STOICHIOMETRY
Mass relationships between
reactants and products in a
chemical reaction
STOICHIOMETRY
Stoichiometry:
looks at what
mass of products (in grams) is
produced when you start with
a certain mass of reactant
Tells you how much you make
when you start with a certain
amount of stuff.
EXAMPLE
 Let’s
say you start with the following
recipe:
2
eggs + 1 cup of flour + 2 tubs of
frosting  1 cake + 3 cupcakes
 Write
this down and answer the
following questions:
QUESTIONS
1.
2.
3.
4.
If you start with 6 eggs, how many
cupcakes would you make?
If you started with 4 cups of flour, how
many cakes could you make?
If you made 6 cupcakes, how many
tubs of frosting did you need?
If you started with 1 ½ cups of flour, how
many cupcakes could you make?
ANSWERS
1.
2.
3.
4.
9 Cupcakes
4 Cakes
4 Tubs of frosting
4 ½ Cupcakes
Question: What method did you use to
answer these questions?
STOICHIOMETRY
 Answer:
You used the ratio between one
ingredient and what you made.
 How

does this relate?
Chemistry is like cooking. Chemical
reactions occur in exact ratios. As a result,
we can also predict how much stuff we
make.
EXAMPLE
 Write
the balanced equation for
the following:
 Sodium
and chlorine mix to form
sodium chloride.
BALANCED EQUATION
 2Na
 In
+ Cl2  2NaCl
this reaction, you have a ratio of
reactants and products.
 They are always the same.
 ANSWER THE FOLLOWING
QUESTIONS?
QUESTIONS
 If
you start with 4 Na, how much NaCl do
you make?
 If
you start with 4 Cl2, how much NaCl do
you make?
 If
you made 5 NaCl, how much Cl2 did
you start with?
ANSWERS
1.
2.
3.


4 NaCl
8 NaCl
2.5 Cl2
In each case you take the ratio from the
chemical reaction
IMPORTANT NOTE: WHAT ARE THE
NUMBERS A RATIO OF???
MOLAR RATIO
The
numbers in front of
the compounds are
ALWAYS RATIOS OF
MOLES.
USING DIMENSIONAL ANALYSIS
 To
solve for the amounts of reactants and
products, we use the balanced chemical
equation
 Just like every other dimensional analysis
problem, you start by writing what you
start with and cancel out the stinking
units.
EXAMPLE
 Let’s

 If
use the previous reaction:
2Na + Cl2  2 NaCl
you start with 3.75 moles of Cl2, how
many moles of NaCl can you make?
 Start with what you know:
 3.75 moles of Cl2
ANSWER
 YOU
USE THE RATIO OF MOLES AS THE
CONVERSION FACTOR
 3.75 moles Cl2 | 2 moles of NaCl =
1 mole of Cl2
 7.50 moles of NaCl
 IMPORTANT NOTE: Not
only do you
have to put the units in dimensional
analysis, you also have to put the
name of the compound.
TRY THESE
1.
H2 + N2  NH3

2.
For the above reaction, balance the
reaction. Then, if you start with 2.5 moles
of H2, how many moles of NH3 do you
make?
Zn + HCl  ZnCl2 + H2

For the above reaction, balance the
reaction. If you start with 5.0 moles of Zn,
how many moles of HCl do you need to
complete the reaction?
ANSWERS
1.
2.


1.7 moles of NH3
10 moles of HCl
NOTE: In each case the units and name
of the compound were included.
NOTE: In #2, you can also find the
amount of other reactants from a
starting reactant.
MOLECULES  MOLES MASS
 Review:
How
do you convert from
molecules to moles?
How
do you convert from
mass to moles?
MOLECULES  MOLES MASS
 Molecules
Moles
1mole = 6.02x1023 molecules
 Mass
Moles
#grams = 1 mole (periodic table)
HOW DOES THIS WORK FOR
US?
Reminder:
Chemical reactions
are ratios of moles.
To use molecules or mass, YOU
MUST FIRST CONVERT TO MOLES
EXAMPLE
 You
begin with the following
reaction:
MgBr2 + 2NaOH  2NaBr + Mg(OH)2
 If
you start with 1.2 x 1024 molecules
of NaOH, how many moles of
Mg(OH)2 will you produce?
STEP 1
 Convert
molecules to moles (ALL
CHEMICAL REACTIONS ARE RATIOS OF
MOLES ONLY!!!)

1.2 x 1024 molecules |
1 mole
__ =
6.02x1023 molecules
 2.0
moles
STEP 2
 MgBr2
+ 2NaOH  2NaBr + Mg(OH)2
 2.0moles
NaOH | 1 moles Mg(OH)2 =
2 moles NaOH
 1 mole of Mg(OH)2
 YOU
CAN SET THIS UP IN ONE DIMENSIOAL
ANALYSIS
EXAMPLE #2
 You
begin with the following
reaction:
2H2 + O2  2H2O
 If
you start with 96.0g of O2, how
many moles of H2O do you
produce?
STEP 1
 Convert
mass to moles first (CHEMICAL
REACTIONS ARE ALWAYS RATIOS OF
MOLES!!!)
 96 g O2 | 1 mole O2 | 2 moles H2O
| 32 g O2 |
1 mole O2
 = 6 moles of H2O
 NOTICE:
EVERY NUMBER HAS A UNIT
AND THE NAME OF THE COMPOUND
TRY THIS
For
the following balance the
equation and then answer the
question:
 AgNO3
If
+ CaBr2  AgBr + Ca(NO3)2
you start with 5.00 g of CaBr2,
how many moles of AgBr do
you make?
ANSWER

5.00g CaBr2 | 1 mole CaBr2 | 2 moles AgBr
| 200g of CaBr2| 1 mole CaBr2
=
.0500 moles of AgBr
NOTE:
AGAIN, YOU CANCEL
OUT THE UNITS AND THE NAME
OF THE COMPOUD!!!
MASS TO MASS
 The
final goal of stoichiometry:
Predict
what mass of
substance is produced or
used by having a
balanced chemical
reaction
SO FAR . . .
 We’ve
looked at going from a starting
number of moles and calculating the
number of moles we produce
 We’ve looked at going from a starting
mass or number of molecules and
calculating the number of moles we
produce
 NOW: We are going to start with a
mass and calculate what mass we
produce
MASS TO MASS
 REMINDER:
A CHEMICAL REACTION
SHOWS A RATIO OF MOLES!!!!
 To go from mass to mass, we must use the
following format:
 Mass  Moles  Moles  Mass
molar mass
chemical reaction
molar mass
EXAMPLE
 Given
the following chemical
reaction:
NaOH + HCl  NaCl + H2O
 If
you start with 345g of HCl, how
many grams of NaCl do you
produce?
STEP 1
 Find
the number of moles of HCl
345g HCl | 1 mole of HCl_
| 36.46 g of HCl
= 9.46 moles of HCl
STEP 2
 Find
the number of moles of NaCl
produced:
9.46 moles of HCl | 1 mole of NaCl
| 1 mole of HCl
= 9.46 moles of NaCl
STEP 3
 Find
the mass of NaCl
9.46 moles of NaCl | 58.44g of NaCl
| 1 mole of NaCl
= 553g of NaCl
 NOTE:
In each conversion, the name of
the compound was included.
ONE DIMENSIONAL ANALYSIS
SETUP

345g HCl | 1 mole of HCl | 1 mole of NaCl | 58.44g of NaCl
| 36.46g of HCl | 1 mole of HCl | 1 mole of NaCl
= 553g of NaCl
TRY THIS
 Balance
 If
the following reaction:
P + O2  PO5
you produce 155g of PO5, what
mass of O2 did you start with (in
grams)?
ANSWER
 2P
+ 5O2  2PO5
155g PO5 | 1mole PO5 | 5moles O2 | 32g O2
| 111g PO5 | 2moles PO5 | 1mole O2
= 112 g O2
CONCEPT MAP
Molecule/Atoms
Molecule/Atoms
1 mole = 6.02x1023
Moles
Balanced chemical equation
Moles
molar mass (#g =1 mole)
Mass
Mass
density (#g = 1mL)
Volume
Volume
TRY THIS
 Translate
and balance:

Iron (III) chloride combines with bromine to
form iron (III) bromide and chlorine.

If you start with 5.66x1023 molecules of
bromine, what volume of iron (III) bromide
do you produce? The density of iron (III)
bromide is 4.50g/mL.
ANSWER
41.2 mL of FeBr3
TRY THIS
 Translate
and balance:

Aluminum combines with hydrobromic acid
to form aluminum bromide and hydrogen
gas.

If you start with 22.4mL of aluminum what
volume of aluminum bromide do you
produce? The density of aluminum is
7.87g/mL. The density of aluminum
bromide is 2.67g/mL.
ANSWER
653 mL of AlBr3
TRY THIS
 Translate
and balance:

Sulfuric acid decomposes to form sulfur
dioxide, oxygen and hydrogen.

If you start with 125 mL of sulfuric acid, how
many molecules of oxygen do you
produce? The density of sulfuric acid is
1.84g/mL.
ANSWER
1.41 X 1024 molecules O2
PERCENT YIELD
 In
theory, you should always produce a
certain mass of product if you start with a
specific mass of reactant.
 In reality, things aren’t perfect:



You could measure inaccurately
Some product could be lost during the
reaction
The reaction doesn’t go to completion
PERCENT YIELD
 As

a result . . .
YOU ALWAYS PRODUCE LESS PRODUCT
THAN IS PREDICTED
 There
is a way to measure how much
product that you did produce:
PERCENT YIELD
CALCULATING PERCENT YIELD
 To
calculate the percent yield, you
first need to calculate the
theoretical amount of product that
should be produced
 Just like how we have been
doing.
 You then have to measure the
actual amount or product made
(this value will be give)
CALCULATING PERCENT YIELD
 Then
you use the following formula:
Actual amount of product
Theoretical amount of product
 The
x 100
closer the number is to 100%, the
closer you are to the theoretical value
EXAMPLE
 You
 If
start with the following reaction:
NaOH + HCl  NaCl + H2O
you start with 75.0g of HCl and produce
35.0g of H2O, what is your percent yield
for H2O?
STEP 1

Calculate the theoretical yield (the amount
of H2O you should produce)
75.0gHCl | 1mole HCl | 1 mole H2O | 18.02g H2O
| 36.34g HCl | 1mole HCl | 1mole H2O
= 37.1g of H2O
STEP 2
 Use
the amount of product measured to
calculate %yield


Theoretical yield = 37.1g
Actual yield = 35.0 g
 %yield
= 35.0g X 100
37.1g
 = 94.3%
TRY THE FOLLOWING
 Using
the following unbalanced
equation, solve the problem:
FeCl3 + NaBr  FeBr3 + NaCl
 If you start with 225g of NaBr and
produce 200g of FeBr3, what is the
percent yield?
SOLUTION


Balanced chemical reaction
FeCl3 + 3NaBr  FeBr3 + 3NaCl
Step 1
225g NaBr | 1mole NaBr | 1mole FeBr3 | 296g FeBr3
| 103g NaBr |3moles NaBr | 1mole FeBr3
= 216g of FeBr3
SOLUTION
 %YIELD
= 92.6%
= 200g FeBr3 x 100
216g FeBr3
LIMITING REACTANT
 If
you start with the mass of 2
reactants, how do you determine
the mass of the product?
 EXAMPLE:
 2H2 + O2  2H2O
 If you have 5.00g of H2 and 65.0g
of O2, which reactant runs out
first?
SOLVING
 To
solve, you must calculate how much
water each reactant could theoretically
produce.




5.00g H2 | 1mole H2 | 2moles H2O | 18g H2O__ =
2.02g H2 | 2moles H2 | 1mole H2O
44.5g of H2O
65.0g O2 | 1mole O2 | 2moles H2O | 18g H2O__
=
32g O2 | 1moles O2 | 1mole H2O
73.1g of H2O
WHAT DOES IT MEAN?
 In



this example:
5.00g of H2 would produce 44.5g of H2O
65.0g of O2 would produce 73.1g of H2O
Therefore, you only produce 44.5g of H2O.
At this point the H2 runs out and you can’t
make any more water.
 Since
the H2 produces less water than the
O2, the H2 will run out before the O2
DEFINITIONS

Since the H2 limits how much water is produced,
we call this reactant the:


LIMITING REACTANT: a reactant that is totally
consumed during a chemical reaction.
Determines the amount of product.
Since we will have extra O2 left over from the
reaction, we call this reactant the:
 EXCESS REACTANT: a reactant that is leftover
when a chemical reaction stops.
LET’S PRACTICE
 Zinc
combines with hydrochloric
acid to from zinc chloride and
hydrogen gas. If you start with 125g
of zinc and 125g of HCl, how much
hydrogen gas do you produce?
What is the limiting reactant? What
is the excess reactant?
ANSWER

Zn + 2HCl  ZnCl2 + H2





125g Zn | 1mole Zn| 1mole H2 | 2g H2 __=
65g Zn | 1mole Zn| 1mole H2
3.85g H2
125g HCl | 1mole HCl | 1mole H2 | 2g H2___ =
36g HCl | 2mole HCl | 1mole H2
3.47g H2
Which is the limiting reactant? Which is the
excess reactant?
WHAT ABOUT THE EXCESS
REACTANT?
 When
you do a reaction, sometimes it is
necessary to determine how much excess
reactant is remaining.
 To calculate this, you use the amount of
limiting reactant to calculate the amount
of excess reactant you actually used
 Let’s look at an earlier example
PREVIOUS EXAMPLE

EXAMPLE:
 2H2 + O2  2H2O
 If you have 5.00g of H2 and 65.0g of O2, which
reactant runs out first?
 5.00g H2 | 1mole H2 | 2moles H2O | 18g H2O__ =
2.02g H2 | 2moles H2 | 1mole H2O
 44.5g of H2O


65.0g O2 | 1mole O2 | 2moles H2O | 18g H2O__ =
32g O2 | 1moles O2 | 1mole H2O
73.1g of H2O
EXAMPLE
 In
this case, H2 was the limiting reactant
and O2 was the excess reactant.



We know that all of the H2 was used up
We have to calculate how much O2 was
used up
Using this, we can calculate how much O2
is remaining
EXAMPLE
 Since
H2 is the limiting factor, we calculate
how much O2 is needed to exactly
combine with the H2


5.00g H2 | 1mole H2 | 1mole O2 | 32g O2__
2.02g H2 | 2moles H2| 1mole O2
= 39.6g O2
 Therefore
you used 39.6g of O2 to
completely react with the H2
ANSWER
 You
started with 65.0g of O2
 You used 39.6g of O2
 65.0g
– 39.6g = 25.4g
 Therefore,
you have 25.4g of O2 remaining
at the end of the reaction.
SUMMARY
1.
2.
3.
4.
5.
6.
Translate and balance reaction
Figure out what you need to solve for
(grams of reactant to grams of product)
Solve for each reactant
Determine limiting reactant
Use the limiting reactant to solve for how
much excess reactant you use
Calculate how much excess reactant is
remaining
LET’S PRACTICE
 Aluminum
combines with chlorine to form
aluminum chloride. If 55.0g of aluminum
reacts with 155g of chlorine, find the
following:
 the limiting reactant
 the mass of product
 mass of excess reactant
 If 170.0g of aluminum chloride is actually
produced, what is the percent yield?
SOLUTION


2Al + 3Cl2  2AlCl3
55.0gAl | 1mole Al | 2moles AlCl3 | 132g AlCl3__ =
27g Al | 2moles Al | 1mole AlCl3
 269g AlCl3 (Al is the excess reactant)

155g Cl2 | 1mole Cl2 | 2moles AlCl3| 132g AlCl3_ =
70g Cl2 | 3moles Cl2 | 1mole AlCl3
 195g AlCl3 (Cl2 is the limiting reactant)

Therefore, you produce 195g AlCl3
SOLUTION

155g Cl2 | 1mole Cl2 | 2moles Al | 27g Al__ =
70g Cl2
| 3mole Cl2 | 1mole Al
 In
the reaction, 39.9g of Al was used
 Therefore:
55.0g – 39.9g = 15.1g Al excess
 Percent
yield = 170g x 100 = 87%
195g