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Chapter 3

Chapter 3. Calculations with Chemical Formulas and Equations. Molar Mass. Sum atomic masses represented by formula atomic masses => gaw molar mass => MM. One Mole of each Substance. Clockwise from top left: 1-Octanol, C 8 H 17 OH; Mercury(II) iodide, HgI 2 ;

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Chapter 3

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  1. Chapter 3 Calculations with Chemical Formulas and Equations Dr. S. M. Condren

  2. Molar Mass Sum atomic masses represented by formula atomic masses => gaw molar mass => MM Dr. S. M. Condren

  3. One Mole of each Substance Clockwise from top left: 1-Octanol, C8H17OH; Mercury(II) iodide, HgI2; Methanol, CH3OH; and Sulfur, S8. Dr. S. M. Condren

  4. Example What is the molar mass of ethanol, C2H5O1H1? MM = 2(gaw)C + (5 + 1)(gaw)H + 1(gaw)O = 2(12.011)C + 6(1.00794)H + 1(15.9994)O = 24.022 + 6.04764 + 15.9994 = 46.069 g/mol Significant figures rule for multiplication Significant figures rule for addition Sequence – multiplication then addition, apply significant figure rules in proper sequence Dr. S. M. Condren

  5. The Mole • a unit of measurement, quantity of matter present • Avogadro’s Number 6.022 x 1023 particles • Latin for “pile” Dr. S. M. Condren

  6. Example How many moles of carbon dioxide molecules are there in 10.00g of carbon dioxide? CO2 MM = 1(gaw)C + 2(gaw)O = 44.01 g/mol #mol CO2 = (10.00g) (1 mol/44.01g) Dr. S. M. Condren

  7. Example How many moles of carbon dioxide molecules are there in 10.00g of carbon dioxide? CO2 MM = 1(gaw)C + 2(gaw)O = 44.01 g/mol #mol CO2 = (10.00g) (1 mol/44.01g) Dr. S. M. Condren

  8. Example How many moles of carbon dioxide molecules are there in 10.00g of carbon dioxide? CO2 MM = 1(gaw)C + 2(gaw)O = 44.01 g/mol #mol CO2 = (10.00) (1 mol/44.01) = 0.2272 mol Dr. S. M. Condren

  9. Combustion Analysis Dr. S. M. Condren

  10. Percentage Composition description of a compound based on the relative amounts of each element in the compound Dr. S. M. Condren

  11. EXAMPLE: What is the percent composition of chloroform, CHCl3, a substance once used as an anesthetic? MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl = (12.011 + 1.00797 + 3*35.453)amu = 119.377amu Dr. S. M. Condren

  12. EXAMPLE: What is the percent composition of chloroform, CHCl3, a substance once used as an anesthetic? MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl = (12.011 + 1.00797 + 3*35.453)amu = 119.377amu 1(gaw) %C = ------------ X 100 MM 1(12.011) %C = -------------- X 100 = 10.061% C 119.377 Dr. S. M. Condren

  13. EXAMPLE: What is the percent composition of chloroform, CHCl3, a substance once used as an anesthetic? MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl = (12.011 + 1.00797 + 3*35.453)amu = 119.377amu 1(1.00797) %H = ---------------- X 100 = 0.844359% H 119.377 Dr. S. M. Condren

  14. EXAMPLE: What is the percent composition of chloroform, CHCl3, a substance once used as an anesthetic? MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl = (12.011 + 1.00797 + 3*35.453)amu = 119.377amu 3(35.453) %Cl = -------------- X 100 = 89.095% Cl 119.377 Dr. S. M. Condren

  15. EXAMPLE: What is the percent composition of chloroform, CHCl3, a substance once used as an anesthetic? MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl = 119.377amu %C = 10.061% C %H = 0.844359% H %Cl = 89.095% Cl Dr. S. M. Condren

  16. Simplest (Empirical) Formula • formula describing a substance based on the smallest set of subscripts Dr. S. M. Condren

  17. Acetylene, C2H2, and benzene, C6H6, have the same empirical formula. Is the correct empirical formula: C2H2 CH C6H6 Dr. S. M. Condren

  18. EXAMPLE: A white compound is formed when phosphorus burns in air. Analysis shows that the compound is composed of 43.7% P and 56.3% O by mass. What is the empirical formula of the compound? Element P O Relative Number of Atoms (%/gaw) % 43.7 56.3 Divide by Smaller 1.41/1.41 = 1.00 3.52/1.41 = 2.50 43.7/30.97 = 1.41 56.3/15.9994 = 3.52 Dr. S. M. Condren

  19. EXAMPLE: A white compound is formed when phosphorus burns in air. Analysis shows that the compound is composed of 43.7% P and 56.3% O by mass. What is the empirical formula of the compound? Relative Number of Atoms Multiply % (%/gaw) Divide by Smaller by Integer P 43.743.7/30.97 = 1.41 1.41/1.41 = 1.00 2*1.00 => 2 O 56.356.3/15.9994 = 3.523.52/1.41 = 2.50 2*2.50 => 5 Empirical Formula => P2O5 Dr. S. M. Condren

  20. EXAMPLE: A sample of a brown-colored gas that is a major air pollutant is found to contain 2.34 g of N and 5.34 g of O. What is the empirical formula of the compound? 2.34 %N = ----------------- X 100 = 30.5% N 2.34 + 5.34 5.34 %O = ----------------- X 100 = 69.5% O 2.34 + 5.34 Dr. S. M. Condren

  21. EXAMPLE: A sample of a brown-colored gas that is a major air pollutant is found to contain 2.34 g of N and 5.34 g of O. What is the empirical formula of the compound?%N = 30.5% N%O = 69.5% O Element N O Relative # Atoms (%/gaw) 30.5/14.0067 = 2.18 69.5/15.9994 = 4.34 % 30.5 69.5 Divide by Smaller 2.18/2.18 = 1.00 4.34/2.18 = 1.99 Dr. S. M. Condren

  22. EXAMPLE: A sample of a brown-colored gas that is a major air pollutant is found to contain 2.34 g of N and 5.34 g of O. What is the empirical formula of the compound?%N = 30.5% N%O = 69.5% O Element N O Relative # Atoms Multiply by Integer 1*1.00=>1 1*1.99=>2 % 30.5 69.5 (%/gaw) 30.5/14.0067 = 2.18 69.5/15.9994 = 4.34 Divide by Smaller 2.18/2.18 = 1.00 4.34/2.18 = 1.99 Empirical Formula => NO2 Dr. S. M. Condren

  23. Molecular Formula • the exact proportions of the elements that are formed in a molecule Dr. S. M. Condren

  24. Molecular Formula from Simplest Formula empirical formula => EF molecular formula => MF MF = X * EF Dr. S. M. Condren

  25. Molecular Formula from Simplest Formula formula mass => FM sum of the atomic weights represented by the formula molar mass = MM = X * FM Dr. S. M. Condren

  26. Molecular Formula from Simplest Formula first, knowing MM and FM X = MM/FM then MF = X * EF Dr. S. M. Condren

  27. EXAMPLE: A colorless liquid used in rocket engines, whose empirical formula is NO2, has a molar mass of 92.0. What is the molecular formula? FM = 1(gaw)N + 2(gaw)O = 46.0 MM 92.0 X = ------- = -------- = 2 FM 46.0 thus MF = 2 * EF Dr. S. M. Condren

  28. What is the correct molecular formula for this colorless liquid rocket fuel? 2NO2 NO2 N2O4 Dr. S. M. Condren

  29. Stoichiometry stoi·chi·om·e·trynoun 1.Calculation of the quantities of reactants and products in a chemical reaction. 2.The quantitative relationship between reactants and products in a chemical reaction. Dr. S. M. Condren

  30. The Mole and Chemical Reactions:The Macro-Nano Connection 2 H2 + O2 -----> 2 H2O 2 H2 molecules 1 O2 molecule 2 H2O molecules 2 H2 moles molecules 1 O2 mole molecules 2 H2O moles molecules 4 g H2 32 g O2 36 g H2O Dr. S. M. Condren

  31. EXAMPLE How much H2O, in moles results from burning an excess of H2 in 3.3 moles of O2? H2 + O2 -----> H2O 2 H2 + O2 -----> 2 H2O Dr. S. M. Condren

  32. EXAMPLE How much H2O, in moles results from burning an excess of H2 in 3.3 moles of O2? H2 + O2 -----> H2O 2 H2 + O2 -----> 2 H2O (3.3 mol O2) (2 mol H2O) #mol H2O = ------------------------------------ (1 mol O2) Dr. S. M. Condren

  33. EXAMPLE How much H2O, in moles results from burning an excess of H2 in 3.3 moles of O2? H2 + O2 -----> H2O 2 H2 + O2 -----> 2 H2O (3.3 mol O2) (2 mol H2O) #mol H2O = ------------------------------------ (1 mol O2) Dr. S. M. Condren

  34. EXAMPLE How much H2O, in moles results from burning an excess of H2 in 3.3 moles of O2? H2 + O2 -----> H2O 2 H2 + O2 -----> 2 H2O (3.3) (2 mol H2O) #mol H2O = ------------------------ = 6.6 mol H2O (1) Dr. S. M. Condren

  35. Combination Reaction PbNO3(aq) + K2CrO4(aq)  PbCrO4(s) + 2 KNO3(aq) Colorless yellow yellow colorless Dr. S. M. Condren

  36. Stoichiometric Roadmap Dr. S. M. Condren

  37. EXAMPLEHow much iron(III) oxide and aluminum powder are required to field weld the ends of two rails together? Assume that the rail is 132 lb/yard and that one inch of the rails will be covered by an additional 10% mass of iron. http://www.cbu.edu/~mcondren/Demos/Thermite-Welding.ppt Dr. S. M. Condren

  38. EXAMPLEHow much iron(III) oxide and aluminum powder are required to field weld the ends of two rails together? Assume that the rail is 132 lb/yard and that one inch of the rails will be covered by an additional 10% mass of iron. The mass of iron in 1 inch of this rail is: #g/in = (132 #/yard) (1 yard/36 in) (454 g/#) = 1.67 X 103 g/in Dr. S. M. Condren

  39. EXAMPLEHow much iron(III) oxide and aluminum powder are required to field weld the ends of two rails together? Assume that the rail is 132 lb/yard and that one inch of the rails will be covered by an additional 10% mass of iron. The mass of iron in 1 inch of this rail is: #g/in = (132) (1/36 in) (454 g) = 1.67 X 103 g/in The mass of iron in a weld adding 10% mass: #g = (1.67 X 103 g) (0.10) = 167 g Dr. S. M. Condren

  40. EXAMPLEHow much iron(III) oxide and aluminum powder are required to field weld the ends of two rails together? Assume that the rail is 132 lb/yard and that one inch of the rails will be covered by an additonal 10% mass of iron. The mass of iron in 1 inch of this rail is: #g/in = (132 #/yard) (1 yard/36 in) (454 g/#) = 1.67 X 103 g/in The mass of iron in a weld adding 10% mass: #g = (1.67 X 103 g) (0.10) = 167 g Balanced chemical equation: Fe2O3 + 2 Al ---> 2 Fe + Al2O3 Dr. S. M. Condren

  41. EXAMPLEHow much iron(III) oxide and aluminum powder are required to field weld the ends of two rails together? Assume that the rail is 132 lb/yard and that one inch of the rails will be covered by an additonal 10% mass of iron. The mass of iron in 1 inch of this rail is: #g/in = (132 #/yard) (1 yard/36 in) (454 g/#) = 1.67 X 103 g/in The mass of iron in a weld adding 10% mass: #g = (1.67 X 103 g) (0.10) = 167 g Balanced chemical equation: Fe2O3 + 2 Al ---> 2 Fe + Al2O3 What mass of Fe2O3 is required for the thermite process? Dr. S. M. Condren

  42. EXAMPLEHow much iron(III) oxide and aluminum powder are required to field weld the ends of two rails together? Assume that the rail is 132 lb/yard and that one inch of the rails will be covered by an additonal 10% mass of iron. The mass of iron in a weld adding 10% mass: #g = (1.67 X 103 g) (0.10) = 167 g Balanced chemical equation: Fe2O3 + 2 Al ---> 2 Fe + Al2O3 What mass of Fe2O3 is required for the thermite process? #g Fe2O3 = (159.7 g Fe2O3) ------------------- (1 mol Fe2O3) (1 mol Fe2O3) ----------------- (2 mol Fe) (1 mol Fe) * -------------- (55.85 g Fe) (167 g Fe) = 238 g Fe2O3 Dr. S. M. Condren

  43. EXAMPLEHow much iron(III) oxide and aluminum powder are required to field weld the ends of two rails together? Assume that the rail is 132 lb/yard and that one inch of the rails will be covered by an additonal 10% mass of iron. The mass of iron in a weld adding 10% mass: #g Fe = 167 g Fe Balanced chemical equation: Fe2O3 + 2 Al ---> 2 Fe + Al2O3 What mass of Fe2O3 is required for the thermite process? #g Fe2O3 = 238 g Fe2O3 What mass of Al is required for the thermite process? #g Al = (26.9815 g Al) ------------------- (1 mol Al) (1 mol Fe) * ---------------- (55.85 g Fe) (2 mol Al) ------------- (2 mol Fe) (167 g Fe) = 80.6 g Al Dr. S. M. Condren

  44. EXAMPLEHow much iron(III) oxide and aluminum powder are required to field weld the ends of two rails together? Assume that the rail is 132 lb/yard and that one inch of the rails will be covered by an additonal 10% mass of iron. The mass of iron in a weld adding 10% mass: #g Fe = 167 g Fe #g Fe2O3 = 238 g Fe2O3 #g Al = 80.6 g Al Dr. S. M. Condren

  45. Limiting Reactant reactant that limits the amount of product that can be produced Dr. S. M. Condren

  46. EXAMPLE What is the number of moles of Fe(OH)3 (S) that can be produced by allowing 1.0 mol Fe2S3, 2.0 mol H2O, and 3.0 mol O2 to react? 2Fe2S3(S) + 6H2O(l) + 3O2(g) -----> 4Fe(OH)3(S) + 6S(S) Dr. S. M. Condren

  47. EXAMPLE What is the number of moles of Fe(OH)3 (S) that can be produced by allowing 1.0 mol Fe2S3, 2.0 mol H2O, and 3.0 mol O2 to react? 2Fe2S3(S) + 6H2O(l) + 3O2(g) -----> 4Fe(OH)3(S) + 6S(S) balanced equation relates: 2Fe2S3(S) <=> 6H2O(l) <=> 3O2(g) Dr. S. M. Condren

  48. EXAMPLE What is the number of moles of Fe(OH)3 (S) that can be produced by allowing 1.0 mol Fe2S3, 2.0 mol H2O, and 3.0 mol O2 to react? 2Fe2S3(S) + 6H2O(l) + 3O2(g) -----> 4Fe(OH)3(S) + 6S(S) balanced equation relates: 2Fe2S3(S) <=> 6H2O(l) <=> 3O2(g) have only: 1Fe2S3(S) <=> 2H2O(l) <=> 3O2(g) Dr. S. M. Condren

  49. EXAMPLE What is the number of moles of Fe(OH)3 (S) that can be produced by allowing 1.0 mol Fe2S3, 2.0 mol H2O, and 3.0 mol O2 to react? 2Fe2S3(S) + 6H2O(l) + 3O2(g) -----> 4Fe(OH)3(S) + 6S(S) balanced equation relates: 2Fe2S3(S) <=> 6H2O(l) <=> 3O2(g) have only: 1Fe2S3(S) <=> 2H2O(l) <=> 3O2(g) not enough H2O to use all Fe2S3 plenty of O2 Dr. S. M. Condren

  50. EXAMPLE What is the number of moles of Fe(OH)3 (S) that can be produced by allowing 1.0 mol Fe2S3, 2.0 mol H2O, and 3.0 mol O2 to react? 2Fe2S3(S) + 6H2O(l) + 3O2(g) -----> 4Fe(OH)3(S) + 6S(S) if use all Fe2S3: (1.0 mol Fe2S3) (4 mol Fe(OH)3) #mol Fe(OH)3 = ------------------------------------------ (2 mol Fe2S3) = 2.0 mol Fe(OH)3 Dr. S. M. Condren

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