Chapter 9
STOICHIOMETRY
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.
Prediction of what is to be produced using molar
EXAMPLE
Let’s say you start with the following recipe:
2 eggs + 1 cup of mix + 2 tubs of frosting 1 cake + 3 cupcakes
Write this down and answer the following
questions:
1. If you start with 6 eggs, how many cupcakes would you
make?
2. If you started with 4 cups of mix, how many cakes could
you make?
3. If you made 6 cupcakes, how many tubs of frosting did
you need?
4. If you started with 1 ½ cups of mix, how many cupcakes
ANSWERS
1.
9 Cupcakes
2.4 Cakes
3.
4 Tubs of frosting
4.4 ½ Cupcakes
STOICHIOMETRY
Answer: You used the ratio between one
ingredient and what you made.
How does this relate?
Chemistry is like baking. Chemical reactions occur in
EXAMPLE
Write the balanced equation for the following:
Sodium and chlorine mix to form sodium chloride.
2Na + Cl
2
2NaCl
In this reaction, you have a ratio of reactants and
products.
They are always the same.
ANSWER THE FOLLOWING QUESTIONS
QUESTIONS
2Na + Cl
2
2NaCl
If you start with 4 Na, how much NaCl do you
make?
4 NaCl
If you start with 4 Cl
2
, how much NaCl do you
make?
8 NaCl
If you made 5 NaCl, how much Cl
2
did you start
with?
2.5 Cl
ANSWERS
In each case you take the ratio from the chemical
reaction
IMPORTANT NOTE: WHAT ARE THE
MOLAR RATIO
The numbers in front of the compounds
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,
EXAMPLE
Let’s use the previous reaction:
2Na + 1Cl
2 2 NaCl
If you start with 3.75 moles of Cl
2
, how many
moles of NaCl can you make?
Start with what you know:
3.75 moles Cl
ANSWER
2
Na +
1
Cl
2
2
NaCl
YOU USE THE RATIO OF MOLES AS THE
CONVERSION FACTOR (from the balance chem eqtn)
3.75 moles Cl
2 X 2 moles of NaCl =
1 1 mole of Cl2
7.50 moles of NaCl
Not only do you have to put the units in dimensional
TRY THESE
1. H2 + N2 NH3
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?
3 H
2 + 1 N2 2 NH3 2.5 mol H
2 x 2 molNH3
3mol H2
2. 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?
Zn + 2 HCl ZnCl
2 + H2(g) = 5 mol Zn x 2 molHCL
1mol Zn
=1.67mols NH3
MOLECULES
MOLES
MASS
(dnw)
Molecules Moles
1mole = 6.022x10
23molecules
6.022x10
23molecules
1mole
Mass
Moles
#grams = 1 mole
1 mole
#grams
Review:
How do you convert from molecules to
moles? mass to moles?
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)
)
HOW DOES THIS WORK FOR US?
dnw
Reminder: Chemical reactions are ratios of moles.
To use molecules or mass, YOU MUST FIRST
CONVERT TO MOLES
EXAMPLE (write)
You begin with the following reaction:
MgBr
2+ 2NaOH 2NaBr + Mg(OH)
2
If you start with 1.2 x 10
24molecules of NaOH, how
STEP 1
Convert molecules to moles (ALL CHEMICAL
REACTIONS ARE RATIOS OF MOLES ONLY!!!)
1.2 x 1024 molecules X 1 mole__ =
6.022x1023 molecules
STEP 2
MgBr
2
+ 2NaOH
2NaBr + Mg(OH)
2DNW
2.0moles NaOH X 1 moles Mg(OH)
2
=
2 moles NaOH
1 mole of Mg(OH)
2
EXAMPLE #2
You begin with the following reaction:
2H
2+ O
2 2H
2O
If you start with 96.0g of O
2
, how many moles of
STEP 1
Convert mass to moles first (CHEMICAL REACTIONS
ARE ALWAYS RATIOS OF MOLES!!!)
x x
= 6 moles of H
2O
NOTICE(dnw): EVERY NUMBER HAS A UNIT AND
THE NAME OF THE COMPOUND 96 g O2
1
1 mole O2 32 g O2
TRY THESE
In your notes
For the following balance the equation and then
answer the question:
AgNO
3
+ CaBr
2 AgBr + Ca(NO
3)
2
If you start with 5.25 g of CaBr
2
, how many
Solution
2AgNO
3 +1CaBr2 2AgBr + Ca(NO3)2
x x
=.0525 moles of AgBr
NOTICE(dnw): EVERY NUMBER HAS A UNIT AND
THE NAME OF THE COMPOUND 5.25g CaBr2
1
1 mole CaBr2 199.88g of CaBr2
2 moles AgBr =
If you start with 3.18 g of AgBr, how many molecules of Ca(NO
3)2 do you make?
2AgNO
3 + CaBr2 2AgBr + Ca(NO3)2
= 5.099 x 1021 molecules of Ca(NO
3)2
NOTICE: EVERY NUMBER HAS A UNIT AND THE
NAME OF THE COMPOUND
3.18g AgBr 1
1 mole AgBr 187.77g of AgBr
1 moles Ca(NO3)2 2 mole AgBr
x x x Avo #molecules
2AgNO3 + CaBr2 2AgBr + Ca(NO3)2
If you have 1.35 x 1030 atoms of CaBr
2 , how many grams of Ca(NO3)2 will be produced??
1.35 x 1030 atoms CaBr 2
1
X 1 moles CaBr X
2
Avo # atoms CaBr2
X
1 moles Ca(NO3)2 1 mole CaBr2
164.1 g Ca(NO3)2 1 moles Ca(NO3)2
=
367,000,000.00 g Ca(NO
3
)
2or
3.674 x 10
8g Ca(NO
TRY THIS
Balance the following reaction:
P
+ O
2
PO
5
If you produce 155g of PO
5
, what mass
ANSWER
2P + 5O
2
2PO
5155g PO5 x 1mole PO5 x 5moles O2 x 32g O2
110.97g PO5 2moles PO5 1mole O2
PERCENT YIELD
I. In theory, you should always produce a certain mass of
product if you start with a specific mass of reactant.
II. In reality, things aren’t perfect:
i. You could measure inaccurately
ii. Some product could be lost during the reaction iii. The reaction doesn’t go to completion
III. As a result . . .
i. YOU ALWAYS PRODUCE LESS PRODUCT THAN
IS PREDICTED
IV. There is a way to measure how much product that you did
produce:
CALCULATING PERCENT YIELD
I. To calculate the percent yield, you first need to calculate
the theoretical amount of product that should be produced
i. Just like how we have been doing.
II. You then have to measure the actual amount or product
made (this value will be give)
III. Then you use the following formula:
Actual amount of product (from the lab) x 100 Theoretical amount of product( from stoich calc)
STEP 1
Calculate the theoretical yield (the amount of
H
2O you should produce)
75.0gHCl x 1mole HCl x| 1 mole H2O | x18.02g H2O 36.46g HCl 1mole HCl | 1mole H2O
= 37.1g of H
2O
EXAMPLE
NaOH + HCl NaCl + H2O
STEP 2
I.
Use the amount of product measured to calculate
%yield
i. Theoretical yield = 37.1g ii. Actual yield = 35.0 g
II.
%yield = 35.0g X 100
III.37.1g
TRY THE FOLLOWING
Using the following unbalanced equation, solve
the problem:
FeCl
3+ NaBr
FeBr
3+ NaCl
If you start with 225g of NaBr and produce
Balanced chemical reaction
FeCl
3+ 3NaBr
FeBr
3+ 3NaCl
Step 1
225g NaBr x 1mole NaBr x 1mole FeBr3 x 296g FeBr3
103g NaBr 3moles NaBr 1mole FeBr3
= 216g of FeBr3
FeCl3 + NaBr FeBr3 + NaCl
SOLUTION
%YIELD =
200g FeBr
3
x 100
216g FeBr
3= 92.6%
LIMITING REACTANT
I. If you start with the mass of 2 reactants, how do you
determine the mass of the product?
II. 2 eggs + 1 cup of mix + 2 tubs of frosting 1 cake + 3 cupcakes
If you have 15 eggs and 8 tubs of frosting how many cupcakes can you make??? Use stoich…
15 eggs x 3 cupcakes = 22.5 cupcakes
2 eggs
8tubs frosting x 3 cupcakes =12 cupcakes
2 tubs frosting
How many cupcakes can you actually make?
LIMITING REACTANT
I. If you start with the mass of 2 reactants, how do you
determine the mass of the product?
II. EXAMPLE:
i. 2H2 + O2 2H2O
ii. If you have 5.00g of H2 and 65.0g of O2, how much H2O can
you make? Which reactant runs out first (limited)?
III. To solve, you must calculate how much water each
reactant could theoretically produce.
5.00g H2 x 1mole H2 x 2moles H2O x 18g H2O__ = 44.5g of H2O
2.02g H2 2moles H2 | 1mole H2O
65.0g O2 x 1mole O2 x 2moles H2O x 18g H2O__ =73.1g of H2O
WHAT DOES IT MEAN?
I.
In this example:
i. 5.00g of H2 would produce 44.5g of H2O ii. 65.0g of O2 would produce 73.1g of H2O
iii. Therefore, you can ONLY produce 44.5g of H2O. At
this point the H2 runs out and you can’t make any more water.
II.
Since the H
2produces less water than the O
2,the H
2LIMITING Vs EXCESS
REACTANT
:
I. Since the H2 limits how much water is produced, we call
this reactant the:
i. LIMITING REACTANT: a reactant that is totally
consumed during a chemical reaction. Determines the
actual amount of product.
II. Since we will have extra O2 left over from the reaction, we
call this reactant the:
i. EXCESS REACTANT: a reactant that is leftover
LET’S PRACTICE
Zinc combines with hydrochloric acid to form zinc
chloride and hydrogen gas. If you start with 125g of
zinc and 125g of HCl, how much hydrogen gas will
you actually produce? What is the limiting
ANSWER
Zn + 2HCl
ZnCl
2+ H
2 125g Zn x 1mole Zn x 1mole H
2 x 2.02g H2 __=3.85g H2
65.39g Zn 1mole Zn 1mole H2
125g HCl x 1mole HCl x 1mole H
2 x 2.02g H2__ =3.47g H2
36.46g HCl 2mole HCl 1mole H2
WHAT ABOUT THE EXCESS REACTANT?
I.
When you do a reaction, sometimes it is necessary
to determine how much excess reactant is
remaining.
II.
To calculate this, you use the amount of limiting
reactant to calculate the amount of excess reactant
you actually used
III.
You then subtract the amount of excess you started
will from how much was used.
PREVIOUS EXAMPLE
EXAMPLE: (DNW already in notes)
2H
2 + O2 2H2O
If you have 5.00g of H
2 and 65.0g of O2, which reactant
runs out first?
5.00g H
2 | 1mole H2 | 2moles H2O | 18g H2O__ =
2.02g H2 | 2moles H2 | 1mole H2O
44.5g of H 2O
65.0g O
2 | 1mole O2 | 2moles H2O | 18g H2O__ =
32g O2 | 1moles O2 | 1mole H2O
EXAMPLE
I. In this case, H2 was the limiting reactant and O2 was the
excess reactant.
i. We know that all of the H2 was used up
ii. We have to calculate how much O2 was used up
iii. Using this, we can calculate how much O2 is remaining
II. Since H2 is the limiting factor, we calculate how much O2
is needed to exactly combine with the H2
5.00g H2 x1mole H2 x 1mole O2 | x 32g O2__= 39.6g O2
2.02g H2 2moles H2 1mole O2
44.5g H2O x1mole H2O x 1mole O2 |x 32g O2__= 39.6g O2
18.02g H2O 2moles H2O 1mole O2
ANSWER
I. You started with 65.0g of O2 II. You used 39.6g of O2
III. 65.0g – 39.6g = 25.4g
IV. Therefore, you have 25.4g of O2 remaining at the end of
the reaction.
SUMMARY
1. Translate and balance reaction
2. Figure out what you need to solve for (grams of reactant
to grams of product)
3. Solve for each reactant
4. Determine limiting reactant
5. Use the limiting reactant to solve for how much excess
reactant you use
LET’S PRACTICE
1. Aluminum combines with chlorine to form aluminum
chloride. If 55.0g of aluminum reacts with 155g of chlorine, find the following:
a) The limiting reactant b) The mass of product
c) Mass of excess reactant(leftovers)
d) If 170.0g of aluminum chloride is actually produced, what
Aluminum combines with chlorine to form aluminum chloride. If 55.0g of aluminum reacts with 155g of chlorine , find the following: 2Al + 3Cl2 2AlCl3
The limiting reactant???
155g Cl2 x 1mol Cl2 x 2mol AlCl3 x 133.33g AlCl3 = 194.27gAlCl3 1 70.90gCl2 3mol Cl2 1mol AlCl3
55g Al x 1mol Al x 2mol AlCl3 x 133.33g AlCl3 = 271.91g AlCl3 1 26.98gAl 2mol Al 1mol AlCl3
The mass of product????
Mass of excess reactant(leftovers)
155g Cl2 x 1mol Cl2 x 2mol Al x 26.98 g Al= 39.32gAl 1 70.90gCl2 3mol Cl2 1mol Al
If 170.0g of aluminum chloride is actually produced, what is the percent yield?
39.32gAl (170/194.27)100= Chlorine 194.27g AlCl3 15.68gAl 87.5%
