practice Problems  Determine the mass in grams of 2.11x10 24 formula units of Na 2 S. 61

62. Ch. 10.4 - Percent Composition PPPPercent composition is the percent, by mass, of each element in a compound. IIIIn general, it’s the mass of the element/mass of the formula: Mass of element x 100 = %mass Mass of compound of element 62

63. Percent Composition  Example: If a 50.0 g sample of H 2 O contains 5.6 g of H and 44.4 g of O the percent composition is:  5.6 g H x 100% = 11.1% H 50.0 g H 2 O 50.0 g H 2 O  44.5 g O x 100% = 88.9% O 50 g H 2 O 50 g H 2 O 63

64. Percent Composition  You can calculate the percent composition by finding the mass of each element in 1 mole of a compound.  Example: Water’s formula is H 2 O which means there are 2 mol of hydrogen and 1 mol oxygen in one mol of water. 64

65. Percent Composition SSSSo, if you have 1 mol of water, you have 18.0 g of water (molar mass) TTTThe 18.0 g of water is made up of 2 mol Hydrogen (2 g) and 1 mol oxygen (16 g). HHHH: 2 g x x 100% = 11.1% 18.0 g OOOO: 16 g x 100% = 88.9% 18 g 65

66. Percent Composition Example CCCCalculate the percent composition of each element in Ca(OH)2. 1.D etermine the mass of each element present in 1 mol of cmpd. Ca: 1 mol x 40.1 g/mol = 40.1 g O: 2 mol x 16.0 g/mol = 32.0 g H: 2 mol x 1.0 g/mol = 2.0 g 2. Determine mass in g of one mole of compound. 74.1 g 66

67. Percent Composition Example (Continued) 3.C alculate percentage of each element Ca: 40.1 g/74.1 g * 100 = 54.1% H: 2.0 g/74.1 g * 100 = 2.7% O: 32.0 g/74.1 g * 100 = 43.2% 67

68. Percent Composition Practice DDDDetermine the % Comp of each element in: 1.K Br 2.F e2O3 3.B arium nitrate 68

69. Empirical Formulas  An empirical formula for a compound is the formula of a substance written with the smallest integer subscripts. In other words, the simplest mole ratios.  You can use the percent composition of a compound to determine is empirical formula. 69

70. Empirical Formula SSSSteps to determine formula: 1.C onsider 100g of compound 2.C onvert percentages of elements to grams 3.D ivide each element’s respective mass by its molar mass to obtain moles 4.D ivide each mole value by the smallest mole value. This give the mole ratio. 5.M ultiply by appropriate number to get whole number subscripts. 70

71. Empirical Formula Example Determining the Empirical Formula from the Masses of Elements. We have determined the mass percentage composition of calcium chloride: 36.0% Ca and 64.0% Cl. What is the empirical formula of calcium chloride? 71

72. Empirical formula Calcium Chloride (36% Calcium; 64% Chlorine) AtomMass %In gramsMolar Mass MolesRatio Ca Cl Ratio Ca : Cl  Empirical Formula 72

73. Empirical Formula Practice  Determine the empirical formula of a compound that is 36.8% nitrogen and 63.2% oxygen. 73

74. More Empirical Formula Practice Determining The Empirical Formula from Percentage Composition. (General) Benzene is a widely used industrial solvent. This compound has been analyzed and found to contain 92.26% carbon and 7.74% hydrogen by mass. What is its empirical formula? Hint: Consider a 100 g sample. The empirical formula is: 74

75. same empirical formula Compounds with different molecular formulas can have the same empirical formula, and such substances will have the same percentage composition. Remember that the molecular formula has the actual number of atoms of each element that make one molecule of that compound. Eg. Compound A = C 2 H 2 Compound B = C 6 H 6 both have the empirical formula = Molecular Formulas 75

76. Molecular Formula from Empirical Formula The molecular formula of a compound is a multiple of its empirical formula. Molecular mass = n x empirical formula mass where n = number of empirical formula units in the molecule. 76

77. Determining the Molecular Formula from the Percent Composition and Molar Mass. We have already determined the mass composition and empirical formula of benzene (CH). In a separate experiment, the molar mass of benzene was determined to be 78.1. What is the molecular formula of benzene Mass of empirical formula = Molecular Formula Example Molar mass benzene _ = Mass of empirical formula 77

78. Molecular Formula Example WWWWhat is the molecular formula of a compound that has an empirical formula of CH3 and a molar mass of 30.0 g/mol. WWWWhat is the mass of each empirical unit? C: 1 x 12.0 = 12.0 H: 3 x 1.0 = 3.0 15.0 How many times units? 30/15 = 2 So molecular formula is…C2H6 78

79. Molecular Formula Practice 1.Empirical Formula is NO 2 ; molar mass is 92.0 g/mol 2.A compound contains 26.76% C, 2.21%H, 71.17% O and has a molar mass of 90.04 g/mol. Determine its molecular formula. 79

80. Ch. 10.5 - Salt Hydrates Ch. 10.5 - Salt Hydrates  Water molecules bound to a compound compound  For example, CaCl 2 2H 2 O –Each molecule of calcium chloride has two water molecules bound to it  See “conceptual” picture next slide 80

81. CaCl 2 2H 2 O Ca +2 Cl - Water -- -- 81

82. Example of Percent Composition of a Hydrate DDDDetermine the percent salt and percent water (by mass) in 1 mol of CaCl22H2O. 1.F ind the mass of 1 mole of the compound. Ca: 1 x 40.1 = 40.1 Cl: 2 x 35.5 = 71.0 = 147.1 g/mol H2O: 2 x 18.0 = 36.0 82

83. Example of Percent Composition of a Hydrate 2.F ind the percent of salt. 111.1 g CaCl2 = = 75.5% CaCl2 147.1 g CaCl22H2O 3.F ind the percent of water. 3 36.0 g H2O = = 24.5% H2O 147.1 g CaCl22H2O Or you can say 100% – 75.5% CaCl2 = 24.5% H2O. 83

84. Determination of Formula Example  A nickel(II) cyanide hydrate, Ni(CN) 2 XH 2 O, contains 39.4% water by mass. What is the formula of the hydrated compound? 1.Assume 100 g of the compound. If it is 39.4% water, it is 100 - 39.4 or 60.6% Ni(CN) 2. So there are If it is 39.4% water, it is 100 - 39.4 or 60.6% Ni(CN) 2. So there are 60.6 g of Ni(CN) 2 84

85. Determination of Formula Example 2.D etermine the number of moles of the salt and the water. 60.6 g Ni(CN)2 = = = = 0.547mol Ni(CN)2 110.7 g/mol 39.4 g H2O = 2.19 mol H2O 18.0 g/mol Ni: 1 x 58.7 = 58.7 C: 2 x 12.0 = 24.0 N: 2 x 14.0 = 28.0 110.7 85

86. Determination of Formula Example 3.D etermine the mol ratio of the water to the salt. 2.19 mol H2O 0.547 mol Ni(CN)2 = 4 mol H2O/mol Ni(CN)2 So the formula is Ni(CN)24H2O 86

87. Moles and Gases (Ch. 13.1-13.2)  Remember Boyle’s Law, Charles’s Law and Combined Gas Law?  Boyle: P 1 V 1 =P 2 V 2  Charles: V 1 /T 1 =V 2 /T 2  Combined: P 1 V 1 = P 2 V 2 T 1 T 2 T 1 T 2 There is another law that relates volume to moles. 87

88. The Gas Laws – Avogadro’s Law  Amedeo Avogadro (1776 – 1856) studied the relationship between volume and amount of gas. 88

89. The Gas Laws – Avogadro’s Law  Avogadro’s Principle: Equal volumes of gas at the same P & T contain equal numbers of particles (molecules or atoms).  Avogadro’s Law – The volume of a gas at constant P & T is directly proportional to the number of moles of gas. Volume α n ; hold P & T constant V = k x n 89

90. Avogadro’s Principle  As an extension of Avogadro’s Principle, 1 mol of a gas at 0˚C (273 K) and 1 atm of pressure occupies a volume of 22.4 Liters (Molar Volume)  0˚C (273 K) and 1 atm of pressure are standard temperature and pressure (STP) 90

91. Avogadro’s Principle  Since  Since volume is related to moles, if you know the volume a gas occupies at STP, you know the # of moles. 22.4 L 1 mol  Example:  Example: If you have 11.2 L of a gas at STP, how many moles are there? 11.2 L x 1 mol mol = 0.500 mole 22.4 L 91

92. Avogadro’s Principle Practice  Given the conditions at STP find:  Volume of 0.881 mol of gas  # of moles of N 2 (g) in a 2.0 L flask  # of atoms of Kr in 28.5 L 92

93. Ch. 13.2 - Ideal Gas Law  The gas laws that we covered all have one thing in common: they relate the volume of the gas to one of the other variables.  Boyle: V  1/P  Charles: V  T  Avogadro: V  n  We can put them all together to get… 93

94. Ideal Gas Law V  nT P  To change from a proportionality, , to an equation, we introduce, R, the proportionality constant or Universal Gas Constant. 94

95. Ideal Gas Law TTTThis makes the previous proportionality relationship into V = R*nT/P Or more commonly PV = nRT the IDEAL GAS EQUATION 95

96. Ideal Gas Law  The most common value of R is 0.0821 L-atm Mol-K Mol-K There are other values of R with different units that are listed in your book, but we’ll use the one above. Note: they all are equivalent, it just depends on what unit (mostly pressure) you are using. 96

97. Relationship to Other Gas Laws P 1 V 1 = P 2 V 2 n 1 T 1 n 2 T 2 Both equations are equal to a constant, which is…R (the gas constant) 97

98. Ideal Gas Equation Example AAAA deodorant can has a volume of 0.175 L and a pressure of 3.80 atm at 22ºC. How many moles of gas are contained in the can? GGGGiven: P=3.80 atm; V = 0.175 L; T=22+273=295 K; R = 0.0821 L-atm/mol-K PPPPV=nRT  Looking for n (moles) So… nnnn = PV = (3.80 atm)(0.175 L) RT (0.0821L*atm/mol*K)*295K ====0.0275 mol 98

99. Ideal Gas Equation Practice  Calculate the volume that a 0.323 mol sample of gas will occupy at 265 K and 0.900 atm.  A 47.3 L container containing 1.62 mol of He is heated until the pressure reaches 1.85 atm. What is the temperature in ⁰C ?  Kr gas in a 18.5 L cylinder exerts a pressure of 8.61 atm at 24.8ºC What is the mass of Kr? 99

100. Gas Density and Molar Mass LLLLet M stand for molar mass (g/mol) WWWWhere n = m/M ; m = mass in grams, n is moles AAAAnd we know PV=nRT SSSSo then PV= (m/M)RT RRRRearranging gives: M = mRT V P FFFFinally we get: M = dRT/P 100 Mass/volume = Density

101. Gas Density & Molar Mass Example 1 WWWWhat is molar mass of a gas that has a density of 1.40 g/L at STP? GGGGiven: T=273; P=1.00 atm; d=1.40 g/L; R=0.0821 L-atm/mol-K KKKKnow M = DRT/P so… =(1.40 g/L)(0.0821 L-atm/mol-K)(273K) 1.0 atm ==== 31.4 g/mol 101

102. Dalton’s Law of Partial Pressures  Dalton’s Law of Partial Pressure (end of Ch. 12.1) states that the total pressure of a mixture of gases is equal to the sum of the pressures of all the gases in the mixture.  In other words… P total = P 1 + P 2 + P 3 …P n 102

103. Dalton’s Law of Partial Pressures John Dalton 1766-1844 103

104. Dalton’s Law of Partial Pressures  partial pressure depends only on: –number of moles of gas –container volume –temperature of the gas.  It does not depend on the identity of the gas.  Because gas molecules are so far apart, they don’t interact with each other. 104

105. Dalton’s Law Example 1  We have a mixture of O 2, CO 2, and N 2 in a vessel at STP. The partial pressure of CO 2 is 0.70 atm and the partial pressure of N 2 is 0.12 atm. What is the partial pressure of O 2 ?  Answer: What is total pressure? 1 atm, so…  1.00 = 0.70 + 0.12 + P(O 2 ) P(O 2 ) = 1.00 – 0.70 -0.12 = 0.18 atm 105

106. Partial Pressure & Mole Fraction  The partial pressure of a gas component in a mixture is dependent on how much (moles) are there:  P a = n A RT/V  The total pressure of a gas mixture is P total = n total RT/V 106

107. Partial Pressure & Mole Fraction 107

108. Partial Pressure & Mole Fraction Mole Fraction 108

109. Mole Fraction Example 1  A mixture of O 2, N 2, and He gases are in an enclosed tank. The total pressure in the tank is 12.3 atm. If there are 10.6 mol of N 2, 3.3 mol O 2 and 1.2 mol He in the tank, what is the partial pressure of each gas? 109

110. Mole Fraction Example 1 110

111. Mole Fraction Example 2 A gas mixture containing 0.538 mol He, 0.315 mol Ne and 0.103 mol Ar is confined to a 7.00 L vessel at 25°C. a)Calculate the partial pressure of each gas in the mixture b)Calculate the total pressure of the mixture. P(He) = 0.538 mol(0.0821 L-atm/mol-K)(298K)/7.0 L = 1.88 atm P(Ne) = 0.315 mol(0.0821 L-atm/mol-K)(298K)/7.0 L = 1.10 atm P(Ar) = 0.103 mol(0.0821 L-atm/mol-K)(298K)/7.0 L = 0.36 atm 111

112. Mole Fraction Example 2  Can do previous problem because the gases are at same T & V.  Therefore, their mole ratios will be proportional.  Greater moles means greater partial pressure! 112

113. Tired of moles? So am I. 113