1. Chapter 12 Stoichiometry
Mr. Mole

2. Molar Mass of Compounds
Molar mass (MM) of a compound - determined by adding up the atomic masses of each element
Ex. Molar mass of CaCl2
MM of Calcium = g/mol
MM of Chlorine = 35.45g g/mol x 2 = 70.9 g/mol
Molar Mass of CaCl2= g/mol Ca g/mol Cl = g/mol CaCl2 20
Ca 
17 Cl 35.45

3. Flowchart Atoms or Molecules Moles Mass (grams) Divide by 6.02 X 1023
Multiply by 6.02 X 1023
Moles
Multiply by atomic/molar mass from periodic table
Divide by atomic/molar mass from periodic table
Mass (grams)

4. Practice Calculate the Molar Mass of calcium phosphide Formula =
Masses elements:
Ca = g/mol * 3
P = g/mol * 2
Molar Mass = g/mol
Ca3P2

5. Calculations
molar mass Avogadro’s number Grams Moles particles 22.4 L Volume Everything must go through Moles!!!

6. Chocolate Chip Cookies!!
1 cup butter 1 cup packed brown sugar 2 eggs 2 1/2 cups all-purpose flour 2 cups semisweet chocolate chips Makes 3 dozen
How many eggs are needed to make 3 dozen cookies?
How many eggs to make 9 dozen cookies?
How much brown sugar would I need if I had 4 eggs? 2 6 2

7. Cookies and Chemistry…Huh!?!?
Just like chocolate chip cookies have recipes, chemists have recipes called equations
Instead of using cups and teaspoons, we use moles
Lastly, instead of eggs, butter, sugar, etc. we use compounds

8. Chemistry Recipes
Balanced reactions tell us how much reactant will react to get a product
like the cookie recipe
Be sure you have a balanced equation before you start!
Ex: 2 Na + Cl2  2 NaCl
Reaction tells us by mixing 2 moles of sodium with 1 mole of chlorine we will get 2 moles of sodium chloride
What if we wanted 4 moles of NaCl? 10 moles? 50 moles?

9. Practice
Write the balanced reaction for hydrogen gas reacting with oxygen gas.
2 H2 + O2  2 H2O
How many moles of reactants are needed?
What if we wanted 4 moles of water?
What if we had 3 moles of oxygen, how much hydrogen would we need to react, and how much water would we get?
What if we had 50 moles of hydrogen, how much oxygen would we need, and how much water produced?
2 H2, 1 O2
4 H2, 2 O2
6 H2, 6 H2O
25 O2, 50 H2O

10. Stoichiometry is… Greek for “measuring elements”
Pronounced “stoy kee ah muh tree”
Defined as: calculations of the quantities in chemical reactions, based on a balanced equation.
There are 4 ways to interpret a balanced chemical equation

11. #1. In terms of Particles An Element is made of atoms
A Molecular compound (made of only nonmetals) is made up of molecules
Ionic Compounds (made of a metal and nonmetal parts) are made of formula units

12. Example: 2H2 + O2 → 2H2O
Two molecules of hydrogen and one molecule of oxygen form two molecules of water.
Another example: 2Al2O3 ® 4Al + 3O2 2
formula units
Al2O3
form 4
atoms Al
and 3
molecules O2
Now read this: 2Na + 2H2O ® 2NaOH + H2

13. #2. In terms of Moles
Coefficients tell us how many moles of each substance
2Al2O3 ® 4Al + 3O2
2 mol Al2O3, 4 mol Al, 3 mol O2
2Na + 2H2O ® 2NaOH + H2
Remember: A balanced equation is a Molar Ratio

14. #3. In terms of Mass Be + 2F  BeF2
The Law of Conservation of Mass applies
We can check mass by using moles.
Be + 2F  BeF2
1 mole Be
9.01 g Be =
9.01 g Be
1 mole Be +
19.00 g F
2 mole F
38.00 g F =
1 mole F
47.01 g Be + 2F
36.04 g H2 + O2
reactants

15. In terms of Mass (for products)
Be + 2F  BeF2
1 moles BeF2
47.01 g BeF2 =
47.01 g BeF2
1 mole BeF2
47.01 g Be + 2F =
47.01 g BeF2
reactant = product
Mass of reactants and products. must be equal

16. 67.2 Liters of reactant ≠ 44.8 Liters of product!
#4. In terms of Volume
At STP, 1 mol of any gas = 22.4 L
2H O ® H2O
(2 x 22.4 L H2) + (1 x 22.4 L O2) ® (2 x 22.4 L H2O)
**Mass and atoms are ALWAYS conserved
however, molecules, formula units, moles, and volumes will not necessarily be conserved!
67.2 Liters of reactant ≠ 44.8 Liters of product!

17. Practice:
Show that the following equation follows the Law of Conservation of Mass (show the atoms balance, and the mass on both sides is equal)
2Al2O3 ® 4Al + 3O2
Atoms: 4 Al and 6 O = Al and 6 O
Mass: Reactant: g/mol = Products: 108 g/mol g/mol

18. Practice Atoms: 2 N and 6 H = 2 N and 6 H Moles: 1 N2 and 3 H2 ≠ 2 NH3
1). Balance the equation and interpret it in terms of atoms, moles and mass. Show that the law of conservation is observed.
N H2--> NH3
Atoms: 2 N and 6 H = N and 6 H
Moles: 1 N2 and 3 H ≠ NH3
Mass: Reactants: g/mol g/mol = Products: g/mol

19. Section 12.2 Chemical Calculations

20. Mole to Mole conversions
2Al2O3 ® 4Al + 3O2
each time we use 2 moles of Al2O3 we will also make 3 moles of O2
2 moles Al2O3
3 mole O2 or
3 mole O2
2 moles Al2O3
This is how we can convert from mols to mols.
This is why we need balanced equations.

21. Mole to Mole conversions
How many moles of O2 are produced when 3.34 moles of Al2O3 decompose?
2Al2O3 ® 4Al + 3O2
3.34 mol Al2O3
3 mol O2 =
5.01 mol O2
2 mol Al2O3
Conversion factor from balanced equation
If you know the amount of ANY chemical in the reaction, you can find the amount of ALL the other chemicals!

22. Practice: 2C2H2 + 5O2 ® 4CO2 + 2H2O 8.95 mol C2H2 9.6 mol O2
If 3.84 moles of C2H2 are burned, how many moles of O2 are needed?
3.84 moles C2H2
5 moles O2
9.6 mol O2
2 moles C2H2
How many moles of C2H2 are needed to produce 8.95 mole of H2O?
2 moles C2H2
8.95 moles H2O
8.95 mol C2H2
2 moles H2O

23. Steps to Calculate Stoichiometric Problems
1. Correctly balance the equation. 2. Convert the given amount into moles. 3. Set up mole ratios. 4. Use mole ratios to calculate moles of desired chemical. 5. Convert moles back into final unit.

25. Mole-Mass Conversions
Most of the time in chemistry, the amounts are given in grams instead of moles
We still use the mole ratio, but now we also use molar mass to get to grams
Example: How many grams of chlorine are required to react with 5.00 moles of sodium to produce sodium chloride?
2 Na + Cl2  2 NaCl
5.00 moles Na 1 mol Cl g Cl2
2 mol Na 1 mol Cl2
= 177g Cl2

26. Practice
Calculate the mass in grams of Iodine required to react completely with 0.50 moles of aluminum.
Al + 3I  AlI3
g I
0.50 mol Al
3 mol I
= g I
1 mol I
1 mol Al

27. Mass-Mole We can also start with mass and convert to moles
We use molar mass and the mole ratio to get to moles of the compound of interest
Calculate the number of moles of ethane (C2H6) needed to produce 10.0 g of water
2 C2H6 + 7 O2  4 CO2 + 6 H20
10.0 g H2O 1 mol H2O 2 mol C2H6
18.0 g H2O 6 mol H20
= mol C2H6

28. Practice
Calculate how many moles of oxygen are required to make 10.0 g of aluminum oxide
2Al + 3O  Al2O3
10.0 g Al2O3
1 mol Al2O3
3 mol O2
g Al2O3
1 mol Al2O3
0.29 g O2

29. Mass-Mass Problem: 4Al + 3O2  2Al2O3 12.3 g Al2O3 are formed
6.50 grams of aluminum reacts with oxygen. How many grams of aluminum oxide are formed?
4Al O2  2Al2O3
6.50 g Al
1 mol Al
2 mol Al2O3
g Al2O3 =
? g Al2O3
26.98 g Al
4 mol Al
1 mol Al2O3
12.3 g Al2O3
are formed
(6.50 x 1 x 2 x ) ÷ (26.98 x 4 x 1) =

30. Another example:
If 10.1 g of Fe are added to a solution of Copper (II) Sulfate, how many grams of solid copper would form?
2Fe + 3CuSO4 ® Fe2(SO4)3 + 3Cu
10.1 g Fe
1 mol Fe
3 mol Cu
63.55 g Cu
55.85 g Fe
2 mol Fe
1 mol Cu
Answer = 17.2 g Cu

31. Volume-Volume Calculations:
How many liters of CH4 at STP are required to completely react with 17.5 L of O2 ?
CH4 + 2O2 ® CO2 + 2H2O
22.4 L O2
1 mol O2
1 mol CH4
1 mol CH4
22.4 L CH4
1 mol O2
22.4 L CH4
17.5 L O2
22.4 L O2
2 mol O2
1 mol CH4
= 8.75 L CH4

32. Stoichiometry Song - Mark Rosengarten

33. Section 12.3 Limiting Reagent & Percent Yield
OBJECTIVES:
Identify the limiting reagent in a reaction.

34. “Limiting” Reagent
If you are given one dozen loaves of bread, a gallon of mustard, and three pieces of salami, how many salami sandwiches can you make?
The limiting reagent is the reactant you run out of first.
The excess reagent is the one you have left over.
The limiting reagent determines how much product you can make

35. How do you find out which is limited?
The chemical that makes the least amount of product is the “limiting reagent”.
Limiting reagent problems will give you 2 amounts of chemicals
You must do two stoichiometry problems; one for each reagent given.
If 2 products are given, pick one and use it for both calculations

36. Cu is the Limiting Reagent, since it produced less product.
If 10.6 g of copper reacts with 3.83 g sulfur, how many grams of the product (copper (I) sulfide) will be formed?
2Cu + S ® Cu2S
Cu is the Limiting Reagent, since it produced less product.
1 mol Cu
1 mol Cu2S
g Cu2S
10.6 g Cu
63.55g Cu
2 mol Cu
1 mol Cu2S
= 13.3 g Cu2S
1 mol S
1 mol Cu2S
g Cu2S
3.83 g S
32.06g S
1 mol S
1 mol Cu2S
= 19.0 g Cu2S

37. Finding the Amount of Excess
By calculating the amount of the reactant needed to completely react with the limiting reactant, we can subtract that amount from the given amount to find the amount of excess.
Can we find the amount of excess potassium in the next problem?

38. Finding Excess Practice
15.0 g of potassium reacts with 15.0 g of iodine. 2 K + I2  2 KI
We found that Iodine is the limiting reactant.
15.0 g I mol I mol K g K
254 g I mol I mol K
= 4.62 g K USED!
15.0 g K – 4.62 g K = g K EXCESS
Given amount of excess reactant
Amount of excess reactant actually used
Note that we started with the limiting reactant! Once you determine the LR, you should only start with it!

39. Another example: 2Al + 3CuSO4 → 3Cu + Al2(SO4)3
If 10.3 g of aluminum are reacted with 51.7 g of CuSO4, how much copper (grams) will be produced?
2Al + 3CuSO4 → 3Cu + Al2(SO4)3
CuSO4 limiting reactant
10.3 g Al
1 mol Al
3 mol Cu
63.55 g Cu
= g Cu
26.98 g Al
2 mol Al
1 mol Cu
51.7 g CuSO4
1 mol CuSO4
3 mol Cu
63.55 g Cu
g CuSO4
3 mol CuSO4
1 mol Cu
= g Cu

40. How much excess reactant do we have from the last problem?
CuSO4 limiting reactant
Excess Al = g – 5.83 g =
51.7 g CuSO4
1 mol CuSO4
26.98 g Al
2 mol Al
g CuSO4
3 mol CuSO4
1 mol Al
= 5.83 g Al actually used
4.47 g Al excess

41. Limiting Reactant: Recap
You can recognize a limiting reactant problem because there is MORE THAN ONE GIVEN AMOUNT.
Convert ALL of the reactants to the SAME product (pick any product you choose.)
The lowest answer is the limiting reactant
The other reactant(s) are in EXCESS.
To find the amount of excess, subtract the amount used from the given amount.
If you have to find more than one product, be sure to start with the limiting reactant. You don’t have to determine which is the LR over and over again!

42. The Concept of:
A little different type of yield than you had in Driver’s Education class.

43. What is Yield?
Yield is the amount of product made in a chemical reaction.
There are three types:
1. Theoretical yield- what the balanced equation tells should be made
2. Actual yield- what you actually get in the lab when the chemicals are mixed
3. Percent yield = Actual Theoretical
x 100%

44. Example:
6.78 g of copper is produced when 3.92 g of Al are reacted with excess copper (II) sulfate.
2Al + 3 CuSO4 ® Al2(SO4)3 + 3Cu
What is the actual yield?
What is the theoretical yield? What is the percent yield?
= 6.78 g Cu
3.92 g Al
1 mol Al
3 mol Cu
63.55 mol Cu
= 13.8 g Cu
26.98 g Al
2 mol Al
1 mol Cu
6.78 g Cu
X 100
= 49.1 %
13.8 g Cu

45. Details on Yield Percent yield tells us how “efficient” a reaction is.
Percent yield can not be bigger than 100 %.
Theoretical yield will always be larger than actual yield!
Why? Due to impure reactants; competing side reactions; loss of product in filtering or transferring between containers; measuring