2. 1 © 2006 Brooks/Cole - Thomson Chemistry and Chemical Reactivity 6th Edition John C. Kotz Paul M. Treichel Gabriela C. Weaver CHAPTER 4 Chemical Equations and Stoichiometry © 2006 Brooks/Cole Thomson Lectures written by John Kotz

3. 2 © 2006 Brooks/Cole - Thomson CHEMICAL REACTIONS Chapter 4 Reactants: Zn + I 2 Product: Zn I 2

4. 3 © 2006 Brooks/Cole - Thomson Chemical Equations Depict the kind of reactants and products and their relative amounts in a reaction. 4 Al(s) + 3 O 2 (g) ---> 2 Al 2 O 3 (s) The numbers in the front are called stoichiometric coefficients The letters (s), (g), and (l) are the physical states of compounds.

5. 4 © 2006 Brooks/Cole - Thomson Reaction of Phosphorus with Cl 2 Notice the stoichiometric coefficients and the physical states of the reactants and products.

6. 5 © 2006 Brooks/Cole - Thomson Reaction of Iron with Cl 2 Notice the stoichiometric coefficients and the physical states of the reactants and products.

7. 6 © 2006 Brooks/Cole - Thomson Chemical Equations 4 Al(s) + 3 O 2 (g) ---> 2 Al 2 O 3 (s) This equation means 4 Al atoms + 3 O 2 molecules ---give---> 2 “molecules” of Al 2 O 3 2 “molecules” of Al 2 O 3 4 moles of Al + 3 moles of O 2 ---give---> 2 moles of Al 2 O 3 2 moles of Al 2 O 3

8. 7 © 2006 Brooks/Cole - Thomson Chemical Equations Because the same atoms are present in a reaction at the beginning and at the end, the amount of matter in a system does not change.Because the same atoms are present in a reaction at the beginning and at the end, the amount of matter in a system does not change. The Law of the Conservation of MatterThe Law of the Conservation of Matter Demo of conservation of matter, See Screen 4.3. 2HgO(s) ---> 2 Hg(liq) + O 2 (g)

9. 8 © 2006 Brooks/Cole - Thomson Because of the principle of the conservation of matter, an equation must be balanced. It must have the same number of atoms of the same kind on both sides. Chemical Equations Lavoisier, 1788

10. 9 © 2006 Brooks/Cole - Thomson Balancing Equations ___ Al(s) + ___ Br 2 (liq) ---> ___ Al 2 Br 6 (s)

11. 10 © 2006 Brooks/Cole - Thomson Balancing Equations ____C 3 H 8 (g) + _____ O 2 (g) ----> _____CO 2 (g) + _____ H 2 O(g) ____B 4 H 10 (g) + _____ O 2 (g) ----> ___ B 2 O 3 (g) + _____ H 2 O(g)

12. 11 © 2006 Brooks/Cole - Thomson STOICHIOMETRYSTOICHIOMETRY - the study of the quantitative aspects of chemical reactions.

13. 12 © 2006 Brooks/Cole - Thomson STOICHIOMETRYSTOICHIOMETRY It rests on the principle of the conservation of matter. 2 Al(s) + 3 Br 2 (liq) ------> Al 2 Br 6 (s)

14. 13 © 2006 Brooks/Cole - Thomson PROBLEM: If 454 g of NH 4 NO 3 decomposes, how much N 2 O and H 2 O are formed? What is the theoretical yield of products? STEP 1 Write the balanced chemical equation NH 4 NO 3 ---> N 2 O + 2 H 2 O

15. 14 © 2006 Brooks/Cole - Thomson 454 g of NH 4 NO 3 --> N 2 O + 2 H 2 O STEP 2 Convert mass reactant (454 g) --> moles STEP 3 Convert moles reactant (5.68 mol) --> moles product

16. 15 © 2006 Brooks/Cole - Thomson 454 g of NH 4 NO 3 --> N 2 O + 2 H 2 O STEP 3 Convert moles reactant --> moles product Relate moles NH 4 NO 3 to moles product expected. 1 mol NH 4 NO 3 --> 2 mol H 2 O Express this relation as the STOICHIOMETRIC FACTOR.

17. 16 © 2006 Brooks/Cole - Thomson 454 g of NH 4 NO 3 --> N 2 O + 2 H 2 O = 11.4 mol H 2 O produced STEP 3 Convert moles reactant (5.68 mol) --> moles product

18. 17 © 2006 Brooks/Cole - Thomson 454 g of NH 4 NO 3 --> N 2 O + 2 H 2 O STEP 4 Convert moles product (11.4 mol) --> mass product Called the THEORETICAL YIELD ALWAYS FOLLOW THESE STEPS IN SOLVING STOICHIOMETRY PROBLEMS!

19. 18 © 2006 Brooks/Cole - Thomson GENERAL PLAN FOR STOICHIOMETRY CALCULATIONS Mass reactant Stoichiometric factor Moles reactant Moles product Mass product

20. 19 © 2006 Brooks/Cole - Thomson 454 g of NH 4 NO 3 --> N 2 O + 2 H 2 O STEP 5 How much N 2 O is formed? Total mass of reactants = total mass of products total mass of products 454 g NH 4 NO 3 = ___ g N 2 O + 204 g H 2 O mass of N 2 O = 250. g

21. 20 © 2006 Brooks/Cole - Thomson CompoundNH 4 NO 3 N 2 OH 2 OCompoundNH 4 NO 3 N 2 OH 2 O Initial (g) Initial (mol) Change (mol) Final (mol) Final (g) 454 g of NH 4 NO 3 --> N 2 O + 2 H 2 O Amounts Table (page 125) Note that matter is conserved!

22. 21 © 2006 Brooks/Cole - Thomson CompoundNH 4 NO 3 N 2 OH 2 OCompoundNH 4 NO 3 N 2 OH 2 O Initial (g)454 g00 Initial (mol)5.68 mol00 Change (mol)-5.68+5.68+2(5.68) Final (mol)05.6811.4 Final (g)0250204 454 g of NH 4 NO 3 --> N 2 O + 2 H 2 O Amounts Table (page 125) Note that matter is conserved!

23. 22 © 2006 Brooks/Cole - Thomson 454 g of NH 4 NO 3 --> N 2 O + 2 H 2 O STEP 6 Calculate the percent yield If you isolated only 131 g of N 2 O, what is the percent yield? This compares the theoretical (250. g) and actual (131 g) yields.

24. 23 © 2006 Brooks/Cole - Thomson 454 g of NH 4 NO 3 --> N 2 O + 2 H 2 O STEP 6 Calculate the percent yield

25. 24 © 2006 Brooks/Cole - Thomson PROBLEM: Using 5.00 g of H 2 O 2, what mass of O 2 and of H 2 O can be obtained? 2 H 2 O 2 (liq) ---> 2 H 2 O(g) + O 2 (g) Reaction is catalyzed by MnO 2 Step 1: moles of H 2 O 2 Step 2: use STOICHIOMETRIC FACTOR to calculate moles of O 2 Step 3: mass of O 2

26. 25 © 2006 Brooks/Cole - Thomson Reactions Involving a LIMITING REACTANT In a given reaction, there is not enough of one reagent to use up the other reagent completely.In a given reaction, there is not enough of one reagent to use up the other reagent completely. The reagent in short supply LIMITS the quantity of product that can be formed.The reagent in short supply LIMITS the quantity of product that can be formed.

27. 26 © 2006 Brooks/Cole - Thomson LIMITING REACTANTS eactants R eactantsProducts 2 NO(g) + O 2 (g) 2 NO 2 (g) Limiting reactant = ___________ Excess reactant = ____________

28. 27 © 2006 Brooks/Cole - Thomson LIMITING REACTANTS Demo of limiting reactants on Screen 4.7

29. 28 © 2006 Brooks/Cole - Thomson Rxn 1: Balloon inflates fully, some Zn left * More than enough Zn to use up the 0.100 mol HCl Rxn 2: Balloon inflates fully, no Zn left * Right amount of each (HCl and Zn) Rxn 3: Balloon does not inflate fully, no Zn left. * Not enough Zn to use up 0.100 mol HCl LIMITING REACTANTS React solid Zn with 0.100 mol HCl (aq) Zn + 2 HCl ---> ZnCl 2 + H 2 123 (See CD Screen 4.8)

30. 29 © 2006 Brooks/Cole - Thomson Rxn 1Rxn 2Rxn 3 Rxn 1Rxn 2Rxn 3 mass Zn (g)7.003.271.31 mol Zn0.1070.0500.020 mol HCl0.1000.1000.100 mol HCl/mol Zn0.93/12.00/15.00/1 Lim ReactantLR = HClno LRLR = Zn LIMITING REACTANTS React solid Zn with 0.100 mol HCl (aq) Zn + 2 HCl ---> ZnCl 2 + H 2 0.10 mol HCl [1 mol Zn/2 mol HCl] = 0.050 mol Zn

31. 30 © 2006 Brooks/Cole - Thomson Reaction to be Studied 2 Al + 3 Cl 2 ---> Al 2 Cl 6

32. 31 © 2006 Brooks/Cole - Thomson PROBLEM: Mix 5.40 g of Al with 8.10 g of Cl 2. What mass of Al 2 Cl 6 can form? Mass reactant Stoichiometric factor Moles reactant Moles product Mass product

33. 32 © 2006 Brooks/Cole - Thomson Step 1 of LR problem: compare actual mole ratio of reactants to theoretical mole ratio.

34. 33 © 2006 Brooks/Cole - Thomson 2 Al + 3 Cl 2 ---> Al 2 Cl 6 Reactants must be in the mole ratio Step 1 of LR problem: compare actual mole ratio of reactants to theoretical mole ratio.

35. 34 © 2006 Brooks/Cole - Thomson Deciding on the Limiting Reactant If There is not enough Al to use up all the Cl 2 2 Al + 3 Cl 2 ---> Al 2 Cl 6 Lim reag = Al

36. 35 © 2006 Brooks/Cole - Thomson If There is not enough Cl 2 to use up all the Al 2 Al + 3 Cl 2 ---> Al 2 Cl 6 Lim reag = Cl 2 Deciding on the Limiting Reactant

37. 36 © 2006 Brooks/Cole - Thomson We have 5.40 g of Al and 8.10 g of Cl 2 Step 2 of LR problem: Calculate moles of each reactant

38. 37 © 2006 Brooks/Cole - Thomson Find mole ratio of reactants This should be 3/2 or 1.5/1 if reactants are present in the exact stoichiometric ratio. Limiting reactant is Cl 2 2 Al + 3 Cl 2 ---> Al 2 Cl 6

39. 38 © 2006 Brooks/Cole - Thomson Mix 5.40 g of Al with 8.10 g of Cl 2. What mass of Al 2 Cl 6 can form? Limiting reactant = Cl 2 Base all calcs. on Cl 2 Limiting reactant = Cl 2 Base all calcs. on Cl 2 moles Cl 2 moles Al 2 Cl 6 mass Cl 2 mass Al 2 Cl 6 2 Al + 3 Cl 2 ---> Al 2 Cl 6

40. 39 © 2006 Brooks/Cole - Thomson CALCULATIONS: calculate mass of Al 2 Cl 6 expected. CALCULATIONS: calculate mass of Al 2 Cl 6 expected. Step 1: Calculate moles of Al 2 Cl 6 expected based on LR. Step 2: Calculate mass of Al 2 Cl 6 expected based on LR.

41. 40 © 2006 Brooks/Cole - Thomson Cl 2 was the limiting reactant. Therefore, Al was present in excess. But how much?Cl 2 was the limiting reactant. Therefore, Al was present in excess. But how much? First find how much Al was required. First find how much Al was required. Then find how much Al is in excess.Then find how much Al is in excess. How much of which reactant will remain when reaction is complete?

42. 41 © 2006 Brooks/Cole - Thomson 2 Al + 3 Cl 2 products 0.200 mol 0.114 mol = LR Calculating Excess Al Excess Al = Al available - Al required = 0.200 mol - 0.0760 mol = 0.124 mol Al in excess

43. 42 © 2006 Brooks/Cole - Thomson Chemical Analysis Active Figure 4.8

44. 43 © 2006 Brooks/Cole - Thomson Chemical Analysis An impure sample of the mineral thenardite contains Na 2 SO 4. Mass of mineral sample = 0.123 g The Na 2 SO 4 in the sample is converted to insoluble BaSO 4. The mass of BaSO 4 is 0.177 g What is the mass percent of Na 2 SO 4 in the mineral?

45. 44 © 2006 Brooks/Cole - Thomson Chemical Analysis Na 2 SO 4 (aq) + BaCl 2 (aq) --> 2 NaCl(aq) + BaSO 4 (s) 0.177 g BaSO 4 (1 mol/233.4 g) = 7.58 x 10 -4 mol BaSO 4 7.58 x 10 -4 mol BaSO 4 (1 mol Na 2 SO 4 /1 mol BaSO 4 ) = 7.58 x 10 -4 mol Na 2 SO 4 7.58 x 10 -4 mol Na 2 SO 4 (142.0 g/1 mol) = 0.108 g Na 2 SO 4 (0.108 g Na 2 SO 4 /0.123 g sample)100% = 87.6% Na 2 SO 4

46. 45 © 2006 Brooks/Cole - Thomson Chemical Analysis An impure sample of the mineral thenardite contains Na 2 SO 4. Sample mass = 0.123 g The Na 2 SO 4 is converted to insoluble BaSO 4. The mass of BaSO 4 is 0.177 g What is the mass percent of Na 2 SO 4 in the mineral?

47. 46 © 2006 Brooks/Cole - Thomson Determining the Formula of a Hydrocarbon by Combustion Active Figure 4.9

48. 47 © 2006 Brooks/Cole - Thomson Using Stoichiometry to Determine a Formula Burn 0.115 g of a hydrocarbon, C x H y, and produce 0.379 g of CO 2 and 0.1035 g of H 2 O. C x H y + some oxygen ---> 0.379 g CO 2 + 0.1035 g H 2 O C x H y + some oxygen ---> 0.379 g CO 2 + 0.1035 g H 2 O What is the empirical formula of C x H y ?

49. 48 © 2006 Brooks/Cole - Thomson Using Stoichiometry to Determine a Formula First, recognize that all C in CO 2 and all H in H 2 O is from C x H y. C x H y + some oxygen ---> 0.379 g CO 2 + 0.1035 g H 2 O Puddle of C x H y 0.115 g 0.379 g CO 2 +O 2 0.1035 g H 2 O 1 H 2 O molecule forms for each 2 H atoms in C x H y 1 CO 2 molecule forms for each C atom in C x H y

50. 49 © 2006 Brooks/Cole - Thomson Using Stoichiometry to Determine a Formula First, recognize that all C in CO 2 and all H in H 2 O is from C x H y. 1. Calculate amount of C in CO 2 8.61 x 10 -3 mol CO 2 --> 8.61 x 10 -3 mol C 2. Calculate amount of H in H 2 O 5.744 x 10 -3 mol H 2 O -- >1.149 x 10 -2 mol H C x H y + some oxygen ---> 0.379 g CO 2 + 0.1035 g H 2 O

51. 50 © 2006 Brooks/Cole - Thomson Using Stoichiometry to Determine a Formula Now find ratio of mol H/mol C to find values of x and y in C x H y. 1.149 x 10 -2 mol H/ 8.61 x 10 -3 mol C = 1.33 mol H / 1.00 mol C = 4 mol H / 3 mol C Empirical formula = C 3 H 4 C x H y + some oxygen ---> 0.379 g CO 2 + 0.1035 g H 2 O