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Counting Atoms by Moles ,[object Object],[object Object],[object Object],[object Object],[object Object],Tro's "Introductory Chemistry", Chapter 6
Chemical Packages—Moles ,[object Object],[object Object],[object Object],[object Object],Tro's "Introductory Chemistry", Chapter 6
[object Object],[object Object],Tro's "Introductory Chemistry", Chapter 6
[object Object],[object Object],[object Object],[object Object],[object Object],Example: A silver ring contains 1.1 x 10 22  silver atoms.  How many moles of silver are in the ring? Tro's "Introductory Chemistry", Chapter 6
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Example: A silver ring contains 1.1 x 10 22  silver atoms.  How many moles of silver are in the ring? Tro's "Introductory Chemistry", Chapter 6 atoms Ag moles Ag
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Example: A silver ring contains 1.1 x 10 22  silver atoms.  How many moles of silver are in the ring? Tro's "Introductory Chemistry", Chapter 6 = 1.8266 x 10 22  moles Ag  = 1.8 x 10 22  moles Ag  ,[object Object]
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Example: A silver ring contains 1.1 x 10 22  silver atoms.  How many moles of silver are in the ring? Tro's "Introductory Chemistry", Chapter 6 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.
Practice—Calculate the Number of Atoms in 2.45 Mol of Copper. Tro's "Introductory Chemistry", Chapter 6
Practice—Calculate the Number of Atoms in 2.45 Mol of Copper, Continued. Tro's "Introductory Chemistry", Chapter 6 Since atoms are small, the large number of atoms makes sense. 1 mol = 6.022 x 10 23  atoms 2.45 mol Cu atoms Cu Check: Solution: Solution Map: Relationships: Given: Find: mol Cu atoms Cu
Relationship Between  Moles and Mass ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Tro's "Introductory Chemistry", Chapter 6
Example 6.2—Calculate the Moles of Sulfur  in 57.8 G of Sulfur. Tro's "Introductory Chemistry", Chapter 6 Since the given amount is much less than 1 mol S, the number makes sense. 1 mol S = 32.07 g 57.8 g S mol S Check: Solution: Solution Map: Relationships: Given: Find: g S mol S
[object Object],[object Object],Tro's "Introductory Chemistry", Chapter 6
Example: Calculate the number of moles of sulfur in 57.8 g of sulfur. ,[object Object],[object Object],Tro's "Introductory Chemistry", Chapter 6
[object Object],[object Object],[object Object],[object Object],Example: Calculate the number of moles of sulfur in 57.8 g of sulfur. Tro's "Introductory Chemistry", Chapter 6
[object Object],[object Object],[object Object],[object Object],[object Object],Example: Calculate the number of moles of sulfur in 57.8 g of sulfur. Tro's "Introductory Chemistry", Chapter 6
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Example: Calculate the number of moles of sulfur in 57.8 g of sulfur. Tro's "Introductory Chemistry", Chapter 6 g S moles S
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Example: Calculate the number of moles of sulfur in 57.8 g of sulfur. Tro's "Introductory Chemistry", Chapter 6 = 1.802307 moles S  = 1.80  moles S  ,[object Object]
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Example: Calculate the number of moles of sulfur in 57.8 g of sulfur. Tro's "Introductory Chemistry", Chapter 6 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.
Practice—Calculate the Moles of Carbon in 0.0265 g of Pencil Lead. Tro's "Introductory Chemistry", Chapter 6
Practice—Calculate the Moles of Carbon in 0.0265 g of Pencil Lead, Continued. Tro's "Introductory Chemistry", Chapter 6 Since the given amount is much less than 1 mol C, the number makes sense. 1 mol C = 12.01 g 0.0265 g C mol C Check: Solution: Solution Map: Relationships: Given: Find: g C mol C
[object Object],[object Object],Tro's "Introductory Chemistry", Chapter 6
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Example: How many aluminum atoms are in an aluminum can with a mass of 16.2 g? Tro's "Introductory Chemistry", Chapter 6 g Al mol Al atoms Al
Molar Mass of Compounds ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Tro's "Introductory Chemistry", Chapter 6
[object Object],[object Object],Tro's "Introductory Chemistry", Chapter 6
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Example: Calculate the mass (in grams) of 1.75 mol of water.  Tro's "Introductory Chemistry", Chapter 6
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Example: Calculate the mass (in grams) of 1.75 mol of water. Tro's "Introductory Chemistry", Chapter 6 mol H 2 O g H 2 O
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Example: Calculate the mass (in grams) of 1.75 mol of water.  Tro's "Introductory Chemistry", Chapter 6 = 31.535 g H 2 O  = 31.5 g H 2 O  ,[object Object]
Practice—How Many Moles Are in 50.0 g of PbO 2 ? (Pb = 207.2, O = 16.00) Tro's "Introductory Chemistry", Chapter 6
Practice—How Many Moles Are in 50.0 G of PbO 2 ? (Pb = 207.2, O = 16.00), Continued Tro's &quot;Introductory Chemistry&quot;, Chapter 6 Since the given amount is less than 239.2 g, the moles being < 1 makes sense. 1 mol PbO 2  = 239.2 g 50.0 g mol PbO 2 moles PbO 2 Check: Solution: Solution Map: Relationships: Given: Find: g PbO 2 mol PbO 2
[object Object],[object Object],Tro's &quot;Introductory Chemistry&quot;, Chapter 6
Example: Find the mass of 4.78 x 10 24  NO 2  molecules. ,[object Object],[object Object],Tro's &quot;Introductory Chemistry&quot;, Chapter 6
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Example: Find the mass of 4.78 x 10 24  NO 2  molecules. Tro's &quot;Introductory Chemistry&quot;, Chapter 6
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Example: Find the mass of 4.78 x 10 24  NO 2  molecules. Tro's &quot;Introductory Chemistry&quot;, Chapter 6 = 365.21 g NO 2   = 365 g NO 2   ,[object Object]
Mole Relationships in Chemical Formulas ,[object Object],Tro's &quot;Introductory Chemistry&quot;, Chapter 6 Moles of compound Moles of constituents 1 mol NaCl 1 mol Na, 1 mol Cl 1 mol H 2 O 2 mol H, 1 mol 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
[object Object],[object Object],Tro's &quot;Introductory Chemistry&quot;, Chapter 6
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Example: Find the mass of carbon in 55.4 g of carvone, (C 10 H 14 O). Tro's &quot;Introductory Chemistry&quot;, Chapter 6
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Example: Find the mass of carbon in 55.4 g of carvone, (C 10 H 14 O). Tro's &quot;Introductory Chemistry&quot;, Chapter 6 = 44.2979 g C  = 44.3 g C  ,[object Object]
Practice—Find the Mass of Sodium in 6.2 g of NaCl (MM: Na = 22.99, Cl = 35.45) Tro's &quot;Introductory Chemistry&quot;, Chapter 6
Practice—Find the Mass of Sodium in 6.2 g of NaCl, Continued Tro's &quot;Introductory Chemistry&quot;, Chapter 6 1 mol NaCl = 58.44 g, 1 mol Na = 22.99 g,  1 mol Na : 1 mol NaCl Since the amount of Na is less than the amount of  NaCl, the answer makes sense.  6.2 g NaCl g Na Check: Solution: Solution Map: Relationships: Given: Find: g NaCl mol NaCl mol Na g Na
Percent Composition ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Tro's &quot;Introductory Chemistry&quot;, Chapter 6
Example 6.9—Find the Mass Percent of Cl in C 2 Cl 4 F 2 . Since the percentage is less than 100 and Cl is much heavier than the other atoms, the number makes sense. C 2 Cl 4 F 2 % Cl by mass Check: Solution: Solution Map: Relationships: Given: Find:
Practice—Determine the Mass Percent Composition of the Following: ,[object Object],Tro's &quot;Introductory Chemistry&quot;, Chapter 6
Practice—Determine the Percent Composition of the Following, Continued: ,[object Object],Tro's &quot;Introductory Chemistry&quot;, Chapter 6
Mass Percent as a  Conversion Factor ,[object Object],[object Object],[object Object],[object Object],Tro's &quot;Introductory Chemistry&quot;, Chapter 6
Empirical Formulas ,[object Object],[object Object],[object Object],Tro's &quot;Introductory Chemistry&quot;, Chapter 6 % A  mass A (g)  moles A 100g  MM A % B  mass B (g)  moles B 100g  MM B moles A moles B
Empirical Formulas, Continued Benzene Molecular formula = C 6 H 6 Empirical formula = CH Glucose Molecular formula = C 6 H 12 O 6 Empirical formula = CH 2 O Hydrogen Peroxide Molecular formula = H 2 O 2 Empirical formula = HO
Practice—Determine the Empirical Formula of Benzopyrene, C 20 H 12 . Tro's &quot;Introductory Chemistry&quot;, Chapter 6
Practice—Determine the Empirical Formula of Benzopyrene, C 20 H 12 , Continued ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Tro's &quot;Introductory Chemistry&quot;, Chapter 6
Finding an Empirical Formula ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Tro's &quot;Introductory Chemistry&quot;, Chapter 6
[object Object],[object Object],[object Object],[object Object],[object Object],Tro's &quot;Introductory Chemistry&quot;, Chapter 6
Example: Find the empirical formula of aspirin with the given mass percent composition. ,[object Object],[object Object],[object Object],[object Object],[object Object],Tro's &quot;Introductory Chemistry&quot;, Chapter 6
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Example: Find the empirical formula of aspirin with the given mass percent composition. Tro's &quot;Introductory Chemistry&quot;, Chapter 6 g C mol C g H mol H pseudo- formula empirical formula mole ratio whole number ratio g O mol O
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Example: Find the empirical formula of aspirin with the given mass percent composition. Tro's &quot;Introductory Chemistry&quot;, Chapter 6
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Example: Find the empirical formula of aspirin with the given mass percent composition. Tro's &quot;Introductory Chemistry&quot;, Chapter 6 C 4.996 H 4.44 O 2.221
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Example: Find the empirical formula of aspirin with the given mass percent composition. Tro's &quot;Introductory Chemistry&quot;, Chapter 6
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Example: Find the empirical formula of aspirin with the given mass percent composition. Tro's &quot;Introductory Chemistry&quot;, Chapter 6 {  }  x 4 C 9 H 8 O 4
Example 6.12—Finding an Empirical Formula from Experimental Data Tro's &quot;Introductory Chemistry&quot;, Chapter 6
[object Object],[object Object],Tro's &quot;Introductory Chemistry&quot;, Chapter 6
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Example:  Find the empirical formula of oxide of titanium with the given elemental analysis. Tro's &quot;Introductory Chemistry&quot;, Chapter 6
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Example: Find the empirical formula of oxide of titanium with the given elemental analysis. Tro's &quot;Introductory Chemistry&quot;, Chapter 6 g Ti mol Ti pseudo- formula empirical formula mole ratio whole number ratio g O mol O
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Example: Find the empirical formula of oxide of titanium with the given elemental analysis. Tro's &quot;Introductory Chemistry&quot;, Chapter 6 5.40 g compound  − 3.24 g Ti = 2.6 g O
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Example: Find the empirical formula of oxide of titanium with the given elemental analysis. Tro's &quot;Introductory Chemistry&quot;, Chapter 6
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Example: Find the empirical formula of oxide of titanium with the given elemental analysis. Tro's &quot;Introductory Chemistry&quot;, Chapter 6 Ti 0.0677 O 0.135
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Example: Find the empirical formula of oxide of titanium with the given elemental analysis. Tro's &quot;Introductory Chemistry&quot;, Chapter 6
Practice—Determine the Empirical Formula of Stannous Fluoride, which Contains 75.7% Sn (118.70) and the Rest Fluorine (19.00). Tro's &quot;Introductory Chemistry&quot;, Chapter 6
Practice—Determine the Empirical Formula of Stannous Fluoride, which Contains 75.7% Sn (118.70) and the Rest Fluorine (19.00), Continued. Tro's &quot;Introductory Chemistry&quot;, Chapter 6 Given: 75.7% Sn, (100 – 75.3) = 24.3% F     in 100 g stannous fluoride there are 75.7 g Sn and 24.3 g F. Find:  Sn x F y Conversion Factors: 1 mol Sn = 118.70 g;  1 mol F = 19.00 g Solution Map: g Sn mol Sn g F mol F pseudo- formula empirical formula mole ratio whole number ratio
Practice—Determine the Empirical Formula of Stannous Fluoride, which Contains 75.7% Sn (118.70) and the Rest Fluorine (19.00), Continued. Tro's &quot;Introductory Chemistry&quot;, Chapter 6 Apply solution map: Sn 0.638 F 1.28 SnF 2
All These Molecules Have the Same Empirical Formula.  How Are the Molecules Different? Tro's &quot;Introductory Chemistry&quot;, Chapter 6 Name Molecular Formula Empirical Formula Glyceraldehyde C 3 H 6 O 3 CH 2 O Erythrose C 4 H 8 O 4 CH 2 O Arabinose C 5 H 10 O 5 CH 2 O Glucose C 6 H 12 O 6 CH 2 O
All These Molecules Have the Same Empirical Formula.  How Are the Molecules Different?, Continued Tro's &quot;Introductory Chemistry&quot;, Chapter 6 Name Molecular Formula Empirical Formula Molar Mass, g Glyceraldehyde C 3 H 6 O 3 CH 2 O 90 Erythrose C 4 H 8 O 4 CH 2 O 120 Arabinose C 5 H 10 O 5 CH 2 O 150 Glucose C 6 H 12 O 6 CH 2 O 180
Molecular Formulas ,[object Object],[object Object],Tro's &quot;Introductory Chemistry&quot;, Chapter 6 Molar mass real formula Molar mass empirical formula =  Factor used to multiply subscripts
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 .  ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Tro's &quot;Introductory Chemistry&quot;, Chapter 6
[object Object],[object Object],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 , Continued. Tro's &quot;Introductory Chemistry&quot;, Chapter 6

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Tro3 lecture 06

  • 1.
  • 2.
  • 3.
  • 4.
  • 5.
  • 6.
  • 7.
  • 8. Practice—Calculate the Number of Atoms in 2.45 Mol of Copper. Tro's &quot;Introductory Chemistry&quot;, Chapter 6
  • 9. Practice—Calculate the Number of Atoms in 2.45 Mol of Copper, Continued. Tro's &quot;Introductory Chemistry&quot;, Chapter 6 Since atoms are small, the large number of atoms makes sense. 1 mol = 6.022 x 10 23 atoms 2.45 mol Cu atoms Cu Check: Solution: Solution Map: Relationships: Given: Find: mol Cu atoms Cu
  • 10.
  • 11. Example 6.2—Calculate the Moles of Sulfur in 57.8 G of Sulfur. Tro's &quot;Introductory Chemistry&quot;, Chapter 6 Since the given amount is much less than 1 mol S, the number makes sense. 1 mol S = 32.07 g 57.8 g S mol S Check: Solution: Solution Map: Relationships: Given: Find: g S mol S
  • 12.
  • 13.
  • 14.
  • 15.
  • 16.
  • 17.
  • 18.
  • 19. Practice—Calculate the Moles of Carbon in 0.0265 g of Pencil Lead. Tro's &quot;Introductory Chemistry&quot;, Chapter 6
  • 20. Practice—Calculate the Moles of Carbon in 0.0265 g of Pencil Lead, Continued. Tro's &quot;Introductory Chemistry&quot;, Chapter 6 Since the given amount is much less than 1 mol C, the number makes sense. 1 mol C = 12.01 g 0.0265 g C mol C Check: Solution: Solution Map: Relationships: Given: Find: g C mol C
  • 21.
  • 22.
  • 23.
  • 24.
  • 25.
  • 26.
  • 27.
  • 28. Practice—How Many Moles Are in 50.0 g of PbO 2 ? (Pb = 207.2, O = 16.00) Tro's &quot;Introductory Chemistry&quot;, Chapter 6
  • 29. Practice—How Many Moles Are in 50.0 G of PbO 2 ? (Pb = 207.2, O = 16.00), Continued Tro's &quot;Introductory Chemistry&quot;, Chapter 6 Since the given amount is less than 239.2 g, the moles being < 1 makes sense. 1 mol PbO 2 = 239.2 g 50.0 g mol PbO 2 moles PbO 2 Check: Solution: Solution Map: Relationships: Given: Find: g PbO 2 mol PbO 2
  • 30.
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  • 38. Practice—Find the Mass of Sodium in 6.2 g of NaCl (MM: Na = 22.99, Cl = 35.45) Tro's &quot;Introductory Chemistry&quot;, Chapter 6
  • 39. Practice—Find the Mass of Sodium in 6.2 g of NaCl, Continued Tro's &quot;Introductory Chemistry&quot;, Chapter 6 1 mol NaCl = 58.44 g, 1 mol Na = 22.99 g, 1 mol Na : 1 mol NaCl Since the amount of Na is less than the amount of NaCl, the answer makes sense. 6.2 g NaCl g Na Check: Solution: Solution Map: Relationships: Given: Find: g NaCl mol NaCl mol Na g Na
  • 40.
  • 41. Example 6.9—Find the Mass Percent of Cl in C 2 Cl 4 F 2 . Since the percentage is less than 100 and Cl is much heavier than the other atoms, the number makes sense. C 2 Cl 4 F 2 % Cl by mass Check: Solution: Solution Map: Relationships: Given: Find:
  • 42.
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  • 45.
  • 46. Empirical Formulas, Continued Benzene Molecular formula = C 6 H 6 Empirical formula = CH Glucose Molecular formula = C 6 H 12 O 6 Empirical formula = CH 2 O Hydrogen Peroxide Molecular formula = H 2 O 2 Empirical formula = HO
  • 47. Practice—Determine the Empirical Formula of Benzopyrene, C 20 H 12 . Tro's &quot;Introductory Chemistry&quot;, Chapter 6
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  • 57. Example 6.12—Finding an Empirical Formula from Experimental Data Tro's &quot;Introductory Chemistry&quot;, Chapter 6
  • 58.
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  • 64.
  • 65. Practice—Determine the Empirical Formula of Stannous Fluoride, which Contains 75.7% Sn (118.70) and the Rest Fluorine (19.00). Tro's &quot;Introductory Chemistry&quot;, Chapter 6
  • 66. Practice—Determine the Empirical Formula of Stannous Fluoride, which Contains 75.7% Sn (118.70) and the Rest Fluorine (19.00), Continued. Tro's &quot;Introductory Chemistry&quot;, Chapter 6 Given: 75.7% Sn, (100 – 75.3) = 24.3% F  in 100 g stannous fluoride there are 75.7 g Sn and 24.3 g F. Find: Sn x F y Conversion Factors: 1 mol Sn = 118.70 g; 1 mol F = 19.00 g Solution Map: g Sn mol Sn g F mol F pseudo- formula empirical formula mole ratio whole number ratio
  • 67. Practice—Determine the Empirical Formula of Stannous Fluoride, which Contains 75.7% Sn (118.70) and the Rest Fluorine (19.00), Continued. Tro's &quot;Introductory Chemistry&quot;, Chapter 6 Apply solution map: Sn 0.638 F 1.28 SnF 2
  • 68. All These Molecules Have the Same Empirical Formula. How Are the Molecules Different? Tro's &quot;Introductory Chemistry&quot;, Chapter 6 Name Molecular Formula Empirical Formula Glyceraldehyde C 3 H 6 O 3 CH 2 O Erythrose C 4 H 8 O 4 CH 2 O Arabinose C 5 H 10 O 5 CH 2 O Glucose C 6 H 12 O 6 CH 2 O
  • 69. All These Molecules Have the Same Empirical Formula. How Are the Molecules Different?, Continued Tro's &quot;Introductory Chemistry&quot;, Chapter 6 Name Molecular Formula Empirical Formula Molar Mass, g Glyceraldehyde C 3 H 6 O 3 CH 2 O 90 Erythrose C 4 H 8 O 4 CH 2 O 120 Arabinose C 5 H 10 O 5 CH 2 O 150 Glucose C 6 H 12 O 6 CH 2 O 180
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Editor's Notes

  1. Formula is C15H24