Steps for solving Stoichiometric Problems
1
STOICHIOMETRY
2
and percent yield calculations
Step 1
Write the balanced equation for the reaction.
Step 2
Identify your known and unknown quantities.
Step 3
Convert from grams of known to moles of known
Step 4
Convert from moles of known to moles of unknown
Step 5
Convert from moles of unknown to grams of unknown
Reactants: Zn + I2
Product: Zn I2
Sample Problem:
3
How much H2O
will be
formed if
454 g of
NH4NO3
decomposes ?
SAMPLE
SAMPLE PROBLEM:
PROBLEM:
If
If 454
454 gg of
of NH
NH44NO
NO33 decomposes,
decomposes, how
how
much
H
O
is
formed?
much H22O is formed?
STEP 1
Write the balanced
chemical equation
 N2O + 2 H2O
NH4NO3
SAMPLE
SAMPLE PROBLEM:
PROBLEM:
If
If 454
454 gg of
of NH
NH44NO
NO33 decomposes,
decomposes, how
how
much
H
O
is
formed?
much H22O is formed?
5
STEP 2
Identify your
known and
unknown quantities.
NH4NO3
4
 N2O + 2 H2O
NH
NH44NO
NO33 
N
N22O
O ++ 22 H
H22O
O
STEP 3
Convert from grams of known
to moles of known
Use the molar
mass
as a CONVERSION
454 g •
FACTOR:
1 mol
= 5.68 mol NH 4NO3
80.04 g
6
NH
>N
-NH44NO
NO33 -->
-->
N22O
O ++ 22 H
H22O
O
7
STEP 4
Convert from moles of known to moles of unknown
Use the mole
ratio
1 mol NH4NO3
:
as a CONVERSION
8
NH
>N
-NH44NO
NO33 -->
-->
N22O
O ++ 22 H
H22O
O
STEP 4
Convert from moles of known to moles of unknown
Use the mole
2 mol H2O
ratio
as a CONVERSION
FACTOR:
5.68 mol NH 4NO 3 •
2 mol H 2 O produced
1 mol NH 4NO 3 used
NH
NH44NO
NO33 
 N
N22O
O ++ 22 H
H22O
O
2 mol H 2O produced
1 mol NH 4 NO 3 used
= 11.4 mol H2O produced
GENERAL
GENERAL PLAN
PLAN FOR
FOR
STOICHIOMETRY
STOICHIOMETRY
CALCULATIONS
CALCULATIONS
9
STEP 5
Convert from moles of unknown
to grams of unknown
Use the molar
FACTOR:
Mass
unknown
Mass
known
mass
as a CONVERSION FACTOR:
18.02 g
11.4 mol H 2 O •
= 204 g H 2 O
1 mol
Molar Mass
known
Moles
known
10
Mole Ratio
Molar Mass
unknown
Moles
unknown
ALWAYS FOLLOW THESE STEPS IN
SOLVING STOICHIOMETRY PROBLEMS!
SAMPLE
SAMPLE PROBLEM:
PROBLEM:
If
If 454
454 gg of
of NH
NH44NO
NO33 decomposes,
decomposes, how
how
much
H
O
is
formed?
much H22O is formed?
11
OUR FINAL ANSWER WAS:
= 204 g H2O produced
Theoretical Yield:
• a.k.a. “predicted yield”
yield”
• is calculated (by stoichometry)
stoichometry)
• The amount of product we “should”
should” get.
Actual Yield:
•aka “Experimental Yield”
Yield”
This is called the THEORETICAL YIELD…
YIELD…
•is measured in the lab
…it is how much water we theoretically should produce.
•What you actually did produce
•Always less than the theoretical yield.
12
13
Percent Yield =
Actual Yield
Theoretical Yield
NH
>N
-NH44NO
NO33 -->
-->
N22O
O ++ 22 H
H22O
O
Sample Problem Con’
Con’t:
x 100
Our theoretical yield was 204 g H2O
If Johnny Q. Chemistry did the reaction and only
collected 186 g H2O, the percent yield would be:
• Always less than 100 %
• Ratio of actual production
186 g H2O
204 g H2O
to theoretical production.
• Not the same as percent error!
Stoichiometry can be used
to Determine a Formula:
14
15
x 100 = 91%
Determining the Formula of a
Hydrocarbon by Combustion
16
If we burn an hydrocarbon fuel with an unknown formula,
“CxHy”, using a known quantity of oxygen gas:
4.0 g CxHy + 4.5 g O2 --->
---> 5.0 g CO2 + 3.5 g H2O
CCR, page 138
The mass of CO2 and H2O produced can be used to determine
the amount of C and H present in the formula.
Reactions Involving a
LIMITING REACTANT
• Usually, there is not enough of one reagent to
use up the other reagent completely.
• The reagent in short supply LIMITS the
quantity of product that can be formed.
• It is called the “limiting reagent”
reagent”
17
LIMITING REACTANTS
(See CD Screen 4.8)
React solid Zn with 0.100
mol HCl (aq)
aq)
Zn + 2 HCl --->
---> ZnCl2 + H2
1
2
3
Rxn 1: Balloon inflates fully, some Zn left
* More than enough Zn to use up the 0.100 mol HCl
Rxn 2: Balloon inflates fully, no Zn left
* Right amount of each (HCl
(HCl and Zn)
Rxn 3: Balloon does not inflate fully, no Zn left.
* Not enough Zn to use up 0.100 mol HCl
18
LIMITING REACTANTS
19
20
LIMITING REACTANTS
React solid Zn with 0.100
mol HCl (aq)
aq)
Zn + 2 HCl --->
---> ZnCl2 + H2
mass Zn (g)
mol Zn
mol HCl
mol HCl/mol
HCl/mol Zn
Lim Reactant
Rxn 1
~ 7.0
0.100
0.100
1.00/1
LR = HCl
Rxn 2
3.27
0.050
0.100
2.00/1
no LR
Rxn 3
1.31
0.020
0.100
5.00/1
LR = Zn
Demo of limiting reactants on Screen 4.7
21
454
>N
-454 gg of
of NH
NH44NO
NO33 -->
-->
N22O
O ++ 22 H
H22O
O
• Because the same atoms
are present in a reaction
at the beginning and at
the end, the amount of
matter in a system does
not change.
• The Law of the
Conservation of
Matter
STEP 6 Calculate the percent yield
% yield =
actual yield
• 100%
theoretical yield
% yield =
131 g
• 100% = 52.4%
250. g
LIMITING REACTANTS
Chemical Equations
Demo of conservation of matter, See
Screen 4.3.
23
24
Reaction to be Studied
2 Al + 3 Cl2 ---> Al2Cl6
Reactants
2 NO(g) + O2 (g)
Products
2 NO2(g)
Limiting reactant = ___________
Excess reactant = ____________
22
25
26
PROBLEM:
PROBLEM: Mix
Mix 5.40
5.40 gg of
of Al
Al with
with 8.10
8.10 gg
of
What mass
mass of
of Al
Al22Cl
Cl66 can
can form?
form?
of Cl
Cl22.. What
Mass
product
Mass
reactant
Moles
reactant
Stoichiometric
factor
Step
Step 11 of
of LR
LR problem:
problem:
compare
compare actual
actual mole
mole ratio
ratio
of
reactants
to
of reactants to
theoretical
theoretical mole
mole ratio.
ratio.
Moles
product
Step
Step 11 of
of LR
LR problem:
problem:
compare
compare actual
actual mole
mole ratio
ratio of
of
reactants
reactants to
to theoretical
theoretical
mole
mole ratio.
ratio.
If
mol Cl2
3
=
2
mol Al
If
3
mol Cl2
<
2
mol Al
There is not enough Cl2 to use
up all the Al
Lim reag = Cl2
3
mol Cl2
>
2
mol Al
There is not enough Al to use up all
the Cl2
Reactants must be in the mole ratio
2 Al + 3 Cl2 --->
---> Al2Cl6
28
2 Al + 3 Cl2 --->
---> Al2Cl6
2 Al + 3 Cl2 ---> Al2Cl6
Deciding on the Limiting
Reactant
Deciding on the Limiting
Reactant
27
Lim reag = Al
29
30
Step
Step 22 of
of LR
LR problem:
problem:
Calculate
Calculate moles
moles of
of each
each reactant
reactant
We have 5.40 g of Al and 8.10 g of Cl2
5.40 g Al •
8.10 g Cl 2 •
1 mol
= 0.200 mol Al
27.0 g
1 mol
= 0.114 mol Cl 2
70.9 g
Find
Find mole
mole ratio
ratio of
of reactants
reactants
31
Limiting
Limiting reactant
reactant == Cl
Cl22
Base
calcs
Base all
all calcs.
calcs.. on
on Cl
Cl22
= 0.57
grams
Cl2
This
would be 3/2, or 1.5/1, if
reactants are present in the
exact stoichiometric ratio.
Limiting reagent is
moles
Cl2
Cl2
CALCULATIONS:
CALCULATIONS: calculate
calculate mass
mass of
of
Al
Al22Cl
Cl66 expected.
expected.
32
2 Al + 3 Cl2 --->
---> Al2Cl6
2 Al + 3 Cl2 --->
---> Al2Cl6
mol Cl2
0.114 mol
=
mol Al
0.200 mol
Mix 5.40 g of Al with 8.10 g of Cl2.
What mass of Al2Cl6 can form?
grams
Al2Cl6
1 mol Al2 Cl6
3 mol Cl 2
moles
Al2Cl6
33
34
How
How much
much of
of which
which reactant
reactant will
will
remain
remain when
when reaction
reaction is
is complete?
complete?
Step 1: Calculate moles of Al2Cl6
expected based on LR.
0.114 mol Cl2 •
• Cl2 was the limiting reactant.
1 mol Al2 Cl6
= 0.0380 mol Al 2Cl6
3 mol Cl 2
• Therefore, Al was present
in excess. But how much?
Step 2: Calculate mass of Al2Cl6 expected
based on LR.
0.0380 mol Al 2Cl6 •
Calculating
Calculating Excess
Excess Al
Al
2 Al + 3 Cl2
products
0.200
0.200 mol
mol 0.114
0.114 mol
mol == LR
LR
0.114 mol Cl 2 •
• First find how much Al was required.
266.4 g Al 2Cl6
= 10.1 g Al 2Cl6
mol
2 mol Al
= 0.0760 mol Al req' d
3 mol Cl 2
• Then find how much Al is in excess.
35
Using Stoichiometry to
Determine a Formula
CxHy + some oxygen --->
--->
0.379 g CO2 + 0.1035 g H2O
First, recognize that all C in CO2 and all H in H2O
is from CxHy.
0.379 g CO2
+O2
Excess Al = Al available - Al required
= 0.200 mol - 0.0760 mol
= 0.124 mol Al in excess
1 CO2 molecule forms for
each C atom in CxHy
Puddle of CxHy
0.115 g
+O2
0.1035 g H2O
1 H2O molecule forms for
each 2 H atoms in CxHy
36
Using Stoichiometry to
Determine a Formula
37
Using Stoichiometry to
Determine a Formula
CxHy + some oxygen --->
--->
0.379 g CO2 + 0.1035 g H2O
CxHy + some oxygen --->
--->
0.379 g CO2 + 0.1035 g H2O
First, recognize that all C in CO2 and all H
in H2O is from CxHy.
Now find ratio of mol H/mol C to find values of
“x” and “y” in CxHy.
1. Calculate amount of C in CO2
1.149 x 10 -2 mol H/
H/ 8.61 x 10-3 mol C
8.61 x 10-3 mol CO2 -->
--> 8.61 x 10-3 mol C
2. Calculate amount of H in H2O
5.744 x 10-3 mol H2O -- >1.149 x 10-2 mol
H
= 1.33 mol H / 1.00 mol C
= 4 mol H / 3 mol C
Empirical formula = C3H4
38