1. Stoichiometry and the Mole
Chapter 9
Stoichiometry and the Mole

2. Part 1: 9.1 Stoichiometry—What Is It?
Study of quantitative aspects of chemistry
Applies to balanced equations only
Answers “How much?”
Number of reactants needed
Number of products produced
Relates mass to molecular and atomic amounts

3. 9.2 The Mole The mole Relates number of atoms/molecules to grams
Conversion unit that is necessary due to small size of individual atoms and molecules
Equal to a set number of particles
Avogadro’s Number 1 mole= 6.02x 1023 particles

4. 9.2 The Mole (Continued)
Consider the chemical equation :
H2 + I2  2HI
This equation expresses amounts in molecules and atoms:
“One molecule of H2 reacts with one molecule of I2 to form 2 molecules of HI.”

5. “One mole of H2 reacts with one mole of I2 to form 2 moles of HI.”
9.2 The Mole (Continued)
Same chemical equation :
H2 + I2  2HI
This equation expresses amounts in moles of substances:
“One mole of H2 reacts with one mole of I2 to form 2 moles of HI.”

6. Mole’s relationship with mass
9.2 The Mole (Continued)
Mole’s relationship with mass
The mole is used to convert particles to mass
Let’s see how!

7. 9.2 The Mole (Continued) Molar mass
Allows conversion between mass and moles
Simply replace atomic mass amu with grams to find the mass of one mole.
The mass of one mole is the molar mass

8. 9.2 The Mole (Continued) Suppose you obtain 393.934 g of gold.
(a) What is the molar mass of gold?
(b) How many moles of gold do you have?
(c) How many atoms of gold do you have?

9. 9.2 The Mole (Continued) Suppose you obtain 393.934 g of gold.
(a) What is the molar mass of gold?
grams per mole
(b) How many moles of gold do you have?
2 moles of Au
(c) How many atoms of gold do you have?
x 1024 atoms of Au

10. Molar mass of molecules
9.2 The Mole (Continued)
Molar mass of molecules
Number of atoms = number of moles
Consider CO2

11. 9.2 The Mole (Continued) Molar mass of CO2
Sum of the molar masses of the atoms in the molecule:
Molar Mass
1 mol C atoms g/mol
1 mol O atoms g/mol
1 mol CO2 molecules g/mol

12. Conversions with molar mass
9.2 The Mole (Continued)
Conversions with molar mass
Moles= given mass/molar mass
Example: 1 mole of water has a mass of g/mol.
If there are 36.g of water, that’s how many moles?

13. 9.2 The Mole (Continued)
How many moles of water are in g of H2O?

14. 9.2 The Mole (Continued)
How many moles of water are in g of H2O?

15. 9.2 The Mole (Continued)
How many grams of water are in moles of water?

16. 9.2 The Mole (Continued)
How many grams of water are in moles of water?

17. 9.2 The Mole (Continued)
Stoichiometry Conversions

18. Finishing Part 1
Final advice..
Do Team Practice Problems part 1

19. Part 2 9.3 Reaction Stoichiometry
According to the equation below…
H2 + I2  2HI
“One mole of H2 reacts with one mole of I2 to form 2 moles of HI.”
What do we say if there are 2 moles of each reactant?

20. 9.3 Reaction Stoichiometry
H2 + I2  2HI
The molar ratio of the reaction is
1: 1: 2
In order to keep that ratio, the products must also be doubled: 2:2:4
So 2 mol rectants form 4 mol of HI
A similar calculation must be performed for ANY change in ANY reactant or product!

21. 9.3: Reaction Stoichiometry
Theoretical yield
Maximum amount of product that can be produced from a certain reactant amount

22. 9.3 Reaction Stoichiometry
Actual yield
Amount actually produced in a chemical reaction
Often is less than theoretical yield due to side reactions going on simultaneously
Is directly measured rather than derived (the mass of the ppt from lab is the actual yield!)

23. 9.3 Reaction Stoichiometry
Percent yield
Percentage of theoretical yield that was measured

24. 9.3 Reaction Stoichiometry
What would be the percent yield if g of HI was actually produced?
Remember, the theoretical yield was g HI

25. 9.3 Reaction Stoichiometry
What would be the percentage yield if g of HI was produced?
The percent yield would be %

26. 9.3 Reaction Stoichiometry Practice
Calculating Actual Yield from a Reaction- STEP BY STEP!
4 Al + 3 O2  2 Al2O3
Step 1: Determine the molar ratio
4: 3: 2
“4 moles of Al react with 3 moles of oxygen to produce 2 moles of aluminum oxide”

27. 9.3 Reaction Stoichiometry
Calculating Yield from a reaction:
4 Al + 3 O2  2 Al2O3
Step 2: Convert moles to masses
Al: 4 moles x g/mol =
O2 : 3 mol x g/mol=
Aluminum Oxide 2mol x g/mol=

28. 9.3 Reaction Stoichiometry
Calculating Yield from a reaction:
4 Al + 3 O2  2 Al2O3
Step 2: Convert moles to masses
Al: 4 moles x g/mol =107.92g
O2 : 3 mol x g/mol=95.94g
Aluminum Oxide: 2mol x (101.93)= g

29. 9.3 Reaction Stoichiometry
Calculating Yield from a reaction:
4 Al + 3 O2  2 Al2O3
Therefore…
4 mol of Al require 3mol O2 to completely react
107.92g Al require 95.94g O2 to completely react
ANOTHER INTERPRETATION…
4 mol of Al react with 3mol O2 to produce 2mol Al2O3
107.92g Al react with 95.94g O2 to produce g Al2O3

30. 9.3 Reaction Stoichiometry (Continued)
Calculating Yield from a reaction:
4 Al + 3 O2  2 Al2O3
Step 3: Calculate % Yield
The theoretical yield = g Al2O3
After the reaction is complete, the actual yield =200.3g
Calculate the % yield

31. 9.3 Reaction Stoichiometry
Calculating Yield from a reaction:
4 Al + 3 O2  2 Al2O3
The theoretical yield = g Al2O3
If the actual yield = 200.3g, calculate the % yield
% yield = 200.9/203.86
% yield = x 100 = 98.55%

32. End of Part 2
Complete Part 2 Team Work Problems

33. 9.4 Dealing with a Limiting Reactant
Reactant that is in short supply
“Limits” the amount of product that can be made
Will be completely used up in a reaction
The reactants that is consumed (runs out) first
Excess reactant
Present in excess in a reaction
Will be left over after the reaction stops
Is done sometimes to make sure a reaction goes to completion (to make sure there’s enough)

34. 9.4 Dealing with a Limiting Reactant
Finding the limiting reactant
Write out the balanced equation
Calculate the number of moles of each reactant
Divide moles by the coefficient in the balanced equation.
The smallest mole-to-coefficient ratio is the limiting reactant! SIMPLE!!

35. 9.4 Dealing with a Limiting Reactant
6 moles of propane are available to react with 29 moles of oxygen…What is the limiting reactant?
Balanced Equation:
C3H O2  3CO H2O

36. 9.4 Dealing with a Limiting Reactant
6 moles of propane react with 29 moles of oxygen…
6mol of C3H8 and 29mol O2
C3H8 + 5 O2  3 CO2 + 4 H2O
When this reaction completes, O2 will run out first!

37. 9.4 Limiting Reactant Look Familiar? 4 Al + 3 O2  2 Al2O3
Find the limiting reactant if 13.3mol of Al react with 11.8mol O2

38. 9.4 Limiting Reactant
Look Familiar?
4 Al + 3 O2  2 Al2O3
Find the limiting reactant if 13.3mol of Al react with 11.8mol O2
Al: 13.3mol/4 = 3.33 ratio
O2: 11.8mol/3 = 3.93 ratio
Aluminum is the limiting reactant
O2 will be in excess after the reaction stops

39. 9.4 Limiting Reactant
What happens with yield when there’s a limiting reactant?
Yield is calculated based on a revised molar ratio…
4 Al + 3 O2  2 Al2O3
Molar ratio = 4: 3: 2
IF 13.3mol of Al reacts (which makes it limiting) (see previous page)…
Theoretical to actual molar ratio of Al is proportional to theoretical to actual molar ratio of Al2O3…
4/13.3mol Al = 2/ x mol Al2O3
x = mol Al2O3 = 6.65
6.65mol Al2O3 are produced by reacting 13.3mol Al
Now convert 6.65mol Al2O3 to mass of Al2O3

40. 9.4 Limiting Reactant
What happens with yield when there’s a limiting reactant?
Yield is calculated based on a revised molar ratio…
4 Al + 3 O2  2 Al2O3
Molar ratio = 4: 3: 2
IF 13.3mol of Al reacts, 6.65 moles of Al2O3 are produced
6.65 mol Al2O3 = 678g Al2O3 which = theoretical yield based on the limiting reactant!

41. 9.4 Limiting Reactant
What happens with yield when there’s a limiting reactant?
Yield is calculated based on a revised molar ratio…
4 Al + 3 O2  2 Al2O3
The actual yield must be measured in the lab!
If the actual yield is only 612g of Al2O3 what is the % yield of Al2O3?
612g/678g = .903 = 90.3% yield (pretty high!)

42. 9.4 Limiting Reactant Calculate the “EXCESS REACTANT”
What’s left over that didn’t react?
The mass of the reactant that was left over once the limiting reactant ran out!

43. End of Part 3
Complete part 3 practice problems

44. Find the limiting reactant and calculate the % yield of HI if…
Practice Problem
Find the limiting reactant and calculate the % yield of HI if…
287.34g of HI are produced when
3.3g of H2 react with 296.4g I2
HINT: Calculate moles of each reactant first!
H2 + I2  2HI

45. Find the limiting reactant and calculate the % yield of HI…
Practice Problem
Find the limiting reactant and calculate the % yield of HI…
3.3g H2 / 2.02g/mol= 1.63mol H2
296.4g I2 / 253.8g/mol= 1.15mol I2
Hydrogen: 1.63 mol/1mol = 1.63
Iodine: 1.15mol/1mol= 1.15
Iodine is the limiting reactant
H2 + I2  2HI

46. Practice Problem
Find the limiting reactant and calculate the % yield of HI…
Molar Ratio = 1:1: 2
Iodine is limiting reactant so revise ratio...
1/1.15 mol I2 = 2/x mol HI
X= 2.3mol HI produced
2.3mol HI x g/mol HI = g HI
Theoretical Yield= g HI
H2 + I2  2HI

47. Find the limiting reactant and calculate the % yield of HI…
Practice Problem
Find the limiting reactant and calculate the % yield of HI…
Theoretical Yield= g HI
Actual Yield HI = g
% Yield = /294.19
% Yield = .9767= 97.67%
H2 + I2  2HI

48. Complete the chapter 8-9 Quiz Finish Lab ! You’re all set
Other Homework…
Start studying for the Final!
Final is for ALL SEMESTER
Will include chapters 8,9 and 12
Other news: NO EXAM 3!

49. End of Part 2
Complete Team Class work Problems
Homework: Start Studying for the Final!

50. Lab Work
Finish Lab by measuring actual yield of ppt
Use the reactant masses to find the limiting reactant as in the previous problem
Calculate the theoretical yield with a revised molar ratio as in previous problem
Calculate the excess reactant

51. 9.4 Dealing with a Limiting Reactant (Continued)

52. 9.5 Combustion Analysis Combustion analysis
Burning a material to find its molecular formula
Combines each components with oxygen
Each oxide is massed to find original composition.

53. 9.5 Combustion Analysis (Continued)
Consider a material that is found to contain 60.0% of its mass as C, 4.48% due to H, and 35.5% of the mass due to oxygen.
How do you find the elemental composition?

54. 9.5 Combustion Analysis (Continued)
Finding formula from percent composition
Assume you have 100 g of material.
Convert masses to number of moles.
Find ratio between components.
Often, this requires dividing by the smallest number of moles.
If not whole number, adjust as needed to receive whole-number subscripts.

55. 9.5 Combustion Analysis (Continued)
If the material is 60.0% C, 4.48% H, and 35.5% oxygen, we know in the 100 g we have 60 g C, 4.48g H, and 35.5g O.
Find the number of
moles of each mass
present.

56. 9.5 Combustion Analysis (Continued)
3. Divide by the smallest number of moles.
(2.22 moles of oxygen)
4. Multiply by 4 to receive whole-number
subscripts.

57. 9.5 Combustion Analysis (Continued)
Determining Formula Review

58. 9.5 Combustion Analysis (Continued)
Empirical formula (simplest formula)
Gives only the ratio between elements present
Example would be CH2O.
Molecular formula
The actual composition of a compound
Examples include C6H12O6 or C2H4O2.

59. 9.5 Combustion Analysis (Continued)
Converting an Empirical Formula
to a Molecular Formula

60. 9.6 Going from Molecular Formula to Percent Composition
Converting molecular formula to a percent composition
Find the mass of each element in one mole.
Divide the mass of each element by molar mass.
This gives the percent composition for each element.

61. 9.6 Going from Molecular Formula to Percent Composition (Continued)
Find the elemental mass percentages for a compound with the formula of C6H12O6.
1. Find the mass of each element in one mole.

62. 9.6 Going from Molecular Formula to Percent Composition (Continued)
Find the elemental mass percentages for a compound with the formula of C6H12O6.
1. Find the mass of each element in one mole.

63. 9.6 Going from Molecular Formula to Percent Composition (Continued)
Find the elemental mass percentages for a compound with the formula of C6H12O6.
2. Divide the mass of each element by molar
mass.

64. 9.6 Going from Molecular Formula to Percent Composition (Continued)
Find the elemental mass percentages for a compound with the formula of C6H12O6.
2. Divide the mass of each element by molar mass.

65. 9.1 Stoichiometry—What Is It? (Continued)
Consider a cheesecake that requires five eggs to prepare. How many eggs would be needed to create two cheesecakes?

66. 9.1 Stoichiometry—What Is It? (Continued)
Consider a cheesecake that requires five eggs to prepare. How many eggs would be needed to create two cheesecakes?
Since one cake requires five eggs, two cakes would require ten eggs.

67. 9.1 Stoichiometry—What Is It? (Continued)
Unit analysis
Find the conversion between units.
5 eggs = 1 cake
Use the conversion factor to find your answer.

68. 9.1 Stoichiometry—What Is It? (Continued)
If one cheesecake requires one cup of sugar and you have three cups of sugar, how many cheesecakes can be made?

69. 9.1 Stoichiometry—What Is It? (Continued)
If one cheesecake requires one cup of sugar and you have three cups of sugar, how many cheesecakes can be made?
If each cake requires one cup of sugar and you have three cups, you can make a total of three cheesecakes.

70. 9.2 The Mole (Continued) Avogadro’s number
Number of atoms/molecules in one mole
x 1023 items
Can be a mole of atoms/molecules/doughnuts
Conversion factors
1 mole = 6.02 x 1023 items

71. 9.2 The Mole (Continued) What is a mole’s size?
Every grain of sand on Earth is roughly one mole.
One mole of water is roughly one swallow.
One mole of sugar would fill your cupped hands.

72. 9.2 The Mole (Continued) So you can say:
1 H2 molecule reacts with 1 I2 molecule to give 2 HI molecules, or
1 dozen H2 molecules react with 1 dozen I2 molecules to give 2 dozen HI molecules
1 mole of H2molecules reacts with 1 mole of I2 molecules to give 2 moles of HI molecules.

73. 9.3 Reaction Stoichiometry (Continued)
Balanced equations can yield conversion factors:
H2 + I2  2 HI

74. 9.3 Reaction Stoichiometry (Continued)
Molecular formulas can also yield conversion factors:

75. 9.3 Reaction Stoichiometry (Continued)
What mass of H2 would be required to make 10.0 g of HI?
g HI  moles HI  moles H2  g H2

76. 9.3 Reaction Stoichiometry (Continued)
What mass of H2 would be required to make 10.0 g of HI?
g HI  moles HI  moles H2  g H2
Once the approach is known, complete the calculation:
You find it will require g of H2.

77. 9.3 % Yield
Calculate the % Yield of HI if g of HI are produced in the reaction shown below (balance equation first!)
H2 + I2  HI

78. Calculate the % Yield of HI (balance equation first!)
1 mol H2 reacts with 1 mol I2 to yield 2 mol HI
Molar ratio: 1:1:2
Convert moles to masses…
2.02g H2 react with 253.8g I2 to yield g HI
% Yield= actual yield/ theoretical yield
% Yield= g/255.82g
% Yield= = % HI
H2 + I2  2HI