John C. Kotz
Paul M. Treichel
John Townsend
http://academic.cengage.com/kotz
Chapter 4
Chemical Equations and Stoichiometry
John C. Kotz • State University of New York, College at Oneonta
2
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3
STOICHIOMETRY
- the study of the
quantitative
aspects of
chemical
reactions.
© 2009 Brooks/Cole - Cengage
STOICHIOMETRY
It rests on the principle of the conservation of matter.
2 Al(s) + 3 Br2(liq) f Al2Br6(s)
© 2009 Brooks/Cole - Cengage
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5
PROBLEM: If 454 g of NH4NO3 decomposes, how
much N2O and H2O are formed? What is the
theoretical yield of products?
STEP 1
Write the balanced
chemical equation
NH4NO3(s) f
N2O(g)
+ 2 H2O(g)
© 2009 Brooks/Cole - Cengage
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454 g of NH4NO3 f N2O + 2 H2O
STEP 2 Convert mass of reactant
(454 g) to amount (mol)
 1 mol 
454 g 
= 5.67 mol NH4NO3

 80.04 g 
STEP 3 Convert amount of reactant
(5.68 mol) to amount (mol) of product.
© 2009 Brooks/Cole - Cengage
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454 g of NH4NO3 f N2O + 2 H2O
STEP 3 Convert moles reactant f moles
product
Relate moles NH4NO3 to moles product
expected.
1 mol NH4NO3 f 2 mol H2O
Express this relation as the
STOICHIOMETRIC FACTOR
2 mol H2O produced
1 mol NH4NO3 used
© 2009 Brooks/Cole - Cengage
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454 g of NH4NO3 f N2O + 2 H2O
STEP 3 Convert moles reactant (5.67 mol) to
amount (mol) of product
5.67 mol NH4NO3
 2 mol H2O produced 
 1 mol NH NO used 


4
3
= 11.3 mol H2O produced
© 2009 Brooks/Cole - Cengage
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454 g of NH4NO3 f N2O + 2 H2O
STEP 4 Convert amount of product (11.4
mol) to mass of product
Called the THEORETICAL YIELD
 18.02 g 
11.3 mol H2O 
= 204 g H2O

 1 mol 
ALWAYS FOLLOW THESE STEPS IN
SOLVING STOICHIOMETRY PROBLEMS!
© 2009 Brooks/Cole - Cengage
GENERAL PLAN FOR
STOICHIOMETRY
CALCULATIONS
Mass
product
Mass
reactant
Moles
reactant
© 2009 Brooks/Cole - Cengage
Stoichiometric
factor
Moles
product
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454 g of NH4NO3 f N2O + 2 H2O
STEP 5 What mass of N2O is formed?
Total mass of reactants =
total mass of products
454 g NH4NO3 = ___ g N2O + 204 g H2O
mass of N2O = 250. g
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454 g of NH4NO3 f N2O + 2 H2O
Amounts Table (from page 159)
• Compound
• Initial (g)
• Initial (mol)
• Change (mol)
• Final (mol)
• Final (g)
© 2009 Brooks/Cole - Cengage
NH4NO3
N2O
H2O
13
454 g of NH4NO3 f N2O + 2 H2O
Amounts Table (from page 159)
• Compound
• Initial (g)
• Initial (mol)
• Change (mol)
NH4NO3
N2O
H2O
454 g
0
0
5.67 mol
0
0
-5.67
+5.67 +2(5.67)
• Final (mol)
0
5.67
11.3
• Final (g)
0
250
204
Note that matter is conserved!
© 2009 Brooks/Cole - Cengage
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454 g of NH4NO3 f N2O + 2 H2O
STEP 6 Calculate the percent yield
If you isolated only 131 g of N2O, what is
the percent yield?
This compares the theoretical (250. g)
and actual (131 g) yields.
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454 g of NH4NO3 f N2O + 2 H2O
STEP 6 Calculate the percent yield
actual yield
% yield =
x 100%
theoretical yield
 131 g 
% yield = 
x 100% = 52.4%

 250. g 
© 2009 Brooks/Cole - Cengage
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PROBLEM: Using 5.00 g of H2O2, what
mass of O2 and of H2O can be obtained?
2 H2O2(liq) f 2 H2O(g) + O2(g)
Reaction is catalyzed by MnO2
Step 1: amount (mol) of H2O2
Step 2: use STOICHIOMETRIC FACTOR
to calculate amount (mol) of O2
Step 3: mass of O2
© 2009 Brooks/Cole - Cengage
Reactions Involving a
LIMITING REACTANT
• In a given reaction, 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.
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LIMITING REACTANTS
Reactants
Products
2 NO(g) + O2(g) f 2 NO2(g)
Limiting reactant = ___________
Excess reactant = ____________
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LIMITING REACTANTS
PLAY MOVIE
See demonstration of limiting reactants in Chemistry Now
© 2009 Brooks/Cole - Cengage
LIMITING REACTANTS
React solid Zn with 0.100
mol HCl (aq)
Zn + 2 HCl f ZnCl2 + H2
PLAY MOVIE
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 and Zn)
Rxn 3: Balloon does not inflate fully, no Zn left.
* Not enough Zn to use up 0.100 mol HCl
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LIMITING REACTANTS
React solid Zn with 0.100 mol
HCl (aq)
Zn + 2 HCl f ZnCl2 + H2
0.10 mol HCl [1 mol Zn/2 mol HCl]
= 0.050 mol Zn
mass Zn (g)
mol Zn
mol HCl
mol HCl/mol Zn
Lim Reactant
© 2009 Brooks/Cole - Cengage
Rxn 1
7.00
0.107
0.100
0.93/1
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
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Reaction to be Studied
2 Al(s) + 3 Cl2(g) f Al2Cl6(s)
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PROBLEM: Mix 5.40 g of Al with 8.10 g of
Cl2. What mass of Al2Cl6 can form?
Mass
product
Mass
reactant
Moles
reactant
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Stoichiometric
factor
Moles
product
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Step 1 of LR problem:
compare actual mole ratio
of reactants to
theoretical mole ratio.
© 2009 Brooks/Cole - Cengage
Step 1 of LR problem:
compare actual mole ratio of
reactants to theoretical
mole ratio.
2 Al(s) + 3 Cl2(g) f Al2Cl6(s)
Reactants must be in the mole ratio
mol Cl2
3
=
mol Al
2
© 2009 Brooks/Cole - Cengage
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Deciding on the Limiting
Reactant
2 Al(s) + 3 Cl2(g) f Al2Cl6(s)
If
mol Cl2
3
>
mol Al
2
There is not enough Al to use up all the Cl2
Limiting reagent = Al
© 2009 Brooks/Cole - Cengage
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Deciding on the Limiting
Reactant
2 Al(s) + 3 Cl2(g) f Al2Cl6(s)
If
mol Cl2
3
<
mol Al
2
There is not enough Cl2 to use up all the Al
Limiting reagent = Cl2
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Step 2 of LR problem: Calculate
moles of each reactant
We have 5.40 g of Al and 8.10 g of Cl2
 1 mol 
5.40 g Al 
= 0.200 mol Al

 27.0 g 
 1 mol 
8.10 g Cl2 
= 0.114 mol Cl2

 70.9 g 
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Find mole ratio of reactants
2 Al(s) + 3 Cl2(g) f Al2Cl6(s)
mol Cl2
0.114 mol
=
= 0.57
mol Al
0.200 mol
Ratio should be 3/2 or 1.5/1 if reactants
are present in the exact stoichiometric
ratio.
Limiting reactant is
© 2009 Brooks/Cole - Cengage
Cl2
Mix 5.40 g of Al with 8.10 g of Cl2. What
mass of Al2Cl6 can form?
2 Al(s) + 3 Cl2(g) f Al2Cl6(s)
Limiting reactant = Cl2
Base all calcs. on Cl2
mass
Cl2
mass
Al2Cl6
1 mol Al2Cl6
moles
Cl2
© 2009 Brooks/Cole - Cengage
2 mol Cl2
moles
Al2Cl6
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CALCULATIONS: calculate mass
of Al2Cl6 expected.
Step 1: Calculate moles of Al2Cl6 expected
based on LR.
0.114 mol Cl2
 1 mol Al2Cl6 
 3 mol Cl  = 0.0380 mol Al2Cl6

2 
Step 2: Calculate mass of Al2Cl6 expected
based on LR.
0.0380 mol Al2Cl6
© 2009 Brooks/Cole - Cengage
 266.4 g Al2Cl6 
= 10.1 g Al2Cl6


mol


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How much of which reactant will
remain when reaction is complete?
• Cl2 was the limiting reactant.
Therefore, Al was present in
excess. But how much?
• First find how much Al was required.
• Then find how much Al is in excess.
© 2009 Brooks/Cole - Cengage
Calculating Excess Al
2 Al + 3 Cl2
0.200 mol
products
0.114 mol = LR
0.114 mol Cl2
 2 mol Al 
 3 mol Cl  = 0.0760 mol Al req'd

2
Excess Al = Al available - Al required
= 0.200 mol - 0.0760 mol
= 0.124 mol Al in excess
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Chemical Analysis
See Active Figure 4.6
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Chemical Analysis
• An impure sample of the mineral thenardite
contains Na2SO4.
• Mass of mineral sample = 0.123 g
• The Na2SO4 in the sample is converted to
insoluble BaSO4.
• The mass of BaSO4 is 0.177 g
• What is the mass percent of Na2SO4 in the
mineral?
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Chemical Analysis
• Na2SO4(aq) + BaCl2(aq)
f 2 NaCl(aq) + BaSO4(s)
• 0.177 g BaSO4 (1 mol/233.4 g)
= 7.58 x 10-4 mol BaSO4
• 7.58 x 10-4 mol BaSO4 (1 mol Na2SO4/1 mol BaSO4)
= 7.58 x 10-4 mol Na2SO4
• 7.58 x 10-4 mol Na2SO4 (142.0 g/1 mol)
= 0.108 g Na2SO4
• (0.108 g Na2SO4/0.123 g sample)100%
= 87.6% Na2SO4
© 2009 Brooks/Cole - Cengage
Determining the Formula of a
Hydrocarbon by Combustion
See Active Figure 4.7
PLAY MOVIE
© 2009 Brooks/Cole - Cengage
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Using Stoichiometry to
Determine a Formula
Burn 0.115 g of a hydrocarbon, CxHy, and
produce 0.379 g of CO2 and 0.1035 g of
H2O.
CxHy + some oxygen f
0.379 g CO2 + 0.1035 g H2O
What is the empirical formula of CxHy?
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Using Stoichiometry to
Determine a Formula
CxHy + some oxygen f
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
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
© 2009 Brooks/Cole - Cengage
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Using Stoichiometry to
Determine a Formula
CxHy + some oxygen f
0.379 g CO2 + 0.1035 g H2O
First, recognize that all C in CO2 and all H in H2O is
from CxHy.
1. Calculate amount of C in CO2
8.61 x 10-3 mol CO2 f 8.61 x 10-3 mol C
2. Calculate amount of H in H2O
5.744 x 10-3 mol H2O f 1.149 x 10-2 mol H
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Using Stoichiometry to
Determine a Formula
CxHy + some oxygen f
0.379 g CO2 + 0.1035 g H2O
Now find ratio of mol H/mol C to find values of x and y
in CxHy.
1.149 x 10 -2 mol H/ 8.61 x 10-3 mol C
= 1.33 mol H / 1.00 mol C
Multiply by 3 to get whole number coefficients.
Therefore, we have 4 mol H / 3 mol C
Empirical formula = C3H4
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Quantitative Aspects
of
Reactions in Solution
Sections 4.5-4.7
PLAY MOVIE
© 2009 Brooks/Cole - Cengage
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Terminology
In solution we need to define the
• SOLVENT
the component whose
physical state is
preserved when
solution forms
• SOLUTE
the other solution component
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Concentration of Solute
The amount of solute in a solution is
given by its concentration.
moles solute
Molarity(M) =
liters of solution
Concentration (M) = [ …]
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1.00 L of water
was used to
make 1.00 L of
solution. Notice
the water left
over.
© 2009 Brooks/Cole - Cengage
CCR, page 174
Preparing a Solution
See Active Figure 4.9
© 2009 Brooks/Cole - Cengage
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PROBLEM: Dissolve 5.00 g of NiCl2•6 H2O
in enough water to make 250. mL of
solution. Calculate molarity.
Step 1: Calculate moles of
NiCl2• 6H2O
 1 mol 
5.00 g 
= 0.0210 mol

 237.7 g 
Step 2: Calculate molarity
0.0210 mol
= 0.0841 M
0.250 L
[NiCl2·6 H2O] = 0.0841 M
PLAY MOVIE
© 2009 Brooks/Cole - Cengage
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The Nature of a CuCl2 Solution:
Ion Concentrations
CuCl2(aq) f
Cu2+(aq) + 2 Cl-(aq)
If [CuCl2] = 0.30 M, then
[Cu2+] = 0.30 M
[Cl-] = 2 x 0.30 M
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USING MOLARITY
What mass of oxalic acid, H2C2O4, is
required to make 250. mL of a
0.0500 M solution?
Because
Conc (M) = moles/volume = mol/V
this means that
moles = M•V
© 2009 Brooks/Cole - Cengage
USING MOLARITY
What mass of oxalic acid, H2C2O4, is
required to make 250. mL of a 0.0500 M
solution?
moles = M•V
Step 1: Calculate amount (mol) of acid
required.
(0.0500 mol/L)(0.250 L) = 0.0125 mol
Step 2: Calculate mass of acid required.
(0.0125 mol )(90.00 g/mol) =
© 2009 Brooks/Cole - Cengage
1.13 g
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51
Preparing Solutions
• Weigh out a solid
solute and dissolve in
a given quantity of
solvent.
• Dilute a concentrated
solution to give one
that is less
concentrated.
© 2009 Brooks/Cole - Cengage
Preparing a Solution by
Dilution
See Figure 4.10
© 2009 Brooks/Cole - Cengage
52
PROBLEM: You have 50.0 mL of
3.0 M NaOH and you want 0.50
M NaOH. What do you do?
Add water to the 3.0 M solution to
lower its concentration to 0.50 M
Dilute the solution!
© 2009 Brooks/Cole - Cengage
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PROBLEM: You have 50.0 mL of 3.0
M NaOH and you want 0.50 M
NaOH. What do you do?
But how much water
do we add?
© 2009 Brooks/Cole - Cengage
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PROBLEM: You have 50.0 mL of 3.0
M NaOH and you want 0.50 M
NaOH. What do you do?
How much water is added?
The important point is that
moles of NaOH in ORIGINAL solution =
moles of NaOH in FINAL solution
© 2009 Brooks/Cole - Cengage
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PROBLEM: You have 50.0 mL of 3.0 M NaOH
and you want 0.50 M NaOH. What do you do?
Amount of NaOH in original solution =
M•V =
(3.0 mol/L)(0.050 L) = 0.15 mol NaOH
Amount of NaOH in final solution must also =
0.15 mol NaOH
Volume of final solution =
(0.15 mol NaOH)(1 L/0.50 mol) = 0.30 L
or 300 mL
© 2009 Brooks/Cole - Cengage
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PROBLEM: You have 50.0 mL of 3.0 M
NaOH and you want 0.50 M NaOH. What
do you do?
Conclusion:
add 250 mL of
water to 50.0
mL of 3.0 M
NaOH to
make 300
mL of 0.50
M NaOH.
© 2009 Brooks/Cole - Cengage
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Preparing Solutions by
Dilution
A shortcut
Cinitial • Vinitial = Cfinal • Vfinal
© 2009 Brooks/Cole - Cengage
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pH, a Concentration Scale
59
pH: a way to express acidity
-- the concentration of H3O+ in solution.
Low pH: high [H3O+]
Acidic solution
Neutral
Basic solution
© 2009 Brooks/Cole - Cengage
High pH: low [H3O+]
pH < 7
pH = 7
pH > 7
The pH Scale
+
pH = log (1/ [H3O ])
= - log [H3O+]
In a neutral solution,
[H3O+] = [OH-] = 1.00 x 10-7 M at 25 oC
pH = - log [H3O+] = -log (1.00 x 10-7) = [0 + (-7)]
= 7.000
See book Appendix A.3 for more on logs
See GO CHEMISTRY module on pH
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[H3
+
O]
and pH
If the [H3O+] of soda is 1.6 x 10-3 M,
the pH is ____
Because pH = - log [H3O+]
then
pH= - log (1.6 x 10-3)
pH = -{log (1.6) + log (10-3)}
pH = -{0.20 - 3.00)
pH = 2.80
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pH and [H3
+
O]
If the pH of Coke is 3.12, it is __________.
Because pH = - log [H3O+] then
log [H3O+] = - pH
Take antilog and get
[H3O+] = 10-pH
[H3O+] = 10-3.12 =
7.6 x 10-4 M
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63
SOLUTION STOICHIOMETRY
Section 4.7
• Zinc reacts with acids to
produce H2 gas.
• Have 10.0 g of Zn
• What volume of 2.50 M
HCl is needed to convert
the Zn completely?
© 2009 Brooks/Cole - Cengage
GENERAL PLAN FOR
STOICHIOMETRY CALCULATIONS
Mass
HCl
Mass
zinc
Moles
zinc
Stoichiometric
factor
Moles
HCl
Volume
HCl
© 2009 Brooks/Cole - Cengage
64
Zinc reacts with acids to produce H2 gas. If you
have 10.0 g of Zn, what volume of 2.50 M HCl is
needed to convert the Zn completely?
Step 1: Write the balanced equation
Zn(s) + 2 HCl(aq) f ZnCl2(aq) + H2(g)
Step 2: Calculate amount of Zn
 1.00 mol Zn 
10.0 g Zn 
= 0.153 mol Zn

 65.39 g Zn 
Step 3: Use the stoichiometric factor
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65
Zinc reacts with acids to produce H2 gas. If you
have 10.0 g of Zn, what volume of 2.50 M HCl is
needed to convert the Zn completely?
Step 3: Use the stoichiometric factor
 2 mol HCl 
0.153 mol Zn 
= 0.306 mol HCl

 1 mol Zn 
Step 4: Calculate volume of HCl req’d
 1.00 L 
0.306 mol HCl 
= 0.122 L HCl

 2.50 mol 
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66
ACID-BASE REACTIONS
Titrations
H2C2O4(aq) + 2 NaOH(aq) f
acid
base
Na2C2O4(aq) + 2 H2O(liq)
Carry out this reaction using a TITRATION.
Oxalic acid,
H2C2O4
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Setup for titrating an acid with a base
See Active Figure 4.14
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69
Titration
PLAY MOVIE
© 2009 Brooks/Cole - Cengage
1. Add solution from the buret.
2. Reagent (base) reacts with
compound (acid) in solution
in the flask.
3. Indicator shows when exact
stoichiometric reaction has
occurred.
4. Net ionic equation
H3O+(aq) + OH-(aq)
f 2 H2O(liq)
5. At equivalence point
moles H3O+ = moles OH-
70
LAB PROBLEM #1: Standardize a
solution of NaOH — i.e., accurately
determine its concentration.
1.065 g of H2C2O4 (oxalic acid) requires
35.62 mL of NaOH for titration to an
equivalence point. What is the
concentration of the NaOH?
© 2009 Brooks/Cole - Cengage
1.065 g of H2C2O4 (oxalic acid) requires
35.62 mL of NaOH for titration to an
equivalence point. What is the concentration
of the NaOH?
Step 1: Calculate amount of H2C2O4
 1 mol 
1.065 g 
= 0.0118 mol

 90.04 g 
Step 2: Calculate amount of NaOH req’d
 2 mol NaOH 
0.0118 mol acid 
= 0.0237 mol NaOH

 1 mol acid 
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72
1.065 g of H2C2O4 (oxalic acid) requires 35.62 mL
of NaOH for titration to an equivalence point. What
is the concentration of the NaOH?
Step 1: Calculate amount of H2C2O4
= 0.0118 mol acid
Step 2: Calculate amount of NaOH req’d
= 0.0237 mol NaOH
Step 3: Calculate concentration of NaOH
0.0237 mol NaOH
= 0.664 M
0.03562 L
[NaOH] = 0.664 M
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LAB PROBLEM #2:
Use standardized NaOH to determine
the amount of an acid in an unknown.
Apples contain malic acid, C4H6O5.
C4H6O5(aq) + 2 NaOH(aq) f
Na2C4H4O5(aq) + 2 H2O(liq)
76.80 g of apple requires 34.56 mL of 0.664 M NaOH for
titration. What is weight % of malic acid?
© 2009 Brooks/Cole - Cengage
76.80 g of apple requires 34.56 mL of 0.664 M
NaOH for titration.
What is weight % of malic acid?
Step 1: Calculate amount of NaOH used.
C • V = (0.664 M)(0.03456 L)
= 0.0229 mol NaOH
Step 2: Calculate amount of acid titrated.
 1 mol acid 
0.0229 mol NaOH 
 2 mol NaOH 
= 0.0115 mol acid
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74
76.80 g of apple requires 34.56 mL of 0.664 M
NaOH for titration.
What is weight % of malic acid?
Step 1: Calculate amount of NaOH used.
= 0.0229 mol NaOH
Step 2: Calculate amount of acid titrated
= 0.0115 mol acid
Step 3: Calculate mass of acid titrated.
 134 g 
0.0115 mol acid 
= 1.54 g

 mol 
© 2009 Brooks/Cole - Cengage
75
76.80 g of apple requires 34.56 mL of 0.663 M
NaOH for titration.
What is weight % of malic acid?
Step 1: Calculate amount of NaOH used.
= 0.0229 mol NaOH
Step 2: Calculate amount of acid titrated
= 0.0115 mol acid
Step 3: Calculate mass of acid titrated.
= 1.54 g acid
Step 4: Calculate % malic acid.
1.54 g
x 100% = 2.01%
76.80 g
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77
Spectrophotometry
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78
An Absorption Spectrophotometer
See Figure 4.16
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79
Spectrophotometry
• Amount of light absorbed by a sample depends on path
length and solute concentration.
Different concs of Cu2+
Concentration
© 2009 Brooks/Cole - Cengage
Same concs but
different path lengths
Path length
80
Spectrophotometry
• BEER-LAMBERT LAW relates amount of light absorbed
and the path length and solute concentration.
A = absorbance
 = molar absorptivity
l = path length
c = concentration
• There is a linear relation between A and c for
a given path length and compound.
• This means you can find unknown solution
concentration if A is measured.
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81
Spectrophotometry
• To use the Beer-Lambert law you must first calibrate the
instrument.
The calibration plot
can be used to find
the unknown conc
of a solution from a
measured A.
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