1. Stoichiometry Milwaukie High School

2. Notes One Unit 4 Making Cookies Making Chemicals Stoichiometry Moles-Moles Calculation A 401, A 402A Read pgs 343-352

3. Making Cookies

4. What does the following recipe tell us? 1 C butter +1 C sugar +2 egg yolks +1/4 C vegetable oil +1/2 C cocoa + 2 tsp vanilla + 2-1/4 C flour + 1/4 tsp salt +1 C Chocolate Powder + 4 Oz semisweet chocolate, 2 TBS cream  32 Mud Puddle Cookies How many cookies can be made with… 64 cookies 160 cookies 4 cups of sugar and 1 Egg? 128 cookies 16 cookies 1-1/8 of C flour and 8 cups of oil? 16 cookies 1024 cookies 8 Cookies  _____ cups butter? 0.25 24 Cookies  _____ tsp vanilla? 1.5 64 Cookies  _____ egg yolks? 4 2 cups of butter and 10 eggs?

5. Making Chemicals

6. Stoichiometry Consider the chemical equation: 4NH 3 + 5O 2  6H 2 O + 4NO There are several numbers involved. What do they all mean? “stoichio” = Greek for element “metry” = measurement Stoichiometry is about measuring the amounts of elements and compounds involved in a reaction. Quantitative study of reactants and products in chemical reactions.

7. Stoichiometry Recall that Chemical formulas represent numbers of atoms 4NH 3 + 5O 2  6H 2 O + 4NO NH 3 1 nitrogen and 3 hydrogen atoms O2O2 2 oxygen atoms H2OH2O 2 hydrogen atoms and 1 oxygen atom NO 1 nitrogen atom and 1 oxygen atom

8. Stoichiometry Recall that Chemical formulas have molar masses: 4NH 3 + 5O 2  6H 2 O + 4NO NH 3 17.0 g/mol O2O2 32.0 g/mol H2OH2O18.0 g/mol NO30.0 g/mol ***To find the molar mass of a chemical formula – add the atomic masses of the elements forming the compound. Use the periodic table to determine the atomic mass of individual elements.***

9. Stoichiometry Recall that Chemical formulas are balanced with coefficients 4NH 3 + 5O 2  6H 2 O + 4NO 4 X NH 3 = 4 nitrogen + 12 hydrogen 5 X O 2 = 10 oxygen 6 X H 2 O= 12 hydrogen + 6 oxygen 4 X NO= 4 nitrogen + 4 oxygen

10. Stoichiometry With Stoichiometry we find out that 4 : 5 : 6 : 4 do more than just multiply atoms. 4NH 3 + 5O 2  6H 2 O + 4NO 4 : 5 : 6 : 4 Are what we call a mole ratio.

11. Stoichiometry 4NH 3 + 5O 2  6H 2 O + 4NO 4 : 5 : 6 : 4 Can mean either: 4 molecules of NH 3 react with 5 molecules of O 2 to produce 6 molecules of H 2 O and 4 molecules of NO OR 4 moles of NH 3 react with 5 moles of O 2 to produce 6 moles of H 2 O and 4 moles of NO

12. Stoichiometry Example (1) Stoichiometry Example (1) moles of reactants  moles of products A reaction takes place between Lithium and water: Li (s) + H 2 O (l)  LiOH (aq) + H 2 (g) Question: How many moles of H 2 will be formed if 6.23 moles of Li combine with water?

13. Stoichiometry Example (1) Continued Stoichiometry Example (1) Continued Question: How many moles of H 2 will be formed if 6.23 moles of Li combine with water? 1. Balance the equation using coefficients 2Li (s) + 2 H 2 O (l)  2 LiOH (aq) + 1H 2 (g) 2. Convert the required amount to moles *do not need this step for this question* 6.23 moles of Li (from question)

14. Stoichiometry Example (1) Continued Stoichiometry Example (1) Continued Question: How many moles of H 2 will be formed if 6.23 moles of Li combine with water? 3. Find the mole ratio using the coefficients. 2Li (s) + 2 H 2 O (l)  2 LiOH (aq) + 1 H 2 (g) 2 moles of Li = 1 mole H 2 4. Convert to desired unit The question asked for moles so 3.12 moles H 2 doesn’t need to be converted and is the final answer.

15. 4NH 3 + 5O 2  6H 2 O + 4NO How many moles of H 2 O are produced if 2.00 moles of O 2 are used? Stoichiometry Question (1) 2.00 mol O 2 2.40 mol H 2 O = Notice that a correctly balanced equation is essential to get the right answer 6 mol H 2 O 5 mol O 2

16. 4 mol NO 6 mol H 2 O 4 NH 3 + 5 O 2  6 H 2 O + 4 NO How many moles of NO are produced in the reaction if 15 mol of H 2 O are also produced? Stoichiometry Question (2) 15 mol H 2 O 10. mol NO =

17. Notes Two Unit Four Mass-Mass Calculation A 402B Group Assignment GA 402 Read pgs 343-352 Do A403 during break!

18. Stoichiometry Read the chemical equation: 4NH 3 + 5O 2  6H 2 O + 4NO “stochio” = Greek for element “metry” = measurement Stoichiometry is about measuring the amounts of elements and compounds involved in a reaction. Quantitative study of reactants and products in chemical reactions.

19. Balanced Equations Have Ratios The coefficients in a chemical equation can be interpreted as the ratio of molecules of each substance. 2 CO (g) + O 2 (g)  2 CO 2 (g) The image and the Balance equation mean: 2 molecules of CO reacts with 1 molecule of oxygen gas to generate 2 molecules of CO 2

20. What Does The Ratio Mean? What Does The Ratio Mean? 2 CO (g) + O 2 (g)  2 CO 2 (g) If I have 6 moles of CO and 6 moles of O 2 can I make 6 moles of CO 2 ? (HINT… think how many CO must reach with each O 2 ? 6 moles CO (g) + 6 moles O 2 (g)

21. Ratio Ratio 2 CO (g) + O 2 (g)  2 CO 2 (g) I can’t make 6 moles of CO 2 because for every one O 2 mole I need 2 CO moles That means I will run out of CO moles after I react with 3 O 2 moles. 12 3

22. Results 6 molecules of CO create 3 molecules CO 2 2 to 1 ratio

23. Results Continued 6 molecules of O 2 create 6 molecules CO 2 1 to 1 ratio

24. Moving Along The Stoichiometry Path We always use the same type of information to make the jumps between steps: grams (x)  moles (x)  moles (y)  grams (y) Molar mass of x Molar mass of y Mole ratio from balanced equation We can start anywhere along this path depending on the question we want to answer ** Notice we can’t directly convert from grams of one compound to grams of another. We must go through moles using the mole ratio.**

25. 2 mol NH 3 1 mol N 2 14.0 g N 2 17.0g NH 3 = 1 mol NH 3 1 mol N 2 28.0 g N 2 Mass-Mass Calculation (1) How many grams of ammonia will be made reacting with 14.0 grams of nitrogen reaction with hydrogen? 1N 2 (g) + 3H 2 (g)  2NH 3 (l)

26. How many grams of hydrogen will be formed by a reaction of 80.5g of lithium with water? 2 Li (s) + 2 H 2 O (l)  2 LiOH (aq) + 1 H 2 (g) 1 mol H 2 2 mol Li 80.5 g Li 11.7 g H 2 = 2.0 g H 2 1 mol H 2 1 mol Li 6.9 g Li Mass-Mass Calculation (2)

27. How many grams of NO 2 are formed by reacting 1.44 grams of NO with oxygen gas? 2NO (g) + O 2(g) → 2NO 2(g) 2 mol NO 2 2 mol NO 1.44 g NO 2.21 g NO 2 = 46.0 g NO 2 1 mol NO 2 1 mol NO 30.0 g NO Mass-Mass Calculation (3)

28. Have We Learned It Yet? Try these on your own - 4 NH 3 + 5 O 2  6 H 2 O + 4 NO a) How many moles of H 2 O can be made using 1.6 mol NH 3 ? b) How many grams of NO can be made from 47 g of NH 3 ?

29. 4 NH 3 + 5 O 2  6 H 2 O + 4 NO a) b) Answers 6 mol H 2 O 4 mol NH 3 1.6 mol NH 3 2.4 mol H 2 O = 4 mol NO 4 mol NH 3 47 g NH 3 83 g NO = 30.0 g NO 1 mol NO 1 mol NH 3 17.0 g NH 3

30. Let’s Put All The Steps Together! Factor label Method: it uses UNITS to cancel and work your way through the problem Given Info (g) 1 Mole Given (mol) Moles Requested MM Requested (g) of Requested (g) MM Given (g) Moles Given 1 Mole Requested = GivenRequestedMole to Mole The given converts grams of given to moles of given. The middle converts moles of given to moles of requested. The requested converts moles of requested to grams of requested. All in one multistep process !!! THIS IS A MASS MASS PROBLEM.

31. Notes Three Unit Four CA 401 Mass-Mass Calculations ChemistryPurgatory

32. Remember Stoich??

33. Steps To Follow Stoich

34. 8 mol CO 2 1 mol C 8 H 8 6.00 g C 8 H 8 20.3 g CO 2 = 44.0 g CO 2 1 mol CO 2 1 mol C 8 H 8 104.0 g C 8 H 8 Mass-Mass Calculation (4) How many grams of carbon dioxide will be made by burning 6.00 grams of Isovanillin? 1C 8 H 8 (l)+ 15O 2 (g)  8 CO 2 (g) + 4 H 2 O(g)

35. 4 mol Ag 2 mol AgO 8.00 g AgO 7.45 g Ag = 107.9 g Ag 1 mol Ag 1 mol AgO 231.8 g AgO Mass-Mass Calculation (5) How many grams of silver will be made by decomposing 8.00 grams of Silver oxide? 2Ag 2 O(s)  1O 2 (g)+ 4Ag(s)

36. Notes Four Unit Four Limiting Reactant Mass-Mass Problem Molar Volume Mass-Volume Problem

37. Limiting Reactants If the moles of the reactants exactly matches the molar ratio, then both reactants are completely consumed Otherwise, one of the reactants is only partially consumed The reactant that is used up is called the limiting reactant, since it is the one that stops the reaction

38. 2 mol NH 3 1 mol N 2 14.0 g N 2 17.0 g NH 3 = 1 mol NH 3 1 mol N 2 28.0 g N 2 Determining Limiting Reactant How many grams of ammonia will be made reacting with 14.0 grams of nitrogen with 7.0 g of hydrogen? 1N 2 (g) + 3 H 2 (g)  2NH 3 (l) 3 mol H 2 2 mol NH 3 7.0 g H 2 89.3 g NH 3 = 17.0 g NH 3 1 mol NH 3 1 mol H 2 2.0 g H 2 Nitrogen Is the Limiting Reactant

39. 1 mol C 6 H 4 Cl 2 1 mol C 6 H 6 24.0 g C 6 H 6 126 g C 6 H 4 Cl 2 = 147.0 g C 6 H 4 Cl 2 1 mol C 6 H 4 Cl 2 1 mol C 6 H 6 28.0 g C 6 H 6 Limiting Reactant Problem How many grams of C 6 H 4 Cl 2 (s) will be made reacting 24.0 grams of benzene with 77.0 g of chlorine? 1 mol C 6 H 4 Cl 2 2 mol Cl 2 77.0 g Cl 2 159 g C 6 H 4 Cl 2 = 147.0 g C 6 H 4 Cl 2 1 mol C 6 H 4 Cl 2 1 mol Cl 2 71.0 g Cl 2 Benzene C 6 H 6 Is the Limiting Reactant 1C 6 H 6 (l) + 2Cl 2 (g)  1 C 6 H 4 Cl 2 (s) + 2HCl(g)

40. Molar Volume @STP 1molN 2(g) = 28.0g/mol22.4L/mol 1molO 2 (g) = 32.0g/mol 22.4L/mol 1molAr(g)=39.9g/mol 22.4L/mol

41. 2 mol NH 3 1 mol N 2 56.0 g N 2 89.6 l NH 3 = 22.4 l NH 3 1 mol NH 3 1 mol N 2 28.0 g N 2 Mass-Volume Calculation How many liters of ammonia will be made reacting 56.0 grams of nitrogen with hydrogen @ STP? 1N 2 (g) + 3 H 2 (g)  2 NH 3 (l)

42. 2 mol NH 3 1 mol N 2 33.6 l N 2 67.2 l NH 3 = 22.4 l NH 3 1 mol NH 3 1 mol N 2 22.4 l N 2 Volume-Volume Calculation How many liters of ammonia will be made reacting 33.6 liters of nitrogen with hydrogen @STP? 1N 2 (g) + 3 H 2 (g)  2NH 3 (l)

43. Extra

44. 18.0 g H 2 O 1 mol H 2 O 6 mol H 2 O 4 mol NH 3 4 NH 3 + 5 O 2  6 H 2 O + 4 NO How many grams of H 2 O are produced if 2.2 mol of NH 3 are combined with excess oxygen? Stoichiometry Question (3) 2.2 mol NH 3 59 g H 2 O =

45. 5 mol O 2 6 mol H 2 O 32.0 g O 2 1 mol O 2 4 NH 3 + 5 O 2  6 H 2 O + 4 NO How many grams of O 2 are required to produce 0.3 mol of H 2 O? Stoichiometry Question (4) 0.3 mol H 2 O 8 g O 2 =

46. 4 NH 3 + 5 O 2  6 H 2 O + 4 NO How many grams of NO is produced if 12 g of O 2 is combined with excess ammonia? 4 mol NO 5 mol O 2 Stoichiometry Question (5) 12 g O 2 9.0 g NO = 30.0 g NO 1 mol NO 1 mol O 2 32.0 g O 2

47. Have we learned it yet? Try these on your own - 4 NH 3 + 5 O 2  6 H 2 O + 4 NO a) How many moles of H 2 O can be made using 1.6 mol NH 3 ? b) What mass of NH 3 is needed to make 0.75 mol NO? c) How many grams of NO can be made from 47 g of NH 3 ?

48. 4 NH 3 + 5 O 2  6 H 2 O + 4 NO a) b) c) Answers 6 mol H 2 O 4 mol NH 3 1.6 mol NH 3 2.4 mol H 2 O = 4 mol NH 3 4 mol NO 0.75 mol NO 13 g NH 3 = 17.0 g NH 3 1 mol NH 3 4 mol NO 4 mol NH 3 47 g NH 3 83 g NO = 30.0 g NO 1 mol NO 1 mol NH 3 17.0 g NH 3