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Chapter 3

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  1. Chapter 3 Calculations with Chemical Formulas and Equations Mr. Watson HST

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

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

  4. Example What is the molar mass of ethanol, C2H5OH?

  5. Example What is the molar mass of ethanol, C2H5O1H1? MM = 2(gaw)C + (5 + 1)(gaw)H + 1(gaw)O

  6. Example What is the molar mass of ethanol, C2H5OH? MM = 2(gaw)C + (5 + 1)(gaw)H + 1(gaw)O = 2(12.011)C + 6(1.00794)H + 1(15.9994)O

  7. Example What is the molar mass of ethanol, C2H5OH? MM = 2(gaw)C + (5 + 1)(gaw)H + 1(gaw)O = 2(12.011)C + 6(1.00794)H + 1(15.9994)O = 46.069 g/mol

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

  9. Example How many moles of carbon dioxide molecules are there in 10.00g of carbon dioxide?

  10. Example How many moles of carbon dioxide molecules are there in 10.00g of carbon dioxide? MM = 1(gaw)C + 2(gaw)O = 44.01 g/mol

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

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

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

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

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

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

  17. Combustion Analysis

  18. Percentage Composition description of a compound based on the relative amounts of each element in the compound

  19. 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

  20. 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

  21. 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

  22. 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

  23. 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

  24. 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

  25. Simplest (Empirical) Formula • formula describing a substance based on the smallest set of subscripts

  26. Acetylene, C2H2, and benzene, C6H6, have the same empirical formula. Is the correct empirical formula: C2H2 CH C6H6

  27. 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 Element % (%/gaw) P 43.7 43.7/30.97 = 1.41 O 56.356.3/15.9994 = 3.52

  28. 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 Element % (%/gaw) Divide by Smaller P 43.7 43.7/30.97 = 1.41 1.41/1.41 = 1.00 O 56.356.3/15.9994 = 3.523.52/1.41 = 2.50

  29. 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 Smallerby Integer P 43.7 43.7/30.97 = 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

  30. 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

  31. 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

  32. 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 Relative # Atoms Element % (%/gaw) N 30.5 30.5/14.0067 = 2.18 O 69.569.5/15.9994 = 4.34

  33. 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 Relative # Atoms % (%/gaw) Divide by Smaller N 30.5 30.5/14.0067 = 2.18 2.18/2.18 = 1.00 O 69.569.5/15.9994 = 4.34 4.34/2.18 = 1.99

  34. 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 Relative # Atoms Multiply % (%/gaw) Divide by Smaller by Integer N 30.5 30.5/14.0067 = 2.18 2.18/2.18 = 1.00 1*1.00=>1 O 69.569.5/15.9994 = 4.34 4.34/2.18 = 1.99 1*1.99=>2

  35. 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 Relative # Atoms Multiply % (%/gaw) Divide by Smaller by Integer N 30.5 30.5/14.0067 = 2.18 2.18/2.18 = 1.00 1*1.00=>1 O 69.569.5/15.9994 = 4.34 4.34/2.18 = 1.99 1*1.99=>2 Empirical Formula => NO2

  36. Molecular Formula • the exact proportions of the elements that are formed in a molecule

  37. Molecular Formula from Simplest Formula empirical formula => EF molecular formula => MF MF = X * EF

  38. Molecular Formula from Simplest Formula formula mass => FM sum of the atomic weights represented by the formula molar mass = MM = X * FM

  39. Molecular Formula from Simplest Formula first, knowing MM and FM X = MM/FM then MF = X * EF

  40. 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

  41. 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

  42. What is the correct molecular formula for this colorless liquid rocket fuel? 2NO NO N2O4

  43. 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.

  44. 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

  45. 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

  46. 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)

  47. 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)

  48. 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)

  49. Combination Reaction PbNO3(aq) + K2CrO4(aq)  PbCrO4(s) + 2 KNO3(aq) Colorless yellow yellow colorless

  50. Stoichiometric Roadmap