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

5. 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.

6. 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.

7. 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.

8. 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.

9. 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.

10. Stoichiometry What is the difference in atom count between the notation CO 2 and the notation 2 CO? A.None as both molecules contain one carbon atom. B.None as both molecules contain two oxygen atoms. C.The first notation shows one carbon atom and two oxygen atoms whereas the second one shows two carbon atoms and two oxygen atoms D.The first notation shows one molecule and the second one shows two molecules.

11. Stoichiometry Sample Exercise 3.2 Balancing Chemical Equations Comment Notice that we moved back and forth, placing a coefficient in front of H 2 O, then NaOH, and finally Na. In balancing equations, we often find ourselves following this pattern of moving back and forth from one side of the arrow to the other, placing coefficients first in front of a formula on one side and Continued then in front of a formula on the other side until the equation is balanced. You can always tell if you have balanced your equation correctly by checking that the number of atoms of each element is the same on the two sides of the arrow. Balance these equations by providing the missing coefficients: Answer: (a) 4, 3, 2; (b) 1, 3, 2, 2; (c) 2, 6, 2, 3 Practice

12. Stoichiometry BaCO 3 (s) BaO(s) + CO 2 (g) 2 Li(s) + F 2 (g) 2 LiF(s) Sample Exercise 3.3 Writing Balanced Equations for Combination and Decomposition Reactions (a) With the exception of mercury, all metals are solids at room temperature. Fluorine occurs as a diatomic molecule. Thus, the reactants are Li(s) and F 2 (g). The product will be composed of a metal and a nonmetal, so we expect it to be an ionic solid. Lithium ions have a 1+ charge, Li +, whereas fluoride ions have a 1– charge, F -. Thus, the chemical formula for the product is LiF. The balanced chemical equation is (b) The chemical formula for barium carbonate is BaCO 3. As noted in the text, many metal carbonates decompose to metal oxides and carbon dioxide when heated. In Equation 3.7, for example, CaCO 3 decomposes to form CaO and CO 2. Thus, we expect BaCO 3 to decompose to BaO and CO 2. Barium and calcium are both in group 2A in the periodic table, which further suggests they react in the same way: Solution Write a balanced equation for (a) the combination reaction between lithium metal and fluorine gas and (b) the decomposition reaction that occurs when solid barium carbonate is heated (two products form, a solid and a gas). Write a balanced equation for (a) solid mercury(II) sulfide decomposing into its component elements when heated and (b) aluminum metal combining with oxygen in the air. Answer: (a) HgS(s) Hg(l) + S(s), (b) 4 Al(s) + 3 O 2 (g) 2 Al 2 O 3 (s) Practice

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

14. 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.

15. 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)

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17. © 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.

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

19. 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.

20. 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)

21. 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

22. Stoichiometry 342.0 amu Sample Exercise 3.5 Calculating Formula Weights (a) By adding the atomic weights of the atoms in sucrose, we find the formula weight to be 342.0 amu: (b) If a chemical formula has parentheses, the subscript outside the parentheses is a multiplier for all atoms inside. Thus, for Ca(NO 3 ) 2 we have Solution Calculate the formula weight of (a) sucrose, C 12 H 22 O 11 (table sugar), and (b) calcium nitrate, Ca(NO 3 ) 2. Calculate the formula weight of (a) Al(OH) 3 and (b) CH 3 OH. Answer: (a) 78.0 amu, (b) 32.0 amu Practice 12 C atoms = 12(12.0 amu) = 144.0 amu 22 H atoms = 22(1.0 amu) = 22.0 amu 11 O atoms = 11(16.0 amu) = 176.0 amu 1 Ca atom = 1(40.1 amu) = 40.1 amu 2 N atoms = 2(14.0 amu) = 28.0 amu 6 O atoms = 6(16.0 amu) = 96.0 amu 164.1 amu

23. 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%

24. Stoichiometry Sample Exercise 3.6 Calculating Percentage Composition Calculate the percentage of carbon, hydrogen, and oxygen (by mass) in C 12 H 22 O 11. Let’s examine this question using the problem-solving steps in the accompanying “Strategies in Chemistry: Problem Solving” essay. Analyze We are given a chemical formula and asked to calculate the percentage by mass of each element. Plan We use Equation 3.10, obtaining our atomic weights from a periodic table. We know the denominator in Equation 3.10, the formula weight of C 12 H 22 O 11, from Sample Exercise 3.5. We must use that value in three calculations, one for each element. Solve Check Our calculated percentages must add up to 100%, which they do. We could have used more significant figures for our atomic weights, giving more significant figures for our percentage composition, but we have adhered to our suggested guideline of rounding atomic weights to one digit beyond the decimal point. Solution

25. Stoichiometry © 2012 Pearson Education, Inc. Moles

26. 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.

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29. How many H 2 O molecules are in a 9.00-g sample of water? A. 0.500 B. 3.01  10 23 C. 2.71  10 24 D. 1.08  10 23

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31. © 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).

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

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34. © 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.

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

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

37. 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.

38. 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

39. 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

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

41. 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.

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

43. Stoichiometry

44. © 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).

45. 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

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

47. LIMITING REACTANT is the substance (reactant) that gets“used up” in a chemical reaction. It “limits” the amount of product that can be made. 5 patties + 8 buns ------> ???? hamburgers 10 wheels + 15 frames ------> ???? bicycles 5 5 (1 to 1 ratio: patties get used up, 3 buns left over) (2 to 1 ratio: wheels get used up, 10 frames left over)

48. 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 ).

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

50. How will we know if it is a “Limiting Reactant” problem? 2 Givens and 1 Find Whichever reactant makes the LEAST amount of product IS the “LIMITING REACTANT”. If you have a problem that you gives the mass of 2 or more of the reactants, you know you have a “limiting reactant” problem.

51. Stoichiometry 3 H 2 + N 2  2 NH 3 © 2012 Pearson Education, Inc. If 10.0g of hydrogen reacts with 10.0g of nitrogen, how much ammonia will be produced ? Start with one reactant and determine if there is enough of the other reactant to react with it. 10.0g H 2 1 mol H 2 2.0g H 2 1 mol N 2 3 mol H 2 28.0g N 2 1 mol N 2 = 46.7g N 2 Since 10.0g of H 2 needs 46.7g N 2 to be completely consumed, there is not enough N 2 to use up the H 2, H 2 will be left over, and the N 2 will be the limiting reactant.

52. Stoichiometry © 2012 Pearson Education, Inc. 10.0g N 2 1 mol N 2 28.0g N 2 3 mol H 2 1 mol N 2 2.0g H 2 1 mol H 2 = 2.14g H 2 It only takes 2.14g of H 2 to completely use up the 10.0g of N 2. All of the N 2 will be used up and H 2 will be left over, and the N 2 will be the limiting reactant. If we started with the 10.0g of the nitrogen….

53. Stoichiometry © 2012 Pearson Education, Inc. Since we have determined that the N 2 is the limiting reactant, we will use it to determine how much product will be made. 10.0g N 2 1 mol N 2 28.0g N 2 2 mol NH 3 1 mol N 2 17.0g NH 3 1 mol NH 3 = 12.1g NH 3

54. 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.

55. 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