1. Warm-up
How many grams are in 3.45 X 104 formula units of iron (II) oxide?

2. You may work with a partner. Answers are on page 996.
In textbook:
Pg 323 #1&3
Pg 324 #5
Pg 328 #15
Pg 329 #17
Pg 331 #19
Pg 335 #35
Pg 337 #41

3. Stoichiometry Chemistry Chapter 11

4. 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)

5. Review Practice Calculate the Molar Mass of calcium phosphate
Formula =
Masses elements:
Ca: 3 Ca’s X 40.1 =
P: 2 P’s X 31.0 =
O: 8 O’s X 16.0 =
Molar Mass =
Ca3(PO4)2
120.3 g
62.0 g
128.0 g
120.3g g g
310.3 g/mol

6. Everything must go through Moles!!!
Calculations
molar mass Avogadro’s number Grams Moles particles
Everything must go through Moles!!!

7. Chocolate Chip Cookies!!
1 cup butter
1/2 cup white sugar
1 cup packed brown sugar
1 teaspoon vanilla extract
2 eggs
2 1/2 cups all-purpose flour
1 teaspoon baking soda
1 teaspoon salt
2 cups semisweet chocolate chips
Makes 3 dozen
How many eggs are needed to make 3 dozen cookies?
How much butter is needed for the amount of chocolate chips used?
How many eggs would we need to make 9 dozen cookies?
How much brown sugar would I need if I had 1 ½ cups white sugar?

8. Cookies and Chemistry…Huh!?!?
Just like chocolate chip cookies have recipes, chemists have recipes as well
Instead of calling them recipes, we call them reaction equations
Furthermore, instead of using cups and teaspoons, we use moles
Lastly, instead of eggs, butter, sugar, etc. we use chemical compounds as ingredients
Ms. Ramsey’s

9. Chemistry Recipes
Looking at a reaction tells us how much of something you need to react with something else to get a product (like the cookie recipe)
Be sure you have a balanced reaction before you start!
Example: 2 Na + Cl2  2 NaCl
This reaction tells us that 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?

10. 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 mol H2 1 mol O2
4 mol H2 2 mol O2
6 mol H2, 6 mol H2O
25 mol O2, 50 mol H2O

11. Mole Ratios
These mole ratios can be used to calculate the moles of one chemical from the given amount of a different chemical
Example: How many moles of chlorine is needed to react with 5 moles of sodium (without any sodium left over)?
2 Na + Cl2  2 NaCl
5 moles Na 1 mol Cl2
2 mol Na
= 2.5 moles Cl2

12. Mole-Mole Conversions
How many moles of sodium chloride will be produced if you react 2.6 moles of chlorine gas with an excess (more than you need) of sodium metal?
2 Na + Cl2  2 NaCl
2.6 moles Cl2 2 mol NaCl
1 mol Cl2
= 5.2 moles NaCl

13. Mole-Mass Conversions
Most of the time in chemistry, the amounts are given in grams instead of moles
We still go through moles and 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 completely 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

14. Practice
Calculate the mass in grams of Iodine required to react completely with 0.50 moles of aluminum.
2 Al + 3 I2  2 AlI3
0.50 moles Al 3 moles I g I2
2 moles Al 1 mole I2
= 190 g I2

15. Mass-Mole
We can also start with mass and convert to moles of product or another reactant
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

16. Practice
Calculate how many moles of oxygen are required to make 10.0 g of aluminum oxide
4 Al + 3 O2  2 Al2O3
10.0g Al2O mol Al2O mol O2
101.96g Al2O mol Al2O3
= mol O2

17. Mass-Mass Conversions
Most often we are given a starting mass and want to find out the mass of a product we will get (called theoretical yield) or how much of another reactant we need to completely react with it (no leftover ingredients!)
Now we must go from grams to moles, mole ratio, and back to grams of compound we are interested in

18. Mass-Mass Conversion
Ex. Calculate how many grams of ammonia (NH3) are produced when you react 2.00g of nitrogen with excess hydrogen.
N2 + 3 H2  2 NH3
2.00g N mol N mol NH g NH3
28.014g N2 1 mol N mol NH3
= 2.43 g NH3

19. Practice
How many grams of calcium nitride are produced when 2.00 g of calcium reacts with an excess of nitrogen?
3 Ca + N2  Ca3N2
2.00g Ca 1 mol Ca 1 mol Ca3N g Ca3N2
40.078g Ca 3 mol Ca mol Ca3N2
= 2.47 g Ca3N2

20. Limiting Reactant: Cookies
1 cup butter
1/2 cup white sugar
1 cup packed brown sugar
1 teaspoon vanilla extract
2 eggs
2 1/2 cups all-purpose flour
1 teaspoon baking soda
1 teaspoon salt
2 cups semisweet chocolate chips
Makes 3 dozen
If we had the specified amount of all ingredients listed, could we make 4 dozen cookies?
What if we had 6 eggs and twice as much of everything else, could we make 9 dozen cookies?
What if we only had one egg, could we make 3 dozen cookies?

21. Limiting Reactant
Most of the time in chemistry we have more of one reactant than we need to completely use up other reactant.
That reactant is said to be in excess (there is too much).
The other reactant limits how much product we get. Once it runs out, the reaction s.
This is called the limiting reactant.

22. Limiting Reactant
To find the correct answer, we have to try all of the reactants. We have to calculate how much of a product we can get from each of the reactants to determine which reactant is the limiting one.
The lower amount of a product is the correct answer.
The reactant that makes the least amount of product is the limiting reactant. Once you determine the limiting reactant, you should ALWAYS start with it!
Be sure to pick a product! You can’t compare to see which is greater and which is lower unless the product is the same!

23. Limiting Reactant: Example
10.0g of aluminum reacts with 35.0 grams of chlorine gas to produce aluminum chloride. Which reactant is limiting, which is in excess, and how much product is produced?
2 Al + 3 Cl2  2 AlCl3
Start with Al:
Now Cl2:
10.0 g Al 1 mol Al mol AlCl g AlCl3
g Al 2 mol Al mol AlCl3
= 49.4g AlCl3
35.0g Cl mol Cl mol AlCl g AlCl3
g Cl2 3 mol Cl mol AlCl3
= 43.9g AlCl3

24. LR Example Continued
We get 49.4g of aluminum chloride from the given amount of aluminum, but only 43.9g of aluminum chloride from the given amount of chlorine. Therefore, chlorine is the limiting reactant. Once the 35.0g of chlorine is used up, the reaction comes to a complete

25. Limiting Reactant Practice
15.0 g of potassium reacts with 15.0 g of iodine. Calculate which reactant is limiting and how much product is made.

26. Limiting Reactant Practice
2 K + I2  2 KI
Potassium:
Iodine:
Iodine is the limiting reactant and we get 19.6 g of potassium iodide
15.0g K 1 mol K mol KI g KI
g K mol K mol KI
= 63.7g KI
15.0 g I mol I mol KI g KI
g I mol I mol KI
= 19.6 g KI

27. Finding the Amount of Excess
By calculating the amount of the excess 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 previous problem?

28. 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, and 19.6 g of potassium iodide are produced.
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!

29. 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 correct answer.
The reactant that gave you 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!

30. Percent Yield and Stoichiometry

31. Percent Yield describes how much product was actually made in the lab versus the amount that theoretically could be made.
Actual Yield  100 = Percent Yield
Theoretical Yield
Percent yield tells you how close you were to the 100% mark.
Reactions do not always work perfectly. Experimental error (spills, contamination) often means that the amount of product made in the lab does not match the ideal amount that could have been made.

32. Theoretical Yield = The maximum amount of product that could be formed from given amounts of reactants.
Actual Yield = The amount of product actually formed or recovered when the reaction is carried out in the laboratory.

33. Example:
A chemist was supposed to produce 75.0 g of aspirin. However, he squandered some of the reactants for his own personal use (he was later fired), and so only actually made 50.0 g. What was his percent yield?
50.0 g
= %
X 100
75.0 g

34. More Examples!
When I was a sophomore in college, we had a lab where we were supposed to isolate caffeine from tea leaves. Most of my caffeine was washed down the drain in a freak accident. Although I should have had 5.0 g of caffeine, I only ended up with g of caffeine and a bad grade on the lab. What was my percent yield?
0.040 g
X 100
= %
5.0 g

35. 2. A very sloppy student did not wait until his NaCl was dry before he weighed it. As a result, his product weighed 2.25 g when it should have been 1.75 g. What was his percent yield?
2.25 g
= 129%
X 100
1.75 g

36. 3. What is the theoretical yield if 5
3. What is the theoretical yield if 5.50 grams of hydrogen react with nitrogen to form ammonia?
H N2  NH3 3 2
5.50 g H2
1 mole H2
g NH3
2 moles NH3
1 mol NH3
2.016 g H2
3 moles H2
= grams NH3 = Theoretical Yield!!!!
Only 20.4 grams of ammonia is actually produced in the lab. What is the percent yield?
20.4 g
= 65.8 % yield
X 100
31.0 g