1. © 2012 Pearson Education, Inc. Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations John D. Bookstaver St. Charles Community College Cottleville, MO Lecture Presentation

2. Stoichiometry © 2012 Pearson Education, Inc. Law of Conservation of Mass “We may lay it down as an incontestable axiom that, in all the operations of art and nature, nothing is created; an equal amount of matter exists both before and after the experiment. Upon this principle, the whole art of performing chemical experiments depends.” --Antoine Lavoisier, 1789

3. Stoichiometry © 2012 Pearson Education, Inc. Chemical Equations Chemical equations are concise representations of chemical reactions.

4. Chemical Equations CH 4 and O 2 are the reactants, and CO 2 and H 2 O are the products the (g) after the formulas tells us the state of the chemical the number in front of each substance tells us the numbers of those molecules in the reaction called the ___________ CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(g) 4 coefficients

5. Stoichiometry © 2012 Pearson Education, Inc. Anatomy of a Chemical Equation CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g)

6. Stoichiometry © 2012 Pearson Education, Inc. Anatomy of a Chemical Equation CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) Reactants appear on the left side of the equation.

7. Stoichiometry © 2012 Pearson Education, Inc. Anatomy of a Chemical Equation CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) Products appear on the right side of the equation.

8. Stoichiometry © 2012 Pearson Education, Inc. Anatomy of a Chemical Equation CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) The states of the reactants and products are written in parentheses to the right of each compound.

9. Stoichiometry © 2012 Pearson Education, Inc. Anatomy of a Chemical Equation CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) Coefficients are inserted to balance the equation.

10. Stoichiometry © 2012 Pearson Education, Inc. Subscripts and Coefficients Give Different Information Subscripts tell the number of atoms of each element in a molecule. Coefficients tell the number of molecules.

11. Stoichiometry © 2012 Pearson Education, Inc. Reaction Types

12. Stoichiometry © 2012 Pearson Education, Inc. Combination Reactions Examples: –2Mg(s) + O 2 (g)  2MgO(s) –N 2 (g) + 3H 2 (g)  2NH 3 (g) –C 3 H 6 (g) + Br 2 (l)  C 3 H 6 Br 2 (l) In combination reactions two or more substances react to form one product.

13. Stoichiometry © 2012 Pearson Education, Inc. In a decomposition reaction one substance breaks down into two or more substances. Decomposition Reactions Examples: –CaCO 3 (s)  CaO(s) + CO 2 (g) –2KClO 3 (s)  2KCl(s) + O 2 (g) –2NaN 3 (s)  2Na(s) + 3N 2 (g)

14. Stoichiometry © 2012 Pearson Education, Inc. Combustion Reactions Examples: –CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g) –C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O(g) Combustion reactions are generally rapid reactions that produce a flame. Combustion reactions most often involve hydrocarbons reacting with oxygen in the air.

15. Stoichiometry © 2012 Pearson Education, Inc. Formula Weights

16. Stoichiometry © 2012 Pearson Education, Inc. Formula Weight (FW) A formula weight is the sum of the atomic weights for the atoms in a chemical formula. So, the formula weight of calcium chloride, CaCl 2, would be Ca: 1(40.08 amu) + Cl: 2(35.453 amu) 110.99 amu Formula weights are generally reported for ionic compounds.

17. Stoichiometry © 2012 Pearson Education, Inc. Molecular Weight (MW) A molecular weight is the sum of the atomic weights of the atoms in a molecule. For the molecule ethane, C 2 H 6, the molecular weight would be C: 2(12.011 amu) 30.070 amu + H: 6(1.00794 amu)

18. Stoichiometry © 2012 Pearson Education, Inc. Percent Composition One can find the percentage of the mass of a compound that comes from each of the elements in the compound by using this equation: % Element = (number of atoms)(atomic weight) (FW of the compound) x 100

19. Percent Composition Percentage of each element in a compound Can be determined from 1.the formula of the compound 2.the experimental mass analysis of the compound

20. Stoichiometry © 2012 Pearson Education, Inc. Percent Composition So the percentage of carbon in ethane is %C = (2)(12.011 amu) (30.070 amu) 24.022 amu 30.070 amu = x 100 = 79.887%

21. Stoichiometry © 2012 Pearson Education, Inc. Moles

22. The Mole mole (mol) = number of particles equal to the number of atoms in 12 g of C-12 one mole of anything is 6.022 x 10 23 units of that thing where 6.022 x 10 23 is known as _________________ (often denoted ___) 1 mol of marbles = 6.022 x 10 23 marbles 1 mol of He = __________ He _____ 1 mol of CO 2 = __________ CO 2 ________ Avogadro’s Number NANA 6.022 x 10 23 atoms molecules

23. Stoichiometry © 2012 Pearson Education, Inc. Avogadro’s Number 6.02 x 10 23 1 mole of 12 C has a mass of 12.000 g.

24. Stoichiometry © 2012 Pearson Education, Inc. Molar Mass By definition, a molar mass is the mass of 1 mol of a substance (i.e., g/mol). –The molar mass of an element is the mass number for the element that we find on the periodic table. –The formula weight (in amu’s) will be the same number as the molar mass (in g/mol).

25. Relationship Between Moles and Mass The mass of one mole of atoms is called the __________ The molar mass of an element, in grams, is numerically equal to the element’s atomic mass, in amu atomic mass of Cu = __________ molar mass of Cu = ___________ molar mass 63.55 amu 63.55 g/mol

26. Stoichiometry © 2012 Pearson Education, Inc. Using Moles Moles provide a bridge from the molecular scale to the real-world scale.

27. Stoichiometry © 2012 Pearson Education, Inc. Mole Relationships One mole of atoms, ions, or molecules contains Avogadro’s number of those particles. One mole of molecules or formula units contains Avogadro’s number times the number of atoms or ions of each element in the compound.

28. Mole and Mass Relationships 1 mole Sulfur 32.06 g 1 mole Carbon 12.01 g

29. Example: How many aluminum atoms are in an aluminum can with a mass of 16.2 g? Converting Between Grams and Number of Atoms = 3.6159 x 10 23 atoms Al Sig. Figs. & Round: = 3.62 x 10 23 atoms Al

30. Example: Find the mass of 4.78 x 10 24 NO 2 molecules? Converting Between Grams and Number of Molecules = 365.21 g NO 2 Sig. Figs. & Round: = 365 g NO 2

31. 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

32. 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. Converting Between Grams of a Compound and Grams of a Constituent Element = 44.2979 g C Sig. Figs. & Round: = 44.3 g C

33. 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

34. Example - 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%

35. Empirical Formulas The simplest, whole-number ratio of atoms in a molecule is called the ___________________ can be determined from percent composition or combining masses The Molecular Formula is a multiple of the Empirical Formula Glucose Molecular Formula = C 6 H 12 O 6 Empirical Formula = CH 2 O Empirical Formula

36. Stoichiometry © 2012 Pearson Education, Inc. Finding Empirical Formulas

37. Stoichiometry © 2012 Pearson Education, Inc. Calculating Empirical Formulas One can calculate the empirical formula from the percent composition.

38. 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 a small whole number to make all whole number subscripts fractional subscriptmultiply by.10 10 N 1 O 2.5 x 2  N 2 O 5.20 5.25 4.33 3.50 2.66 3.75 4

39. Stoichiometry © 2012 Pearson Education, Inc. Calculating Empirical Formulas The compound para-aminobenzoic acid (you may have seen it listed as PABA on your bottle of sunscreen) is composed of carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and oxygen (23.33%). Find the empirical formula of PABA.

40. Stoichiometry © 2012 Pearson Education, Inc. Calculating Empirical Formulas Assuming 100.00 g of para-aminobenzoic acid, C:61.31 g x = 5.105 mol C H: 5.14 g x= 5.09 mol H N:10.21 g x= 0.7288 mol N O:23.33 g x = 1.456 mol O 1 mol 12.01 g 1 mol 14.01 g 1 mol 1.01 g 1 mol 16.00 g

41. Stoichiometry © 2012 Pearson Education, Inc. Calculating Empirical Formulas Calculate the mole ratio by dividing by the smallest number of moles: C:= 7.005  7 H:= 6.984  7 N:= 1.000 O:= 2.001  2 5.105 mol 0.7288 mol 5.09 mol 0.7288 mol 1.458 mol 0.7288 mol

42. Stoichiometry © 2012 Pearson Education, Inc. Calculating Empirical Formulas These are the subscripts for the empirical formula: C 7 H 7 NO 2

43. 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% Finding an Empirical Formula from Experimental Data C 4.996 H 4.44 O 2.221 { } x 4 C9H8O4C9H8O4

44. Molecular Formulas The molecular formula is the real formula (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

45. 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 molar mass of the empirical formula 5 C = 60.05 g, 8 H = 8.064 g C 5 H 8 = 68.11 g 2. Divide the given molar mass of the compound by the molar mass of the empirical formula Round to the nearest whole number

46. 3.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

47. 3.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

48. Stoichiometry © 2012 Pearson Education, Inc. Combustion Analysis Compounds containing C, H, and O are routinely analyzed through combustion in a chamber like the one shown in Figure 3.14. –C is determined from the mass of CO 2 produced. –H is determined from the mass of H 2 O produced. –O is determined by difference after the C and H have been determined.

49. Copyright  2011 Pearson Education, Inc. Example Combustion of a 0.8233 g sample of a compound containing only carbon, hydrogen, and oxygen produced the following: CO 2 = 2.445 g H 2 O = 0.6003 g Determine the empirical formula of the compound 49

50. Copyright  2011 Pearson Education, Inc. Example Find the empirical formula of compound with the given amounts of combustion products Write down the given quantity and its units Given:compound = 0.8233 g CO 2 = 2.445 g H 2 O = 0.6003 g 50 Tro: Chemistry: A Molecular Approach, 2/e

51. Copyright  2011 Pearson Education, Inc. Example: Find the empirical formula of compound with the given amounts of combustion products Write down the quantity to find and/or its units Find: empirical formula, C x H y O z Information Given: 0.8233 g compound, 2.445 g CO 2, 0.6003 g H 2 O 51

52. Copyright  2011 Pearson Education, Inc. Write a conceptual plan g CO 2, H 2 O mol ratio empirical formula Example: Find the empirical formula of compound with the given amounts of combustion products mol CO 2, H 2 O mol C, H g C, H gOgO mol O mol C, H, O Information Given: 0.8233 g compound, 2.445 g CO 2, 0.6003 g H 2 O Find: empirical formula, C x H y O z pseudo formula 52

53. Copyright  2011 Pearson Education, Inc. Example: Find the empirical formula of compound with the given amounts of combustion products Collect needed relationships 1 mole CO 2 = 44.01 g CO 2 1 mole H 2 O = 18.02 g H 2 O 1 mole C = 12.01 g C 1 mole H = 1.008 g H 1 mole O = 16.00 g O 1 mole CO 2 = 1 mole C 1 mole H 2 O = 2 mole H Information Given: 0.8233 g compound, 2.445 g CO 2, 0.6003 g H 2 O Find: empirical formula, C x H y O z CP: g CO 2 & H 2 O  mol CO 2 & H 2 O  mol C & H  g C & H  g O  mol O  mol ratio  empirical formula 53

54. Copyright  2011 Pearson Education, Inc. Example: Find the empirical formula of compound with the given amounts of combustion products Apply the conceptual plan calculate the moles of C and H Information Given: 0.8233 g compound, 2.445 g CO 2, 0.6003 g H 2 O Find: empirical formula, C x H y O z CP: g CO 2 & H 2 O  mol CO 2 & H 2 O  mol C & H  g C & H  g O  mol O  mol ratio  empirical formula Rel: MM of CO 2, H 2 O, C, H, O; mol element : 1 mol compound 54

55. Copyright  2011 Pearson Education, Inc. Example: Find the empirical formula of compound with the given amounts of combustion products Apply the conceptual plan calculate the grams of C and H Information Given: 0.8233 g compound, 2.445 g CO 2, 0.6003 g H 2 O Find: empirical formula, C x H y O z CP: g CO 2 & H 2 O  mol CO 2 & H 2 O  mol C & H  g C & H  g O  mol O  mol ratio  empirical formula Rel: MM of CO 2, H 2 O, C, H, O; mol element : 1 mol compound 55

56. Copyright  2011 Pearson Education, Inc. Apply the conceptual plan calculate the grams and moles of O Example: Find the empirical formula of compound with the given amounts of combustion products Information Given: 0.8233 g compound, 2.445 g CO 2, 0.6003 g H 2 O Find: empirical formula, C x H y O z CP: g CO 2 & H 2 O  mol CO 2 & H 2 O  mol C & H  g C & H  g O  mol O  mol ratio  empirical formula Rel: MM of CO 2, H 2 O, C, H, O; mol element : 1 mol compound 56

57. Copyright  2011 Pearson Education, Inc. Apply the conceptual plan write a pseudoformula C 0.05556 H 0.06662 O 0.00556 Example: Find the empirical formula of compound with the given amounts of combustion products Information Given: 0.8233 g compound, 2.445 g CO 2, 0.6003 g H 2 O 0.05556 mol C, 0.6673 g C, 0.06662 mol H, 0.06715 g H, 0.0889 g O, 0.00556 mol O Find: empirical formula, C x H y O z CP: g CO 2 & H 2 O  mol CO 2 & H 2 O  mol C & H  g C & H  g O  mol O  mol ratio  empirical formula Rel: MM of CO 2, H 2 O, C, H, O; mol element : 1 mol compound 57

58. Copyright  2011 Pearson Education, Inc. Apply the conceptual plan find the mole ratio by dividing by the smallest number of moles Example: Find the empirical formula of compound with the given amounts of combustion products Information Given: 0.8233 g compound, 2.445 g CO 2, 0.6003 g H 2 O 0.05556 mol C, 0.6673 g C, 0.06662 mol H, 0.06715 g H, 0.0889 g O, 0.00556 mol O Find: empirical formula, C x H y O z CP: g CO 2 & H 2 O  mol CO 2 & H 2 O  mol C & H  g C & H  g O  mol O  mol ratio  empirical formula Rel: MM of CO 2, H 2 O, C, H, O; mol element : 1 mol compound 58

59. Copyright  2011 Pearson Education, Inc. Apply the conceptual plan multiply subscripts by factor to give whole number, if necessary write the empirical formula 11210 OHC Example: Find the empirical formula of compound with the given amounts of combustion products Information Given: 0.8233 g compound, 2.445 g CO 2, 0.6003 g H 2 O 0.05556 mol C, 0.6673 g C, 0.06662 mol H, 0.06715 g H, 0.0889 g O, 0.00556 mol O Find: empirical formula, C x H y O z CP: g CO 2 & H 2 O  mol CO 2 & H 2 O  mol C & H  g C & H  g O  mol O  mol ratio  empirical formula Rel: MM of CO 2, H 2 O, C, H, O; mol element : 1 mol compound 59

60. Copyright © 2011 Pearson Education, Inc. Quantities in Chemical Reactions The amount of every substance used and made in a chemical reaction is related to the amounts of all the other substances in the reaction Law of Conservation of Mass Balancing equations by balancing atoms The study of the numerical relationship between chemical quantities in a chemical reaction is called stoichiometry 60

61. Copyright © 2011 Pearson Education, Inc. Reaction Stoichiometry The coefficients in a balanced chemical equation specify the relative amounts in moles of each of the substances involved in the reaction 2 C 8 H 18 (l) + 25 O 2 (g)  16 CO 2 (g) + 18 H 2 O(g) 2 molecules of C 8 H 18 react with 25 molecules of O 2 to form 16 molecules of CO 2 and 18 molecules of H 2 O 2 moles of C 8 H 18 react with 25 moles of O 2 to form 16 moles of CO 2 and 18 moles of H 2 O 2 mol C 8 H 18 : 25 mol O 2 : 16 mol CO 2 : 18 mol H 2 O 61

62. Stoichiometry © 2012 Pearson Education, Inc. Stoichiometric Calculations The coefficients in the balanced equation give the ratio of moles of reactants and products.

63. Stoichiometry © 2012 Pearson Education, Inc. Stoichiometric Calculations Starting with the mass of Substance A, you can use the ratio of the coefficients of A and B to calculate the mass of Substance B formed (if it’s a product) or used (if it’s a reactant).

64. Stoichiometry © 2012 Pearson Education, Inc. Stoichiometric Calculations Starting with 1.00 g of C 6 H 12 O 6 … we calculate the moles of C 6 H 12 O 6 … use the coefficients to find the moles of H 2 O… and then turn the moles of water to grams. C 6 H 12 O 6 + 6 O 2  6 CO 2 + 6 H 2 O

65. Copyright © 2011 Pearson Education, Inc. 65 Making Molecules Mass-to-Mass Conversions Mass-to-Mass Conversions we know there is a relationship between the mass and number of moles of a chemical molar mass = g / mol the molar mass of the chemicals in the reaction and the balanced chemical equation allow us to convert from the amount of any chemical in the reaction to the amount of any other

66. Copyright © 2011 Pearson Education, Inc. Practice According to the following equation, how many moles of water are made in the combustion of 0.10 moles of glucose? C 6 H 12 O 6 + 6 O 2  6 CO 2 + 6 H 2 O 66

67. Copyright © 2011 Pearson Education, Inc. Practice  How many moles of water are made in the combustion of 0.10 moles of glucose? 0.6 mol H 2 O = 0.60 mol H 2 O because 6x moles of H 2 O as C 6 H 12 O 6, the number makes sense C 6 H 12 O 6 + 6 O 2 → 6 CO 2 + 6 H 2 O  1 mol C 6 H 12 O 6 : 6 mol H 2 O 0.10 moles C 6 H 12 O 6 moles H 2 O Check: Solution: Conceptual Plan: Relationships: Given: Find: mol H 2 Omol C 6 H 12 O 6 67

68. Copyright © 2011 Pearson Education, Inc. Example: Estimate the mass of CO 2 produced in 2007 by the combustion of 3.5 x 10 15 g gasolne Assuming that gasoline is octane, C 8 H 18, the equation for the reaction is 2 C 8 H 18 (l) + 25 O 2 (g)  16 CO 2 (g) + 18 H 2 O(g) The equation for the reaction gives the mole relationship between amount of C 8 H 18 and CO 2, but we need to know the mass relationship, so the conceptual plan will be g C 8 H 18 mol CO 2 g CO 2 mol C 8 H 18 68

69. Copyright © 2011 Pearson Education, Inc. Example: Estimate the mass of CO 2 produced in 2007 by the combustion of 3.5 x 10 15 g gasoline because 8x moles of CO 2 as C 8 H 18, but the molar mass of C 8 H 18 is 3x CO 2, the number makes sense 1 mol C 8 H 18 = 114.22g, 1 mol CO 2 = 44.01g, 2 mol C 8 H 18 :16 mol CO 2 3.4 x 10 15 g C 8 H 18 g CO 2 Check: Solution: Conceptual Plan: Relationships: Given: Find: g C 8 H 18 mol CO 2 g CO 2 mol C 8 H 18 69

70. Copyright © 2011 Pearson Education, Inc. Example : How many grams of glucose can be synthesized from 37.8 g of CO 2 in photosynthesis? because 6x moles of CO 2 as C 6 H 12 O 6, but the molar mass of C 6 H 12 O 6 is 4x CO 2, the number makes sense 1 mol C 6 H 12 O 6 = 180.2g, 1 mol CO 2 = 44.01g, 1 mol C 6 H 12 O 6 : 6 mol CO 2 37.8 g CO 2, 6 CO 2 + 6 H 2 O  C 6 H 12 O 6 + 6 O 2 g C 6 H 12 O 6 Check: Solution: Conceptual Plan: Relationships: Given: Find: g 44.01 mol 1 6126 6 6 6 6 2 6 6 2 2 2 OHC g 5.82 OHC mol 1 OHC g 180.2 CO mol 6 OHC 1 CO g 44.01 CO mol 1 CO g 7.83   2 6126 CO mol 6 OHC 1 g C 6 H 12 O 6 mol CO 2 g CO 2 mol C 6 H 12 O 6 mol 1 g 180.2 70

71. Copyright © 2011 Pearson Education, Inc. Practice — How many grams of O 2 can be made from the decomposition of 100.0 g of PbO 2 ? 2 PbO 2 (s) → 2 PbO(s) + O 2 (g) (PbO 2 = 239.2, O 2 = 32.00) 71

72. Copyright © 2011 Pearson Education, Inc. Practice — How many grams of O 2 can be made from the decomposition of 100.0 g of PbO 2 ? 2 PbO 2 (s) → 2 PbO(s) + O 2 (g) because ½ moles of O 2 as PbO 2, and the molar mass of PbO 2 is 7x O 2, the number makes sense 1 mol O 2 = 32.00g, 1 mol PbO 2 = 239.2g, 1 mol O 2 : 2 mol PbO 2 100.0 g PbO 2, 2 PbO 2 → 2 PbO + O 2 g O 2 Check: Solution: Conceptual Plan: Relationships: Given: Find: g O 2 mol PbO 2 g PbO 2 mol O 2 72

73. Stoichiometry © 2012 Pearson Education, Inc. Limiting Reactants

74. Stoichiometry © 2012 Pearson Education, Inc. Limiting Reactants The limiting reactant is the reactant present in the smallest stoichiometric amount. –In other words, it’s the reactant you’ll run out of first (in this case, the H 2 ).

75. Stoichiometry © 2012 Pearson Education, Inc. Limiting Reactants In the example below, the O 2 would be the excess reagent.

76. Copyright © 2011 Pearson Education, Inc. The Limiting Reactant For reactions with multiple reactants, it is likely that one of the reactants will be completely used before the others When this reactant is used up, the reaction stops and no more product is made The reactant that limits the amount of product is called the limiting reactant sometimes called the limiting reagent the limiting reactant gets completely consumed Reactants not completely consumed are called excess reactants The amount of product that can be made from the limiting reactant is called the theoretical yield 76

77. Copyright © 2011 Pearson Education, Inc. Limiting and Excess Reactants in the Combustion of Methane CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(g) Our balanced equation for the combustion of methane implies that every one molecule of CH 4 reacts with two molecules of O 2 77

78. Copyright © 2011 Pearson Education, Inc. Limiting and Excess Reactants in the Combustion of Methane CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(g) since less CO 2 can be made from the O 2 than the CH 4, so the O 2 is the limiting reactant 78 If we have five molecules of CH 4 and eight molecules of O 2, which is the limiting reactant?

79. Stoichiometry © 2012 Pearson Education, Inc. Theoretical Yield The theoretical yield is the maximum amount of product that can be made. –In other words, it’s the amount of product possible as calculated through the stoichiometry problem. This is different from the actual yield, which is the amount one actually produces and measures.

80. Copyright © 2011 Pearson Education, Inc. Theoretical and Actual Yield As we did with the pizzas, in order to determine the theoretical yield, we should use reaction stoichiometry to determine the amount of product each of our reactants could make The theoretical yield will always be the least possible amount of product the theoretical yield will always come from the limiting reactant Because of both controllable and uncontrollable factors, the actual yield of product will always be less than the theoretical yield 80

81. Stoichiometry © 2012 Pearson Education, Inc. Percent Yield One finds the percent yield by comparing the amount actually obtained (actual yield) to the amount it was possible to make (theoretical yield): Percent yield =x 100 actual yield theoretical yield

82. Copyright © 2011 Pearson Education, Inc. Practice — How many moles of Si 3 N 4 can be made from 1.20 moles of Si and 1.00 moles of N 2 in the reaction 3 Si + 2 N 2  Si 3 N 4 ? 82

83. Copyright © 2011 Pearson Education, Inc. Limiting reactant Practice — How many moles of Si 3 N 4 can be made from 1.20 moles of Si and 1.00 moles of N 2 in the reaction 3 Si + 2 N 2  Si 3 N 4 ? 2 mol N 2 : 1 Si 3 N 4 ; 3 mol Si : 1 Si 3 N 4 1.20 mol Si, 1.00 mol N 2 mol Si 3 N 4 Solution: Conceptual Plan: Relationships: Given: Find: mol N 2 mol Si 3 N 4 mol Simol Si 3 N 4 Pick least amount Limiting reactant and theoretical yield Theoretical yield 83

84. Copyright © 2011 Pearson Education, Inc. Example Finding limiting reactant, theoretical yield, and percent yield

85. Copyright © 2011 Pearson Education, Inc. Example: When 28.6 kg of C are allowed to react with 88.2 kg of TiO 2 in the reaction below, 42.8 kg of Ti are obtained. Find the limiting reactant, theoretical yield, and percent yield. 85

86. Copyright © 2011 Pearson Education, Inc. Example: When 28.6 kg of C reacts with 88.2 kg of TiO 2, 42.8 kg of Ti are obtained. Find the limiting reactant, theoretical yield, and percent yield TiO 2 (s) + 2 C(s)  Ti(s) + 2 CO(g) Write down the given quantity and its units Given:28.6 kg C 88.2 kg TiO 2 42.8 kg Ti produced 86

87. Copyright © 2011 Pearson Education, Inc. Example: Find the limiting reactant, theoretical yield, and percent yield TiO 2 (s) + 2 C(s)  Ti(s) + 2 CO(g) Write down the quantity to find and/or its units Find: limiting reactant theoretical yield percent yield Information Given: 28.6 kg C, 88.2 kg TiO 2, 42.8 kg Ti 87

88. Copyright © 2011 Pearson Education, Inc. Example: Find the limiting reactant, theoretical yield, and percent yield TiO 2 (s) + 2 C(s)  Ti(s) + 2 CO(g) Write a conceptual plan Information Given:28.6 kg C, 88.2 kg TiO 2, 42.8 kg Ti Find: lim. rct., theor. yld., % yld. kg TiO 2 kg C } smallest amount is from limiting reactant smallest mol Ti 88

89. Copyright © 2011 Pearson Education, Inc. Example: Find the limiting reactant, theoretical yield, and percent yield TiO 2 (s) + 2 C(s)  Ti(s) + 2 CO(g) Collect needed relationships 1000 g = 1 kg Molar Mass TiO 2 = 79.87 g/mol Molar Mass Ti = 47.87 g/mol Molar Mass C = 12.01 g/mol 1 mole TiO 2 : 1 mol Ti (from the chem. equation) 2 mole C : 1 mol Ti (from the chem. equation) Information Given:28.6 kg C, 88.2 kg TiO 2, 42.8 kg Ti Find: lim. rct., theor. yld., % yld. CP: kg rct  g rct  mol rct  mol Ti pick smallest mol Ti  TY kg Ti  %Y Ti 89

90. Copyright © 2011 Pearson Education, Inc. Apply the conceptual plan Information Given:28.6 kg C, 88.2 kg TiO 2, 42.8 kg Ti Find: lim. rct., theor. yld., % yld. CP: kg rct  g rct  mol rct  mol Ti pick smallest mol Ti  TY kg Ti  %Y Ti Rel: 1 mol C=12.01g; 1 mol Ti =47.87g; 1 mol TiO 2 = 79.87g; 1000g = 1 kg; 1 mol TiO 2 : 1 mol Ti; 2 mol C : 1 mol Ti Example: Find the limiting reactant, theoretical yield, and percent yield TiO 2 (s) + 2 C(s)  Ti(s) + 2 CO(g) smallest moles of Ti limiting reactant 90

91. Copyright © 2011 Pearson Education, Inc. Apply the conceptual plan theoretical yield Information Given:28.6 kg C, 88.2 kg TiO 2, 42.8 kg Ti Find: lim. rct., theor. yld., % yld. CP: kg rct  g rct  mol rct  mol Ti pick smallest mol Ti  TY kg Ti  %Y Ti Rel: 1 mol C=12.01g; 1 mol Ti =47.87g; 1 mol TiO 2 = 79.87g; 1000g = 1 kg; 1 mol TiO 2 : 1 mol Ti; 2 mol C : 1 mol Ti Example: Find the limiting reactant, theoretical yield, and percent yield TiO 2 (s) + 2 C(s)  Ti(s) + 2 CO(g) 91

92. Copyright © 2011 Pearson Education, Inc. Apply the conceptual plan Information Given:28.6 kg C, 88.2 kg TiO 2, 42.8 kg Ti Find: lim. rct., theor. yld., % yld. CP: kg rct  g rct  mol rct  mol Ti pick smallest mol Ti  TY kg Ti  %Y Ti Rel: 1 mol C=12.01g; 1 mol Ti =47.87g; 1 mol TiO 2 = 79.87g; 1000g = 1 kg; 1 mol TiO 2 : 1 mol Ti; 2 mol C : 1 mol Ti Example: Find the limiting reactant, theoretical yield, and percent yield TiO 2 (s) + 2 C(s)  Ti(s) + 2 CO(g) 92

93. Copyright © 2011 Pearson Education, Inc. Check the solutions limiting reactant = TiO 2 theoretical yield = 52.9 kg percent yield = 80.9% Because Ti has lower molar mass than TiO 2, the T.Y. makes sense and the percent yield makes sense as it is less than 100% Information Given:28.6 kg C, 88.2 kg TiO 2, 42.8 kg Ti Find: lim. rct., theor. yld., % yld. CP: kg rct  g rct  mol rct  mol Ti pick smallest mol Ti  TY kg Ti  %Y Ti Rel: 1 mol C=12.01g; 1 mol Ti =47.87g; 1 mol TiO 2 = 79.87g; 1000g = 1 kg; 1 mol TiO 2 : 1 mol Ti; 2 mol C : 1 mol Ti Example: Find the limiting reactant, theoretical yield, and percent yield TiO 2 (s) + 2 C(s)  Ti(s) + 2 CO(g) 93

94. Copyright © 2011 Pearson Education, Inc. Practice — How many grams of N 2 (g) can be made from 9.05 g of NH 3 reacting with 45.2 g of CuO? 2 NH 3 (g) + 3 CuO(s) → N 2 (g) + 3 Cu(s) + 3 H 2 O(l) If 4.61 g of N 2 are made, what is the percent yield? 94

95. Copyright © 2011 Pearson Education, Inc. Practice — How many grams of N 2 (g) can be made from 9.05 g of NH 3 reacting with 45.2 g of CuO? 2 NH 3 (g) + 3 CuO(s) → N 2 (g) + 3 Cu(s) + 3 H 2 O(l) If 4.61 g of N 2 are made, what is the percent yield? 1 mol NH 3 = 17.03g, 1 mol CuO = 79.55g, 1 mol N 2 = 28.02 g 2 mol NH 3 : 1 mol N 2, 3 mol CuO : 1 mol N 2 9.05 g NH 3, 45.2 g CuO g N 2 Conceptual Plan: Relationships: Given: Find: g NH 3 mol N 2 mol NH 3 g N 2 g CuOmol N 2 mol CuO Choose smallest 95

96. Copyright © 2011 Pearson Education, Inc. because the percent yield is less than 100, the answer makes sense Check: Solution: Theoretical yield 96 Practice — How many grams of N 2 (g) can be made from 9.05 g of NH 3 reacting with 45.2 g of CuO? 2 NH 3 (g) + 3 CuO(s) → N 2 (g) + 3 Cu(s) + 3 H 2 O(l) If 4.61 g of N 2 are made, what is the percent yield?