Tuesday, Sept. 24th: “A” Day
Wednesday, Sept. 25th: “B” Day
Agenda
 Ch. 8 Tests
 Begin Ch. 9: Stoichiometry
 Start Sec 9.1: “Calculating Quantities in Reactions”
Stoichiometry, mole ratios
Moles
Moles
Moles
Mass
 In-Class :
Worksheet pg. 14-15 (moles
moles)
Mass to Mass Stoichiometry worksheet
**You MUST show work for credit!**
Ch. 8 Test
“Chemical Equations and Reactions”
Class
3A
4B
Average Score Average
(out of 60)
Percentage
53.09
88.48%
51.23
85.38%
Balanced Equations Show Proportions
The proportions of the ingredients in a muffin
recipe let you adjust the amounts to make
enough muffins even if you don’t have
balanced amounts initially.
A balanced chemical equation is very similar
to a recipe in that the coefficients in the
balanced equation show the proportions of the
reactants and products.
Consider the reaction for the synthesis of
water…
Balanced Equations Show Proportions
2 H 2 + O2
2 H2O
The coefficients show that two molecules of
hydrogen react with one molecule of oxygen
and form two molecules of water.
Calculations that involve chemical reactions
use the proportions from balanced chemical
equations to find the quantity of each reactant
and product involved.
Relative Amounts in Equations can be
Expressed in Moles
The coefficients in a balanced equation also
represent the moles of each substance.
For example, the equation below shows that
2 mol C8H18 react with 25 mol O2 to form
16 mol CO2 and 18 mol H2O.
2 C8H18 + 25 O2  16 CO2 + 18 H2O
Relative Amounts in Equations can be
Expressed in Moles
You can determine how much of a reactant is
needed to produce a given quantity of product,
or how much of a product is formed from a
given quantity of reactant.
The branch of chemistry that deals with
quantities of substances in chemical reactions
is known as stoichiometry.
Stoichiometry: the proportional relationship
between two or more substances during
a chemical reaction.
The Mole Ratio is the Key
The coefficients in a balanced chemical
equation show the relative numbers of moles
of the substances in the reaction.
As a result, you can use the coefficients in
conversion factors called mole ratios.
Mole ratios bridge the gap and can convert
from moles of one substance to
moles of another.
The Mole Ratio is the Key!
The mole ratio is…..
THE SECRET MAGIC STEP!!
when solving stoichiometry problems.
All stoichiometry problems can be
solved by following these 3 steps:
1. Change the units of what you know into moles.
2. Use the mole ratio (secret magic step) to
convert moles of the substance you were given
into moles of the substance you want.
3. Change out of moles into the unit that you
need for your final answer.
That’s all there is to it!
Sample Problem A, pg. 304
Consider the Haber process for the commercial
preparation of ammonia, NH3:
N2 + 3 H2
2 NH3
Question: How many moles of hydrogen, H2, are
needed to prepare 312 moles of ammonia, NH3?
1. Start with what you know: 312 moles NH3
2. Use the “secret magic step” to convert moles of
NH3 into moles of what you want, H2.
312 mol NH3 X 3 mol H2 =
2 mol NH3
468 mol H2
Practice #1, pg. 304
Calculate the amounts requested if 1.34 mol
H2O2 completely react according to the following
equation: 2 H2O2
2 H2O + O2
a. Moles of oxygen formed
1.34 mol H2O2 X 1 mol O2 = 0.670 mol O2
2 mole H2O2
b. Moles of water formed
1.34 mol H2O2 X 2 mol H2O = 1.34 mol H2O
2 mol H2O2
Practice #2, pg. 304
Calculate the amounts requested if 3.30 mol
Fe2O3 completely react according to the
following equation:
Fe2O3 + 2 Al 2 Fe + Al2O3
a) Moles of aluminum needed
6.60 mol Al
b) Moles of iron formed
6.60 mol Fe
c) Mole of aluminum oxide formed
3.30 mol Al2O3
Problems Involving Mass, Volume, or
Particles
Substances are usually measured by mass or
volume. (Can you go to the balance and
measure out 5 moles of NaCl?)
As a result, before using the mole ratio
(Secret Magic Step) you will need to convert
units for mass and volume into moles.
Mass
Moles
The conversion factor for converting between
mass and moles is simply the molar mass of
the substance.
You do remember molar mass, don’t you?.....
Molar Mass Review
 What is the molar mass of table salt, NaCl?
Na: 22.99 g/mol
+ Cl: 35.45 g/mol
58.44 g/mol
 What is the molar mass of sulfuric acid, H2SO4?
H: 2 (1.01 g/mol) = 2.02 g/mol
S:
= 32.07 g/mol
+ O: 4 (16.00 g/mol) = 64.00 g/mol
98.09 g/mol
OK, I feel better now that you guys remember this
stuff. Let’s try a sample problem………
Remember the 3 steps:
1. Change the units of what you know into moles.
2. Use the mole ratio (secret magic step) to
convert moles of the substance you were given
into moles of the substance you want.
3. Change out of moles into the unit that you
need for your final answer.
Sample Problem B, pg. 307
What mass of NH3 can be made from 1,221 g H2 and
excess N2?
N2 + 3 H2 2 NH3
***Remember, hydrogen is a diatomic element***
1. Start with what you know: 1,221 g H2
2. Convert to moles using molar mass:
1,221 g H2 X 1 mol H2 = 604.5 mol H2
2.02 g H2
3. Use mole ratio to convert moles H2
moles NH3
604.5 mol H2 X 2 mol NH3 = 403.0 mol NH3
3 mol H2
Sample Problem B cont…
4. Finally, use the molar mass to change moles of
NH3 into grams of NH3:
403.0 mol NH3 X 17.04 g NH3 = 6,867 g NH3
1 mol NH3
That’s it…you did it!
Practice #1, pg. 307
Use the equation below to answer the questions
that follow:
Fe2O3 + 2Al
2Fe + Al2O3
a) How many grams of Al are needed to
completely react with 135 g Fe2O3?
b) How many grams of Al2O3 can form when 23.6 g
Al react with excess Fe2O3?
c) How many grams of Fe2O3 react with excess Al
to make 475 g Fe?
d) How many grams of Fe will form when 97.6 g
Al2O3 form?
Update your graphic organizer:
1. What do you use to change from
Moles
Moles
Mole Ratio
“Secret Magic Step”
2. What do you use to change from
Moles
Mass
Molar Mass
(g/mol)
In-Class/Homework Assignment
Worksheet pg. 14-15 (moles
moles)
Mass to Mass Stoichiometry worksheet
We will finish section 9.1 next time….
***You must SHOW ALL WORK on
these worksheets for credit***