1. Stoichiometry
Chapter 9

2. Stoichiometry – that portion of chemistry dealing with numerical relationships in chemical reactions
It involves the calculation of quantities of substances involved in chemical equations.

3. Interpreting Chemical Equations
Write the balanced chemical equation for the reaction between nitrogen and hydrogen to form ammonia (NH3).
N2 (g) + 3H2 (g) NH3 (g)

4. N2 (g) + 3H2 (g) 2NH3 (g) Interpret equation in terms of molecules
One molecule of nitrogen reacts with three molecules of hydrogen to yield two molecules of ammonia

5. N2 (g) + 3H2 (g) 2NH3 (g) Interpret equation in terms of moles
One mole of nitrogen reacts with three moles of hydrogen to yield two moles of ammonia.

6. N2 (g) + 3H2 (g) 2NH3 (g) Interpret equation in terms of mass.
1 mole N2 has a mass of 2 x 14g = 28 g
3 moles H2 has a mass of 3(2x1g) = 6 g
2 moles NH3 has a mass of 2(14g + 3g) = 34 g
34 g of reactants = 34 g of product

7. N2 (g) + 3H2 (g) NH3 (g)
Assuming STP, interpret equation in terms of the volumes of these three gases.
22.4 L of nitrogen reacts with 67.2 L (3x 22.4 L) of hydrogen to produce 44.8 L (2 x 22.4 L) of ammonia.

8. N2 (g) + 3H2 (g) NH3 (g)
What is conserved (not changed) in a chemical reaction?
# of molecules? no
# of moles?
Amount of matter (mass or number of atoms)?
yes
Volume of substances?

9. Law of Conservation of Mass
The Law of Conservation of Mass indicates that in an ordinary chemical reaction,
Matter cannot be created or destroyed.
No change in total mass occurs in a reaction.
Mass of products is equal to mass of reactants.

10. Conservation of Mass 2Ag (s) + S (s) Ag2S (s)
Reactants Products
2 moles Ag moles S = mole Ag2S
2 (107.9 g) (32.1 g) = (247.9 g)
= g

11. Learning Check
2H2S(g) + 3O2 (g) SO2 (g) + 2H2O(g)
Interpret equation for the number of molecules and moles and the volume of gases at STP.
Show that the balanced equation obeys the law of the conservation of mass.

12. Solution (ques # 1) 2H2S(g) + 3O2 (g) 2SO2 (g) + 2H2O(g)
2 molecules of H2S react with 3 molecules of O2 to produce 2 molecules of SO2 and 2 molecules of H2O
2 moles H2S react with 3 moles O2 to produce 2 moles SO2 and 2 moles H2O
44.8L [2x22.4] H2S react with 67.2L [3x22.4] O2 to produce 44.8L SO2 and 44.8L H2O

13. Solution (ques #2) Mass of reactants 2 mol H2S + 3 mol O2
(2 mol X 34.1g) + (3 mol X 32g)
mol mol
(68.2 g) ( 96 g) = g
Mass of products
2 mol SO mol H2O
(2 mol X 64.1g) + (2 mol X 18g)
mol mol
( g) (36 g ) = g

14. Sample problems (grams to moles and vice-versa)
How many moles of ice melt (MgCl2) do we have in a 10 lb bag
(10 lbs is 4,535 grams)?
How many grams are in .2 moles of MgCl2

15. A mole–mole factor is a ratio of the moles (from the coefficients) for any two substances in an equation

16. 4Fe(s) + 3O2(g) 2Fe2O3(s) Fe and O2: 4 moles Fe and 3 moles O2
3 moles O moles Fe
Fe and Fe2O3: moles Fe and 2 moles Fe2O3
2 moles Fe2O moles Fe
O2 and Fe2O3: 3 moles O and 2 moles Fe2O3
2 moles Fe2O moles O2

17. Learning Check Consider the following equation: 3H2(g) + N2(g) 2NH3(g)
A. A mole–mole factor for H2 and N2 is
1) 3 moles N ) 1 mole N ) 1 mole N2
1 mole H moles H moles H2
B. A mole–mole factor for NH3 and H2 is
1) 1 mole H ) 2 moles NH ) 3 moles N2
2 moles NH moles H moles NH3 17

18. Solution Consider the following equation: 3H2(g) + N2(g) 2NH3(g)
A. A mole–mole factor for H2 and N2 is
2) 1 mole N2
3 moles H2
B. A mole–mole factor for NH3 and H2 is
2) 2 moles NH3 18

20. How many moles of Fe are needed for the reaction of 12.0 moles of O2?
Given:
4Fe(s) + 3O2(g) Fe2O3(s)
How many moles of Fe are needed for the reaction of 12.0 moles of O2? 20

21. Solution STEP 1 STEP 2 Plan: moles of O2 moles of Fe
Given: 12 moles of O2
4Fe(s) + 3O2(g) Fe2O3(s)
Need: moles of Fe
STEP 2
Plan: moles of O2 moles of Fe 21

22. STEP 3
Write mole–mole factors from coefficients:
4 moles of Fe = 3 moles of O2
4 moles Fe and 3 moles O2
3 moles O2 4 moles Fe

23. Solution (continued) STEP 4 Set up problem to cancel moles of O2:
12.0 mol O2 x 4 mol Fe = moles of Fe
3 moles O2 23

24. Assignment Sample Problem 9-3 Basic Stoichiometry (Mass to Mass
1 a & b

25. Moles to Grams
Suppose we want to determine the mass (g) of NH3 that can form from 2.50 moles N2.
N2(g) + 3H2(g) NH3(g)

26. N2(g) + 3H2(g) 2NH3(g) The plan needed would be
moles N moles NH grams NH3
The factors needed would be:
mole factor of NH3/N2
the molar mass of NH3

27. N2(g) + 3H2(g) 2NH3(g) The setup for the solution would be:
2.50 mole N2 x 2 moles NH3 x g NH3
1 mole N mole NH3
given mole-mole factor molar mass
= g NH3

28. Learning Check How many grams of O2 are needed to produce
0.400 mole Fe2O3 in the following reaction?
4Fe(s) O2 (g) Fe2O3(s)
1) g O2
2) g O2
3) g O2

29. Solution
2) g O2
0.400 mole Fe2O3 x 3 mole O2 x g O2 = g O2
2 mole Fe2O mole O2
mole factor molar mass

30. Mass to Mass Calculations
You can not use a balanced equation to obtain a direct mass to mass relationship.
Change the known mass to moles using molecular mass conversion factor ( molecular mass of a substance in grams = 1 mole)

31. Then use the coefficients in the equation to determine the mole to mole factor between the known mass and the unknown mass.
Once you know the moles of the unknown you can then use the molar mass of the unknown to convert moles to grams.

32. Steps Change known mass to moles using molar mass
Determine moles of unknown by using the coefficients of the equation to set up a mole-mole factor
Once the moles of unknown is determined, use its molar mass to determine mass of unknown.

33. Mass of a Reactant
The reaction between H2 and O2 produces 13.1 g water.
How many grams of O2 reacted?
2H2(g) ? O2(g) 2H2O(g)
The plan and factors would be
g H2O mole H2O mole O g O2
molar mole-mole molar
mass H2O factor mass O2

34. The reaction between H2 and O2 produces 13. 1 g water
The reaction between H2 and O2 produces 13.1 g water. How many grams of O2 reacted,
2H2(g) ? O2(g) 2H2O(g)
The setup would be:
13.1 g H2O x 1 mole H2O x 1 mole O2 x g O2
18.0 g H2O moles H2O 1 mole O2
given molar mole-mole molar
mass H2O factor mass O2
= g O2

35. Learning Check 2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(g) 1) 88.6 g C2H2
Acetylene gas C2H2 burns in the oxyacetylene torch for welding. How many grams of C2H2 are burned if the reaction produces 75.0 g CO2?
2C2H2(g) O2(g) CO2(g) + 2H2O(g)
1) g C2H2
2) g C2H2
3) g C2H2

36. Solution 3) 22.2 g C2H2 2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(g)
75.0 g CO2 x 1 mole CO2 x 2 moles C2H2 x g C2H2
44.0 g CO moles CO mole C2H2
molar mole-mole molar
mass CO factor mass C2H2
= g C2H2

37. Mass of Product
When 18.6 g ethane gas C2H6 burns in oxygen, how many grams of CO2 are produced?
2C2H6(g) O2(g) CO2(g) + 6H2O(g)
18.6 g ? g
The plan and factors would be
g C2H mole C2H mole CO g CO2
molar mole-mole molar
mass C2H factor mass CO2

38. Calculating the Mass of Product
The setup would be
18.6 g C2H6 x 1 mole C2H6 x 4 moles CO2 x g CO2
30.1 g C2H moles C2H mole CO2
molar mole-mole molar
mass C2H factor mass CO2
= g CO2

39. Learning Check
How many grams H2O are produced when 35.8 g C3H8 react by the following equation?
C3H8(g) + 5O2(g) CO2(g) + 4H2O(g)
g H2O
g H2O
g H2O

40. Solution C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g) = 58.4 g H2O (2)
35.8 g C3H8 x 1 mole C3H8 x 4 mole H2O x g H2O
44.1 g C3H mole C3H mole H2O
molar mole-mole molar
mass C3H factor mass H2O
= g H2O (2)

41. Limiting Reactants and Percent Yield
Chapter 9 section 3

42. Theoretical, Actual, and Percent Yield
theoretical yield - the maximum amount of product, which is calculated using the balanced equation.
actual yield - the amount of product obtained when the reaction takes place 42

43. percent yield- the ratio of actual yield to theoretical yield x 100%
Percent yield = actual yield (g) x theoretical yield (g)

44. 44

45. Calculating Percent Yield
Suppose you have prepared cookie dough to make 5 dozen cookies. The phone rings and you answer. While you talk, a sheet of 12 cookies burns, and you have to throw them out. The rest of the cookies you make are okay. What is the percent yield of edible cookies?
Theoretical yield: 60 cookies possible
Actual yield: cookies to eat
Percent yield: 48 cookies x 100% = 80.%yield cookies 45

46. Problem
With a limited amount of oxygen, the reaction of carbon and oxygen produces carbon monoxide.
2C(g) + O2(g) CO(g)
What is the percent yield if 40.0 g of CO are produced when 30.0 g of O2 are used? 46

47. Solution STEP 1 Given: 40.0 g of CO produced (actual)
30.0 g of O2 used
Need: percent yield of CO 47

48. 2C(g) + O2(g) CO(g)
STEP 2
Write a plan to calculate (theoretical) % yield of CO:
g O moles O2 moles CO g CO

49. 2C(g) + O2(g) 2CO(g) STEP 3 Write conversion factors:
Grams O2 to moles O2
1 mole O2 and g O2
32.0 g O mole O2
Moles O2 to moles CO
1 mole O2 and 2 moles CO
2 moles CO mole O2
moles of CO = g of CO
1 mole CO and g CO
28.0 g CO mole CO 49

50. 2C(g) + O2(g) CO(g)
STEP 4
Setup to calculate theoretical yield in g of O2:
30.0 g O2 x 1 mole O2 x 2 moles CO x g CO
32.0 g O mole O mole CO
= g of CO (theoretical) 50

51. Setup to calculate percent yield:
Percent yield of CO = g of CO (actual) x 100%
g of CO (theoretical)
40.0 g CO (actual) x = 76.2% yield 52.5 g CO (theoretical)

52. Learning Check When N2 and 5.00 g of H2 are mixed, the reaction
produces 16.0 g of NH3. What is the percent yield
for the reaction?
N2(g) + 3H2(g) NH3(g)
1) 31.3% of NH3
2) 56.9% of NH3
3) 80.0% of NH3 52

53. Solution STEP 1 Given: 16.0 g of NH3 produced (actual)
5.00 g of H2 used
Need: percent yield of NH3 53

54. g H2 moles H2 moles NH3 g of NH3
STEP 2
Write a plan to calculate theoretical % yield of NH3:
g H moles H moles NH g of NH3

55. g of H2 moles of moles of g of NH3 H2 NH3
STEP 3 Write conversion factors:
g of H2 to moles of H2
1 mole H2 and g H2
2.02 g H mole H2
moles of H2 to moles of NH3
1 mole H and 2 moles NH3
2 moles NH mole H2
moles of NH3 to g of NH3
1 mole NH3 and g NH3
17.0 g NH mole NH3

56. Solution (continued)
STEP 4 Setup to calculate theoretical yield of g of NH3:
5.00 g H2 x 1 mole H2 x 2 moles NH3 x 17.0 g NH3
2.02 g H moles H mole NH3
= g of NH3 (theoretical) 56

57. Setup to calculate percent yield:
Percent yield = actual x 100
theoretical
Percent yield = g NH3 x 100 = 56.9%
28.1 g NH3

58. Limiting Reactant
A limiting reactant – the reactant in a chemical reaction that is used up.
When the reactant is used up, the chemical reaction stops.
Product stops being formed. 58

59. Reacting Amounts
In a table setting, there is 1plate, 1 fork, 1 knife, and 1 spoon.
How many table settings are possible from 5 plates,
6 forks, 4 spoons, and
7 knives?
What is the limiting item? 59

60. Only 4 place settings are possible. Initially Used Left over
Plates
Forks
Spoons
Knives
The limiting item is the spoon. 60

61. Example 1 of an Everyday Limiting Reactant
How many peanut butter sandwiches could be made from 8 slices of bread and 1 jar of peanut butter?
With 8 slices of bread, only 4 sandwiches could be made.
What is the limiting item? 61

62. 62

63. Limiting Reactant (moles)
When 4.00 moles of H2 is mixed with 2.00 moles of
Cl2, how many moles of HCl can form?
H2(g) Cl2(g) 2HCl(g)
4.00 moles moles ??? Moles
Calculate the moles of product that each reactant, H2 and Cl2, could produce.
The limiting reactant is the one that produces the smaller number of moles of product. 63

64. 4.00 moles H2 x 2 moles HCl = 8.00 moles of HCl 1 moles H2
HCl from H2
4.00 moles H2 x 2 moles HCl = 8.00 moles of HCl
1 moles H2
HCl from Cl2
2.00 moles Cl2 x 2 moles HCl = 4.00 moles of HCl
1 mole Cl2
4.00 moles of HCl is the smaller number of moles
produced. At that point, Cl2 will be used up.
The limiting reactant is Cl2. 64

65. Learning Check
If 4.80 moles Ca are mixed with moles N2, which is the limiting reactant?
3Ca(s) + N2(g)  Ca3N2(s)

66. Solution
moles of Ca3N2 from Ca
4.80 moles Ca x 1 mole Ca3N2 = 1.60 moles Ca3N moles Ca
moles of Ca3N2 from N2
2.00 moles N2 x 1 mole Ca3N2 = moles Ca3N2
1 mole N2
Ca is used up when 1.60 mole Ca3N2 forms. Thus, the limiting reactant is Ca.

67. Limiting Reactants Using Mass
Problem:
What is the mass of ammonia (NH3) that can be produced when g of H2 and 24.0 g of N2 react?
3H2 (g) + N2 (g) NH3 (g)

68. Solution Step 1 Given: 8.00g H2 24.0g N2 3H2 + N2 2NH3
Need: mass NH3 produced by 8.00gH2
mass NH3 produced by 24.0g N2

69. Step 2: set up & solve step 1 H2
3H2 (g) + N2 (g) NH3 (g)
Step 2: set up & solve step 1 H2
8gH2 X 1 mole H2 X 2 moles NH3 = 2.6 moles NH3
2 g H moles H2 N2
24g N2 X 1 mole N2 X 2 moles NH3 = 1.7 moles NH3
28g N mole N2

70. The limiting reactant is N2
3H2 (g) + N2 (g) NH3 (g)
Step 3
Determine the limiting reactant- reactant that produces the least amount of product
The limiting reactant is N2

71. Step 4
Convert the number of moles produced by the limiting reactant into grams
1.7 moles NH3 X 17g NH3 = g NH3
1 mole NH3

72. Learning Check What is the mass of water that can be produced when
8.00 g of H2 and 24.0 g of O2 react?
2H2 (g) + O2 (g) H2O (l)
1) 8.0 g of H2O
2) g of H2O
3) 72 g of H2O 72

73. Solution Step 1 & 2 Moles of H2O from H2: Moles of H2O from O2:
8.00 g H2 x 1 mole H2 x 2 moles H2O = 4.0 moles of H2O
2.0 g H moles H2
Moles of H2O from O2:
24.0 g O2 x 1 mole O2 x 2 moles H2O = 1.50 moles of H2O
32.0 g O mole O2

74. Step 4 – mass of product produced by limiting reactant
Limiting reactant is O2
Step 4 – mass of product produced by limiting reactant
1.50 moles H2O x g H2O = g of H2O
1 mole H2O