1. USING THE REACTION EQUATION LIKE A RECIPE
STOICHIOMETRY
USING THE REACTION EQUATION LIKE A RECIPE

2. Nearly everything we use is manufactured from chemicals.
USING EQUATIONS
Nearly everything we use is manufactured from chemicals.
Soaps, shampoos, conditioners, cd’s, cosmetics, medications, and clothes.
For a manufacturer to make a profit the cost of making any of these items can’t be more than the money paid for them.
Chemical processes carried out in industry must be economical, this is where balanced equations help.

3. Equations are a chemist’s recipe.
USING EQUATIONS
Equations are a chemist’s recipe.
Equations tell chemists what amounts of reactants to mix and what amounts of products to expect.
When you know the quantity of one substance in a rxn, you can calculate the quantity of any other substance consumed or created in the rxn.
Quantity meaning the amount of a substance in grams, liters, molecules, or moles.

4. USING EQUATIONS
The calculation of quantities in chem-ical reactions is called stoichiometry.
Imagine you are in charge of manu-facturing for Rugged Rider Bicycle Company.
The business plan for Rugged Rider requires the production of 128 custom-made bikes each day.
You are responsible for insuring that there are enough parts at the start of each day.

5. F +S+2W+H+2P  FSW2HP2 USING EQUATIONS
Assume that the major components of the bike are the frame (F), the seat (S), the wheels (W), the handlebars (H), and the pedals (P).
The finished bike has a “formula” of FSW2HP2.
The balanced equation for the production of 1 bike is.
F +S+2W+H+2P  FSW2HP2

6. F S 2W + H P
FSW HP 2

7. # of wheels = ? wheels USING EQUATIONS
Now in a 5 day workweek, Rugged Riders is scheduled to make 640 bikes. How many wheels should be in the plant on Monday morning to make these bikes?
What do we know?
Number of bikes = 640 bikes
1 FSW2HP2=2W (balanced equation)
What is unknown?
# of wheels = ? wheels

8. The connection between wheels and bikes is 2 wheels per bike
The connection between wheels and bikes is 2 wheels per bike. We can use this information as a conversion factor to do the calculation.
2 W
640 FSW2HP2 =
1280 wheels
1 FSW2HP2
We can make the same kinds of connections from a chemical rxn eqn.
N2(g) + 3H2(g)  2NH3(g)
The key is the “coefficient ratio”.

9. 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.
The coefficients of the balanced chemical equation indicate the numbers of moles of reactants and products in a chemical rxn.
1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.
N2 and H2 will always react to form ammonia in this 1:3:2 ratio of moles.
So if you started with 10 moles of N2 it would take 30 moles of H2 and would produce 20 moles of NH3

10. Using the coefficients, from the balan-ced rxn equation to make connections between reactants and products, is the most important information that a rxn equation provides.
Using this information, you can calculate the amounts of the reactants involved and the amount of product you might expect.
Any calculation done with the next process is a theoretical number, the real world isn’t always perfect.

11. Using the coefficients of balanced rxn equations and our knowledge of mole conversions we can perform powerful calculations. A.K.A. stoichiometry.
A balanced rxn equation is essential for all calculations involving amounts of reactants and products.
If you know the number of moles of 1 substance, the balanced equation allows you to calculate the number of moles of all other substances in a rxn equation.

12. MOLE – MOLE EXAMPLE
The following rxn shows the synthesis of aluminum oxide.
3O2(g) + 4Al(s)  2Al2O3(s)
3O2(g) + 4Al(s)  2Al2O3(s)
If you only had 1.8 mol of Al how much product could you make?
Given: 1.8 moles of Al
Unknown: ____ moles of Al2O3

13. MOLE – MOLE EXAMPLE
Solve for the unknown:
3O2(g) + 4Al(s)  2Al2O3(s)
2 mol Al2O3
1.8 mol Al
= 0.90mol
Al2O3
4 mol Al
Mole Ratio

14. MOLE – MOLE EXAMPLE 2
The following rxn shows the synthesis of aluminum oxide.
3O2(g) + 4Al(s)  2Al2O3(s)
If you wanted to produce 24 mol of product how many mol of each reactant would you need?
Given: 24 moles of Al2O3
Unknown: ____ moles of Al
____ moles of O2

15. MOLE – MOLE EXAMPLE 2
Solve for the unknowns:
3O2(g) + 4Al(s)  2Al2O3(s)
4 mol Al
24 mol Al2O3
= 48 mol Al
2 mol Al2O3
3 mol O2
24 mol Al2O3
= 36 mol O2
2 mol Al2O3

16. MASS – MASS CALCULATIONS
No lab balance measures moles directly, generally mass is the unit of choice.
From the mass of 1 reactant or product, the mass of any other reactant or product in a given chemical equation can be calculated, provided you have a balanced rxn equation.
As in mole-mole calculations, the unknown can be either a reactant or a product.

17. MASS – MASS CALCULATIONS 1
Acetylene gas (C2H2) is produced by adding water to calcium carbide (CaC2).
CaC2 + 2H2O  C2H2 + Ca(OH)2
How many grams of C2H2 are produced by adding water to 5.00 g CaC2?

18. MASS – MASS CALCULATIONS 1
What do we know?
Given mass = 5.0 g CaC2
Mole ratio: 1 mol CaC2 = 1 mol C2H2
MM of CaC2 = 64.0 g CaC2
MM of C2H2 = 26.0g C2H2
What are we asked for?
grams of C2H2 produced

19. MASS – MASS CALCULATIONS 1
mass A  moles A  moles B  mass B
5.0 g CaC2
1 mol CaC2
1 mol C2H2
64.0 g CaC2
1mol CaC2
26.0 g C2H2
1mol C2H2
= 2.03 g C2H2

20. MASS – MASS CALCULATIONS 2
You’ve recently learned that Copper will replace silver ions out of solution. You’re eyes light up with this money making opportunity. However, you decide it might be best if you did some preliminary calculations to determine to the feasibility of this get rich scheme. Copper is not very hard to find, however the largest size of Silver nitrate found in the Flinn Catalog is the 500 g size and it costs $ Currently Silver sells for $31.15/ounce on the stock market. How much money could you sell your manufactured Silver for?

21. Single replacement reaction
Cu + 2AgNO3  2Ag + Cu(NO3)2

22. MASS – MASS CALCULATIONS 2
Cu + 2AgNO3  2Ag + Cu(NO3)2
Cu + 2AgNO3  2Ag + Cu(NO3)2
What do we know?
Given mass = 500 g of AgNO3
Mole ratio: 2 mol AgNO3 = 2 mol Ag
MM of AgNO3: g = 1mol
MM of Ag: g = 1mol
Price of Silver: $31.15 = 1 ounce
Conversion g to oz: g = 1 oz

23. MASS – MASS CALCULATIONS 2
500 g AgNO3
1mol AgNO3
2 mol Ag
169.8gAgNO3
2 mol AgNO3
107.87g Ag
1 oz
$31.15
1mol Ag
28.23 g
1 oz
= $350.49

24. A balanced reaction equation indicates the relative numbers of moles of reactants and products.
We can expand our stoichiometric calculations to include any unit of measure that is related to the mole.
The given quantity can be expressed in numbers of particles, units of mass, or volumes of gases at STP.
The problems can include mass-volume, volume-volume, and particle-mass calculations.

25. In any of these problems, the given quantity is first converted to moles.
Then the mole ratio from the balanced eqn is used to convert from the moles of given to the number of moles of the unknown
Then the moles of the unknown are converted to the units that the problem requests.
The next slide summarizes these steps for all typical stoichiometric problems

26. PARTICLES B
MOLE
6.02X10 A
22.4 L 23
MASS
VOLUME
STOICH MAP
MOLAR MASS
RATIO

27. 2H2O  2H2 + O2 MORE MOLE EXAMPLES
How many molecules of O2 are produced when a sample of 29.2 g of H2O is decomposed by electrolysis according to this balanced equation:
2H2O  2H2 + O2

28. What are we asked for? MORE MOLE EXAMPLES What do we know?
Mass of H2O = 29.2 g H2O
2 mol H2O = 1 mol O2 (from balanced equation)
MM of H2O = 18.0 g H2O
1 mol O2 = 6.02x1023 molecules of O2
What are we asked for?
molecules of O2

29. mass A  mols A  mols B 
molecules B
29.2 g H2O
1 mol H2O
1 mol O2
18.0 g H2O
2 mol H2O
6.02x1023 molecules O2
PARTICLES B
MOLE
6.02X10 A
22.4 L 23
MASS
VOLUME
MOLAR MASS
RATIO
1 mol O2
= 4.88 x 1023
molecules O2

30. MORE MOLE EXAMPLES
The last step in the production of nitric acid is the reaction of NO2 with H2O.
3NO2+H2O2HNO3+NO
How many liters of NO2 must react with water to produce 5.00x1022 molecules of NO?

31. MORE MOLE EXAMPLES What do we know? What are we asked for?
Molecules NO = 5.0x1022 molecules NO
1 mol NO = 3 mol NO2 (from balanced equation)
1 mol NO = 6.02x1023 molecules NO
1 mol NO2 = 22.4 L NO2
What are we asked for?
Liters of NO2

32. molecules A mols A mols B volume B
1 mol NO
5.0x1022 molecules NO
3 mol NO2
1 mol NO
6.02x1023 molecules NO
22.4 L NO2
1 mol NO2
PARTICLES B
MOLE
6.02X10 A
22.4 L 23
MASS
VOLUME
MOLAR MASS
RATIO
= 5.58 L NO2

33. Aspirin can be made from a chemical rxn between the reactants salicylic acid and acetic anhydride. The products of the rxn are acetyl-salicylic acid (aspirin) and acetic acid (vinegar). Our factory makes 125, count bottles of Bayer Aspirin/day. Each bottle contains 100 tablets, and each tablet contains 325mg of aspirin. How much in kgs + 10% for production problems, of each reactant must we have in order to meet production?
C7H6O3 + C4H6O3  C9H8O4 + HC2H3O2
Salicylic acid
Acetic anhydride
aspirin
vinegar

34. Mass of salicylic acid in kgs + 10%
What do we know?
Make 125,000 aspirin bottles/day
100 aspirin/bottle
325 mg aspirin/tablet
Mole ratio of aspirin to salicylic acid (1:1) and acetic anhydride (1:1)
MM aspirin = g
MM C7H6O3 = g
MM C4H6O3 = g
What are we asked for?
Mass of salicylic acid in kgs + 10%
Mass of acetic anhydride in kgs + 10%

35. 100 tablets 1 bottle 1 tablet 1000 mg 180.16g = 22,549.4 mol aspirin
325mg asp.
125,000 bottles
1 bottle
1 tablet
1 g
1mol asp.
1000 mg
180.16g
= 22,549.4 mol aspirin

36. Salicylic Acid: 1 mol asp 1 mol C7H6O3 = 3068.97 kg salicylic
136.10g C7H6O3
22,549.4 mol aspirin
1 mol asp
1 mol C7H6O3
1 kg
= kg salicylic
acid +( kg)
1000 g
= 3380 kg of salicylic acid

37. Acetic Anhydride: 1 mol asp 1 mol C4H6O3 = 2301.39 kg
102.06g C4H6O3
22,549.4 mol aspirin
1 mol asp
1 mol C4H6O3
1 kg
= kg
Acetic anhydride kg
1000 g
= 2530 kg Acetic anhydride

38. Limiting Reagents and Percent Yield

39. What Is a Limiting Reagent?
Many cooks follow a recipe when making a new dish.
When a cook prepares to cook he/she needs to know that sufficient amounts of all the ingredients are available.
Let’s look at a recipe for the formation of a double cheeseburger:

40. 1 hamburger bun
1 tomato slice
1 lettuce leaf
2 slices of cheese
2 burger patties

41. If you want to make 5 double cheese burgers:
How many hamburger buns do you need? 5
How many hamburger patties do you need? 10
How many slices of cheese do you need? 10
How many slices of tomato do you need? 5

42. How many double cheeseburgers can you make if you start with:
1 bun, 2 patties, 2 slices of cheese, 1 tomato slice, 1 lettuce leaf 1
2 buns, 4 patties, 4 slices of cheese, 2 tomato slices, 2 lettuce leaves 2
1 mole of buns, 2 moles of patties, 2 moles of cheese, 1 mole of tomato slices, 1 mole of lettuce
1 mole 1
10 buns, 20 patties, 2 slices of cheese, 10 tomato slices, 10 lettuce leaves

43. We can’t make anymore than 1 double cheeseburger with our ingredients.
The slices of cheese limits the number of cheeseburgers we can make.
If one of our ingredients gets used up during our preparation it is called the limiting reactant/reagent (LR)
The LR limits the amount of product we can form; in this case double cheeseburgers.
It is equally impossible for a chemist to make a certain amount of a desired compound if there isn’t enough of one of the reactants.

44. As we’ve been learning, a balanced chemical rxn is a chemist’s recipe.
Which allows the chemist to predict the amount of product formed from the amounts of ingredients available
Let’s look at the reaction equation for the formation of ammonia:
N2(g) + 3H2(g)  2NH3(g)
When 1 mole of N2 reacts with 3 moles of H2, 2 moles of NH3 are produced.
How much NH3 could be made if 2 moles of N2 were reacted with 3 moles of H2?
2 mols of ammonia

45. The amount of H2 limits the amount of NH3 that can be made.
N2(g) + 3H2(g)  2NH3(g)
The amount of H2 limits the amount of NH3 that can be made.
From the amount of N2 available we can make 4 moles of NH3
From the amount of H2 available we can only make 2 moles of NH3.
H2 is our limiting reactant here.
It runs out before the N2 is used up.
Therefore, at the end of the reaction there should be N2 left over.
When there is reactant left over it is said to be in excess.

46. How much N2 will be left over after the reaction?
In our rxn it takes 1 mol of N2 to react all of 3 mols of H2, so there must be mol of N2 that remains unreacted.
We can use our new stoich calculation skills to determine 3 possible types of LR type calculations.
Determine which of the reactants will run out first (limiting reactant)
Determine amount of product
Determine how much excess reactant is wasted

47. Limiting Reactant Problems:
Given the following reaction:
8Cu + S8  8CuS
What is the limiting reactant when 82.0 g of Cu reacts with 25.0 g S?
What is the maximum amount of CuS that can be formed?
How much of the other reactant is wasted?

48. Our 1st goal is to calculate how much S would react if all of the Cu was reacted.
From that we can determine the limiting reactant (LR).
Then we can use the Limiting Reactant to calculate the amount of product formed and the amount of excess reactant left over.
82g Cu
mol Cu
mol S
g S

49. 8Cu + S8  8CuS 1molCu 1mol S8 256.8 g S8 82.0gCu 63.5gCu 8molCu
So if all of our 82.0g of Copper were completely reacted it would require grams of Sulfur.
Since we initially had only 25 g of S, we will run out of the S, the limiting reactant) & we will have Copper left over unreacted

50. Sulfur being our Limiting Reactant is then used to determine how much product is produced.
The amount of sulfur we initially start with limits the amount of product we can make.
________
1mol S8
8 molCuS
95.6 gCuS
25.0g S8
1molCuS
256.8 g S8
1 mol S8
= g CuS

51. So the reaction between 82. 0g of Cu and 25
So the reaction between 82.0g of Cu and 25.0g of S8 can only produce 74.45g of CuS.
To determine how much Copper remains unreacted:
25.0 g S8 1mol S mol Cu gCu
256.8 g S8 1mol S molCu
= g Cu used, so
82.0 – = g Cu unreacted

52. Example 2
Hydrogen gas can be produced in the lab by the rxn of Magnesium metal with HCl according to the following rxn
equation: Mg + 2HCl  MgCl2 + H2
What is the LR when 6.0 g HCl reacts with 5.0 g Mg? What is the maximum amount of H2 that can be formed? And how much of the other reactant is wasted?

53. 5.0g Mg  mol Mg  2mol HCl  g HCl 1molMg 2molHCl 36.5gHCl 5.0g Mg
So if 5.0g of Mg were used up it would take 15.0g HCl, but we only had 6.0g of HCl to begin with.
Therefore, the 6.0g of HCl will run out before the 5.0g of Mg, so HCl is our Limiting Reactant.

54. 6.0g HCl
2mol HCl 
1mol H2 
g H2
1molHCl
1molH2
2.0gH2
6.0g HCl
1molH2
36.5gHCl
2molHCl
= g H2 produced
6.0g HCl
2mol HCl 
1mol Mg 
g Mg
1molHCl
1molMg
24.3gMg
6.0g HCl
36.5gHCl
2molHCl
1molMg
= g Mg
- 5.0 g Mg
= 3.01g Mg extra

55. Textbook method & example p. 313 PP 1
Some rocket engines use a mixture of hydrazine, N2H4. and hydrogen peroxide, H2O2, as the propellant. The reaction is given by the following equation.
N2H4(l) + 2H2O2(l)  N2(g) + 4H2O(g)
a. Which is the limiting reactant in this reaction when mol N2H4 is mixed with mol H2O2?
b. How much of the excess reactants, in moles, remains unchanged?
c. How much of each product, in moles, is formed?

56. 0.750 mol N2H2 x 1 mol N2 = mol N2
1 mol N2H2
0.500 mol H2O2 x 1 mol N2 = mol N2 (part C)
2 mol H2O2
Since H2O2 makes less product (N2), it is the limiting reagent
B mol H2O2 x 1 mol N2H4 = 0.25 mol N2H4
(0.75 – 0.25 = 0.50 mol N2H4 remains)
C mol H2O2 x 4 mol H2O = 1.0 mol H2O

57. Calculating Percent Yield
In theory, when a teacher gives an exam to the class, every student should get a grade of 100%.
Your exam grade, expressed as a per-cent, is a quantity that shows how well you did on the exam compared with how well you could have done if you had answered all questions correctly

58. This calc is similar to the percent yield calc that you do in the lab when the product from a chemical rxn is less than you expected based on the balanced eqn.
You might have assumed that if we use stoich to calculate that our rxn will produce 5.2 g of product, that we will actually recover 5.2 g of product in the lab.
This assumption is as faulty as assuming that all students will score 100% on an exam.

59. When an equation is used to calculate the amount of product that is possible during a rxn, a value representing the theoretical yield is obtained.
The theoretical yield is the maximum amount of product that could be formed from given amounts of reactants.
In contrast, the amount of product that forms when the rxn is carried out in the lab is called the actual yield.
The actual yield is often less than the theoretical yield.

60. What causes a percent yield to be less than 100%?
The percent yield is the ratio of the actual yield to the theoretical yield as a percent
It measures the efficiency of the reaction
actual yield
Percent yield=
x 100
theoretical yield
What causes a percent yield to be less than 100%?

61. Reactions don’t always go to completion; when this occurs, less than the expected amount of product is formed.
Impure reactants and competing side rxns may cause unwanted products to form.
Actual yield can also be lower than the theoretical yield due to a loss of product during filtration or transferring between containers.
If a wet precipitate is recovered it might weigh heavy due to incomplete drying, etc.

62. What is the percent yield if 33.1 g of CaCO3 is produced?
Calcium carbonate is synthesized by heating, as shown in the following equation: CaO + CO2  CaCO3
What is the theoretical yield of CaCO3 if 24.8 g of CaO is heated with 43.0 g of CO2?
What is the percent yield if 33.1 g of CaCO3 is produced?
Determine which reactant
is the limiting and then decide
what the theoretical yield is.

63. 24.8gCaO molCaO mol CO2 gCO2 LR = 19.5gCO2 24.8gCaO molCaO
mol CaCO3
gCaCO3
24.8 g
CaO
1mol CaO
1molCaCO3
100g CaCO3
56g CaO
1mol CaO
1molCaCO3
= 44.3 g CaCO3

64. CaO is our LR, so the reaction should theoretically produce 44
CaO is our LR, so the reaction should theoretically produce 44.3 g of CaCO3 (How efficient were we?)
Our percent yield is:
_____________
33.1 g CaCO3
Percent yield=
x 100
44.3 g CaCO3
Percent yield = 74.7%