Download The Mole

Warm-up

How many grams are in 3.45 X 104
formula units of iron (II) oxide?
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

Stoichiometry
Chemistry Chapter 11
Atoms or
Molecules
Flowchart
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)
Review Practice

Calculate the Molar Mass of
calcium phosphate
 Formula =
Ca3(PO4)2

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 =
120.3 g
62.0 g
128.0 g
120.3g
310.3
+ 62.0g
g/mol
+128.0g
Calculations
molar mass
Grams
Avogadro’s number
Moles
particles
Everything must go through
Moles!!!
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?
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
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!

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?
 Example:
Practice

Write the balanced reaction for hydrogen gas
reacting with oxygen gas.




2 H2 + O2  2 H2O
2 mol H2
How many moles of reactants are needed? 1 mol O2
4 mol H
What if we wanted 4 moles of water? 2 mol O2
2
What if we had 3 moles of oxygen, how much hydrogen
would we need to react and how much water would we
get?
6 mol H2, 6 mol H2O
What if we had 50 moles of hydrogen, how much oxygen
would we need and how much water produced?
25 mol O2, 50 mol H2O
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
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
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 Cl2
2 mol Na
70.90g Cl2
1 mol Cl2
= 177g Cl2
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 I2
253.8g I2
2 moles Al 1 mole I2
= 190 g I2
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 = 0.185
18.0 g H2O 6 mol H20
mol C2H6
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 Al2O3
1 mol Al2O3
3 mol O2
101.96g Al2O3
2 mol Al2O3
= 0.147 mol O2
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

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 N2 1 mol N2 2 mol NH3 17.031g NH3

28.014g N2 1 mol N2
= 2.43 g NH3
1 mol NH3
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 Ca3N2
40.078g Ca 3 mol Ca
= 2.47 g Ca3N2
148.248g Ca3N2
1 mol Ca3N2
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?
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.
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!
Limiting
Limiting
Reactant


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:
10.0 g Al
1 mol Al
26.982 g Al

Reactant: Example
2 mol AlCl3 133.341 g AlCl3
2 mol Al
1 mol AlCl3
= 49.4g AlCl3
Now Cl2:
35.0g Cl2
1 mol Cl2
2 mol AlCl3 133.341 g AlCl3
70.906 g Cl2 3 mol Cl2
1 mol AlCl3
= 43.9g AlCl3
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
.
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.
Limiting Reactant Practice
2 K + I2  2 KI
 Potassium:
15.0g K 1 mol K
39.098 g K

2 mol KI
166.002 g KI
2 mol K
1 mol KI
Iodine:
15.0 g I2
1 mol I2
2 mol KI
253.808 g I2 1 mol I2

= 63.7g KI
166.002 g KI
1 mol KI
= 19.6 g KI
Iodine is the limiting reactant and we get
19.6 g of potassium iodide
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?

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 I2
1 mol I2
2 mol K
39.1 g K
254 g I2
1 mol I2
1 mol K
= 4.62 g K
USED!
15.0 g K – 4.62 g K = 10.38 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!
Limiting Reactant: Recap
1.
2.
3.
4.
5.
6.
7.
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!
Percent Yield and
Stoichiometry



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.

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.
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
75.0 g
X 100
= 66.7 %
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 0.040 g of caffeine and a
bad grade on the lab. What was my percent yield?
0.040 g
5.0 g
X 100
= 0.80 %
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
X 100 = 129%
1.75 g
3. What is the theoretical yield if 5.50 grams
of hydrogen react with nitrogen to form
ammonia?
3 H2 + N2  2 NH3
5.50 g H2 1 mole H2 2 moles NH3
2.016 g H2
3 moles H2
17.031 g NH3
1 mol NH3
= 31.0 grams NH3 = Theoretical Yield!!!!
Only 20.4 grams of ammonia is actually produced in the lab.
What is the percent yield?
20.4 g
X 100 = 65.8 %
31.0 g
yield