Stoichiometric Calculations
1
Review of Fundamental
Concepts
Formula Weight
It is assumed that you can calculate the
formula or molecular weights of compounds
from respective atomic weights of the
elements forming these compounds. The
formula weight (FW) of a substance is the
sum of the atomic weights of the elements
from which this substance is formed from.
2
Find the formula weight of
CaSO4.7H2O
Element
Ca
S
11 O
14 H
FWt
3
Atomic weight
40.08
32.06
11x16.00
14x1.00

The Mole Concept
The mole is the major word we will use
throughout the course. The mole is defined as
gram molecular weight which means that:
Mole
1 mol H2
1 mol O2
1mol O
1mol NaCl
1 mol Na2CO3
4
Grams
2.00 g
32.00 g
16.00 g
58.5 g
106.00 g
Assuming approximate atomic weights of 1.00,
16.00, 23.00, 35.5, and 12.00 atomic mass
units for hydrogen, oxygen, sodium, chlorine
atom, and carbon, respectively.
The number of moles contained in a specific
mass of a substance can be calculated as:
mol = g substance/FW substance
The unit for the formula weight is g/mol
5
In the same manner, the number of mmol
of a substance contained in a specific
weight of the substance can be
calculated as
mmol = mol/1000
Or,
mmol = mg substance/FW substance
6
The number of mmol of Na2WO4 (FW
= 293.8 mg/mmol) present in 500
mg of Na2WO4 can be calculated as
? mmol of Na2WO4 = 500 mg/293.8
(mg/mmol) = 1.70 mmol
7
The number of mg contained in 0.25
mmol of Fe2O3 (FW = 159.7 mg/mmol)
can be calculated as
? mg Fe2O3 = 0.25 mmol Fe2O3 x 159.7
(mg/mmol) = 39.9 mg
Therefore, either the number of mg of a
substance can be obtained from its
mmols or vice versa.
8
Calculations Involving Solutions
Molarity
Molarity of a solution can be defined as the
number of moles of solute dissolved in 1 L of
solution. This means that 1 mol of solute will
be dissolved in some amount of water and
the volume will be adjusted to 1 L. The
amount of water may be less than 1 L as the
final volume of solute and water is exactly 1
L.
9
Calculations Involving Molarity
Molarity = mol/L = mmol/mL
This can be further formulated as
Number of moles = Molarity X volume in Liters,
or
mol = M (mol/L) x VL
Number of mmol = Molarity x volume in mL , or
mmol = M (mmol/mL) x VmL
10
Find the molarity of a solution resulting from
dissolving 1.26 g of AgNO3 (FW = 169.9
g/mol) in a total volume of 250 mL solution.
First find mmol AgNO3 = 1.26*103 mg AgNO3 /
169.9 mg/mmol = 7.42 mmol
Molarity = mmol/mL
M = 7.42 mmol/250 mL = 0.0297 mmol/mL
11
We can find the molarity directly in one
step using dimensional analysis
? mol AgNO3 / L = (1.26 g AgNO3 / 250
mL) x ( mol AgNO3/169.9 g AgNO3) x
(1000 mL/1L) =0.0297 M
12
Let us find the number of mg of NaCl per mL of
a 0.25 M NaCl solution
First we should be able to recognize the
molarity as 0.25 mol/L or 0.25 mmol/mL. Of
course, the second term offers what we need
directly
? mg NaCl in 1 mL = (0.25 mmol NaCl/mL) x
(58.5 mg NaCl/mmol NaCl) = 14.6 mg NaCl/mL
13
Find the number of grams of Na2SO4 required to
prepare 500 mL of 0.1 M solution.
First, we find mmoles needed from the relation
mmol = M (mmol/mL) x VmL
mmol Na2SO4 = 0.1 mmol/mL x 500 mL = 50 mmol
mmol = mg substance/FW substance
?mg Na2SO4 = 50 mmol x 142 mg/mmol = 7100 mg or
7.1 g
14
One can use dimensional analysis to find
the answer in one step as
? g Na2SO4 = (0.1 mol Na2SO4/1000 mL) x
500 mL x (142 g Na2SO4/mol) = 7.1 g
•
15
When two or more solutions are mixed,
one can find the final concentration of
each ion. However, you should always
remember that the number of moles ( or
mmoles ) is additive. For example:
Find the molarity of K+ after mixing
100 mL of 0.25 M KCl with 200 mL
of 0.1 M K2SO4.
16
The idea here is to calculate the total mmol
of K+ and divide it by volume in mL.
mmol K+ = mmol K+ from KCl + mmol K+
from K2SO4
= 0.25 mmol/ml x 100 mL + 2x0.1
mmol/mL x 200 mL = 65 mmol
Molarity = 65 mmol/(200 + 100) mL = 0.22 M
Note that the concentration of K+ in
0.1M K2SO4 is 2x0.1M (i.e. 0.2 M)
17
Lecture 9
Stoichiometric Calculations
Normality
Density Calculations
18
Normality
We have previously talked about molarity as a
method for expressing concentration. The
second expression used to describe
concentration of a solution is the normality.
Normality can be defined as the number of
equivalents of solute dissolved in 1 L of
solution. Therefore, it is important for us to
define what we mean by the number of
equivalents, as well as the equivalent weight
of a substance as a parallel term to formula
weight.
19
An equivalent is defined as the weight of
substance giving an Avogadro’s
number of reacting units. Reacting
units are either protons or hydroxides
(in acid base reactions) or electrons (in
oxidation reduction reactions). For
example, HCl has one reacting unit (H+)
when reacting with a base like NaOH
but sulfuric acid has two reacting units
(two protons) when reacting completely
with a base.
20
Therefore, we say that the equivalent
weight of HCl is equal to its formula
weight and the equivalent weight of
sulfuric acid is one half its formula
weight. In the reaction where Mn(VII), in
KMnO4, is reduced to Mn(II) five
electrons are involved and the
equivalent weight of KMnO4 is equal to
its formula weight divided by 5.
21
N = eq/L or N = meq/mL
Number of eq = Normality x VL = (eq/L) x L
Number of meq = Normality x VmL
= (meq/mL) x mL
Also, number of equivalents = wt(g)/equivalent
weight (g/eq)
meq = mg/eqw
22
Equivalent weight = FW/n
meq = mg/eqw substitute for eqw = FW/n gives:
meq = mg/(FW/n), but mmol = mg/FW, therefore:
meq = n * mmol
dividing both sides by volume in mL, we get:
N=nM
Where n is the number of reacting units ( protons,
hydroxides, or electrons ) and if you are forming
factors always remember that a mole contains n
equivalents. The factor becomes (1 mol/n eq) or (n
eq/1 mol).
23
Find the equivalent weights of NH3 (FW =
17.03), H2C2O4 (FW = 90.04) in the reaction
with excess NaOH, and KMnO4 (FW = 158.04)
when Mn(VII) is reduced to Mn(II).
Solution
Ammonia reacts with one proton only
Equivalent weights of NH3 = FW/1 = 17.03 g/eq
24
Two protons of oxalic acid react with the base
Equivalent weights of H2C2O4 = FW/2 =
90.04/2 = 45.02 g/eq
Five electrons are involved
reduction of Mn(VII) to Mn(II)
in
the
Equivalent weights of KMnO4 = FW/5 =
158.04/5 = 31.608 g/eq
25
Find the normality of the solution containing
5.300 g/L of Na2CO3 (FW = 105.99), carbonate
reacts with two protons.
Normality is the number of equivalents per
liter, therefore we first find the number of
equivalents
eq wt = FW/2 = 105.99/2 = 53.00
eq = Wt/eq wt = 5.300/53.00 = 0.1000
N = eq/L = 0.1000 eq/1L = 0.1000 N
26
The problem can be worked out simply
as below
? eq Na2CO3 /L = (5.300 g Na2CO3 /L ) x (1
mol Na2CO3 /105.99 g Na2CO3 ) x (2 eq
Na2CO3 /1 mol Na2CO3) = 0.1 N
A further option is to find the number of
moles first followed by multiplying the
result by 2 to obtain the number of
equivalents.
27
The other choice is to find the molarity
first and the convert it to normality
using the relation:
N=nM
No of mol = 5.300 g/(105.99 g/mol)
M = mol/L = [5.300 g/(105.99 g/mol)]/ 1L
N = n M = 2 x [5.300 g/(105.99 g/mol)]/ 1L
= 0.1000
28
Find the normality of the solution containing
5.267 g/L K2Cr2O7 (FW = 294.19) if Cr6+ is
reduced to Cr3+.
The same as the previous example
N = eq/L, therefore we should find the number
of eq where eq = wt/eq wt, therefore we
should find the equivalent weight; where eq
wt = FW/n. Here; each Contributes three
electrons and since the dichromate contains
two Cr atoms we have 6 reacting units
29
Eq wt = (294.19 g/mol)/(6 eq/mol)
Eq = 5.267 g/ (294.19 g/mol)/(6 eq/mol)
N = eq/L = (294.19 g/mol)/(6 eq/mol)/1L = 0.1074
eq/L
Using the dimensional analysis we may write
? eq K2Cr2O7 /L = (5.267 g K2Cr2O7 /L) x (mol
K2Cr2O7 /294.19 g K2Cr2O7 ) x (6 eq K2Cr2O7
/mol K2Cr2O7 ) = 0.1074 eq/L
30
Again one can choose to calculate the
molarity then convert it to normality
mol = 5.267 g/(294.19 g/mol)
M = mol/L = [5.267 g/(294.19 g/mol)]/L
N=nM
N = (6 eq/mol)x [5.267 g/(294.19 g/mol)]/L
= 0.1074 eq/L
31
Density Calculations
In this section, you will learn how to find
the molarity of solution from two pieces
of information (density and
percentage). Usually the calculation is
simple and can be done using several
procedures. Look at the examples
below:
32
Example
What volume of concentrated HCl (FW =
36.5g/mol, 32%, density = 1.1g/mL) are
required to prepare 500 mL of 2.0 M
solution.
Always start with the density and find
how many grams of solute in each mL
of solution.
Density = g solution/mL
33
Remember that only a percentage of the
solution is solute .
mg HCl/ml = 1.1 x 0.32 x103 mg HCl / mL
The problem is now simple as it requires
conversion of mg HCl to mmol since the
molarity is mmol per mL
M = mmol HCl/mL = 1.1x0.32 x103 mg HCl/(36.5
mg/mmol) = 9.64 M
34
Now, we can calculate the volume required
from the relation
MiVi (before dilution) = MfVf (after dilution)
9.64 x VmL= 2.0 x 500mL
VmL = 10.4 mL
This means that 10.4 mL of the concentrated
HCl should be added to distilled water and
the volume should then be adjusted to 500
mL
35
How many mL of concentrated H2SO4
(FW = 98.1 g/mol, 94%, d = 1.831 g/mL)
are required to prepare 1 L of 0.100 M
solution?
mg H2SO4 / mL = 1.831*0.94*103 mg /mL
Now we only need to convert mg to mmol
M = mmol/mL = [(1.831 x 0.94 x 103 mg) /
(98.1 mg/mmol)] / mL = 17.5 M
36
To find the volume required to prepare the
solution
MiVi (before dilution) = MfVf (after dilution)
17.5 x VmL = 0.100 x 1000 mL
VmL = 5.71 mL which should be added to
distilled water and then adjusted to 1 L.
37
An Easy Short-Cut
M=
Density * percentage * 103
Formula Weight
The percentage is a fraction: (i.e. a 35%
is written as 0.35)
38
Analytical Versus Equilibrium
Concentration
When we prepare a solution by weighing
a specific amount of solute and
dissolve it in a specific volume of
solution, we get a solution with specific
concentration. This concentration is
referred to as analytical concentration.
However, the concentration in solution
may be different from the analytical
concentration, especially when partially
dissociating substances are used.
39
An example would be clear if we consider
preparing 0.1 M acetic acid (weak acid) by
dissolving 0.1 mol of the acid in 1 L solution.
Now, we have an analytical concentration of
acetic acid (HOAc) equals 0.1 M. But what is
the actual equilibrium concentration of
HOAc?
We have
HOAc = H+ + OAcThe analytical concentration ( CHOAc ) = 0.1 M
CHOAc = [HOAc]undissociated + [OAc-]
The equilibrium concentration = [HOAc]undissociated.
40
For good electrolytes which are 100%
dissociated in water the analytical and
equilibrium concentrations can be calculated
for the ions, rather than the whole species.
For example, a 1.0 M CaCl2 in water results in 0
M CaCl2, 1.0 M Ca2+, and 2.0 M Cl- since all
calcium chloride dissociates in solution.
For species x we express the analytical concentration
as Cx and the equilibrium concentration as [x].
41
Lecture 10
Stoichiometric Calculations, cont…..
Expressing Concentrations
42
Dilution Problems
In many cases, a dilution step or steps are
involved in analytical procedures. One
should always remember that in any dilution
the number of mmoles of the initial
(concentrated) solution is equal to the
number of mmoles of the diluted solution.
This means:
MiVi (concentrated) = MfVf (dilute)
43
Prepare 200 mL of 0.12 M KNO3 solution from
0.48 M solution.
MiVi (concentrated) = MfVf (dilute)
0.48 x VmL = 0.12 x 200
VmL = (0.12 x 200)/0.48 = 50 mL
Therefore, 50 mL of 0.48 M KNO3 should be
diluted to 200 mL to obtain 0.12 M solution
44
A 5.0 g Mn sample was dissolved in 100 mL
water. If the percentage of Mn (At wt = 55
g/mol) in the sample is about 5%. What
volume is needed to prepare 100 mL of
approximately 3.0x10-3 M solution.
First we find approximate mol Mn in the sample
= 5.0 x (5/100) g Mn/(55 g/mol) = 4.5x10-3 mol
Molarity of Mn solution = (4.5x10-3 x 103
mmol)/100 mL = 4.5x10-2 M
45
The problem can now be solved easily using
the dilution relation
MiVi (concentrated) = MfVf (dilute)
4.5x10-2 x VmL = 3.0x10-3 x 100
VmL = (3.0x10-3 x 100)/4.5x10-2 = 6.7 mL
Therefore, about 6.7 mL of the Mn sample
should be diluted to obtain an approximate
concentration of 3.0x10-3 M solution.
46
What volume of 0.4 M Ba(OH)2 should be
added to 50 mL of 0.30 M NaOH in order to
obtain a solution that is 0.5 M in OH-.
We have to be able to see that the mmol OHcoming from Ba(OH)2 and NaOH will equal
the number of mmol of OH- in the final
solution, which is
mmol OH- from Ba(OH)2 + mmol OH- from NaOH
= mmol OH- in final solution
47
The mmol OH- from Ba(OH)2 is molarity
of OH- times volume and so are other
terms. Molarity of OH- from Ba(OH)2 is
0.8 M (twice the concentration of
Ba(OH)2, and its volume is x mL. Now
performing the substitution we get
0.8 * x + 0.30 * 50 = 0.5 * (x + 50)
x = 33 mL
48
Expressing Concentrations
Solid solutes in solid
samples
Solid solutes in
solutions
Liquid solutes in
solutions
Part per hundred, %
Part per thousand, ppt
Part per million, ppm
Part per billion, ppb
49
For Solid Solutes in solid samples
% (w/w) = [weight solute (g)/weight sample (g)] x 100
ppt (w/w) = [weight solute (g)/weight sample (g)] x 1000
ppm (w/w) = [weight solute (g)/weight sample (g)] x 106
ppb (w/w) = [weight solute (g)/weight sample (g)] x 109
A ppm can be represented by several terms like the
one above, (mg solute/kg sample), ( g solute/106g
sample), etc..
50
If the solute is dissolved in solution we have
% (w/v) = [weight solute (g)/volume sample (mL)] x 100
ppt (w/v) = [weight solute (g)/volume sample (mL)] x
1000
ppm (w/v) = [weight solute (g)/volume sample (mL)] x
106
ppb (w/v) = [weight solute (g)/volume sample (mL)] x
109
Also a ppm can be expressed as above or as (g
solute/106 mL solution), (mg solute/L solution), or
(mg/mL), etc..
51
For Liquid Solutes
% (v/v) = [volume solute (mL)/volume sample (mL)] x
100
ppt (v/v) = [volume solute (mL)/volume sample (mL)] x
1000
ppm (v/v) = [volume solute (mL)/volume sample (mL)] x
106
ppb (v/v) = [volume solute (mL)/volume sample (mL)] x
109
A ppm can be expressed as above or as (mL/L), (mL/103
L), etc..
52
A 2.6 g sample was analyzed and found to
contain 3.6 mg zinc. Find the concentration of
zinc in ppm and ppb.
A ppm is microgram solute per gram sample,
therefore
ppm Zn = 3.6 mg Zn/2.6 g sample = 1.4 ppm
A
ppb is
therefore
nanogram
solute/gram
sample,
ppb Zn = 3.6 x103 ng Zn/2.6 g sample = 1400
ppb
53
A 25.0 mL sample was found to contain 26.7 mg
glucose. Express the concentration as ppm
and mg/dL glucose.
Solution
A ppm is defined as mg/mL, therefore
ppm = 26.7 mg/(25.0x10-3 mL) = 1.07x103 ppm
54
Or one can use dimensional analysis
considering always a ppm as mg/L as below
? mg/L glucose = (26.7 mg/25.0 mL) x (10-3
mg/mg) x (106 mL/L) = 1.07x103 ppm
Now let us find mg glucose per deciliter
?mg glucose/dL = = (26.7 mg/25.0 mL) x (10-3
mg/mg) x (106 mL/L) x (L/10dL) = 107 mg/dL
55
Find the molar concentration of a 1.00 ppm Li
(at wt = 6.94 g/mol) and Pb (at wt = 207
g/mol).
Solution
A 1.00 ppm is 1.00 mg/L, therefore change this
1.00 mg into mmol to obtain molarity.
? mmol Li/mL = (1.00 mg Li/103 mL) x ( 1 mmol
Li/6.94 mg Li) = 1.44x10-4 M
? mmol Pb/mL = (1.00 mg Pb/103 mL) x ( 1
mmol Pb/207 mg Pb) = 4.83x10-6 M
56
Find the number of mg Na2CO3 (FW = 106
g/mol) required to prepare 500 mL of 9.20
ppm Na solution.
The idea is to find mg sodium ( 23.0 mg/mmol)
required and then get the mmoles sodium
and relate it to mmoles sodium carbonate
followed by calculation of the weight of
sodium carbonate.
? mg Na = 9.20 mg/L x 0.5 L = 4.6 mg Na
mmol Na = 4.60 mg Na/23.0 mg/mmol
Na2CO3 = 2Na+
57
mmol Na2CO3 = 1/2 * mmol Na = 4.60/46.0 mmol
= 0.100
? mg Na2CO3 = 0.100 mmol Na2CO3 x (106 mg
Na2CO3/ mmol Na2CO3) = 10.6 mg
One can work such a problem in one step as
below
? mg Na2CO3 = (9.2 mg Na/1000mL) x 500 mL x
(1mmol Na/23.0 mg Na) x (1 mmol Na2CO3/2
mmol Na) x (106 mg Na2CO3/1 mmol Na2CO3)
= 10.6 mg
58
Lecture 11
Stoichiometric Calculations, cont…..
Volumetric Analysis
59
Stoichiometric Calculations:
Volumetric Analysis
In this section we look at calculations involved
in titration processes as well as general
quantitative reactions. In a volumetric
titration, an analyte of unknown
concentration is titrated with a standard in
presence of a suitable indicator. For a
reaction to be used in titration the following
characteristics should be satisfied:
60
1. The stoichiometry of the reaction
should be exactly known. This means
that we should know the number of
moles of A reacting with 1 mole of B.
2. The reaction should be rapid and
reaction between A and B should
occur immediately and instantly after
addition of each drop of titrant (the
solution in the burette).
3. There should be no side reactions. A
reacts with B only.
61
4. The reaction should be quantitative. A
reacts completely with B.
5. There should exist a suitable indicator
which has distinct color change.
6. There should be very good agreement
between the equivalence point
(theoretical) and the end point
(experimental). This means that Both
points should occur at the same
volume of titrant or at most a very
close volume.
62
Standard solutions
A standard solution is a solution of known and
exactly defined concentration. Usually
standards are classified as either primary
standards or secondary standards. There are
not too many secondary standards available
to analysts and standardization of other
substances is necessary to prepare
secondary standards. A primary standard
should have the following properties:
63
1. Should have a purity of at least
99.98%
2. Stable to drying, a necessary step to
expel adsorbed water molecules
before weighing
3. Should have high formula weight as
the uncertainty in weight is decreased
when weight is increased
4. Should be non hygroscopic
5. Should possess the same properties
as that required for a titration
64
Remember!!
NaOH and HCl are not primary standards
and therefore should be standardized
using a primary or secondary standard.
NaOH absorbs CO2 from air, highly
hygroscopic, and usually of low purity.
HCl and other acid in solution are not
standards as the percentage written on
the reagent bottle is a claimed value
and should not be taken as guaranteed.
65
Molarity Volumetric Calculations
Volumetric calculations involving molarity are
rather simple. The way this information is
presented in the text is not very helpful.
Therefore, disregard and forget about all
equations and relations listed in rectangles
in the text, you will not need it. What you
really need is to use the stoichiometry of the
reaction to find how many mmol of A as
compared to the number of mmoles of B.
66
Examples
A 0.4671 g sample containing NaHCO3 (FW =
84.01 mg/mmol) was dissolved and titrated
with 0.1067 M HCl requiring 40.72 mL. Find
the percentage of bicarbonate in the sample.
We should write the equation in order to
identify the stoichiometry
NaHCO3 + HCl g NaCl + H2CO3
67
Now it is clear that the number of mmol of bicarbonate
is equal to the number of mmol HCl
mmol NaHCO3 = mmol HCl
mmol = M x VmL
mmol NaHCO3 = (0.1067 mmol/ml ) x 40.72 mL = 4.345
mmol
Now get mg bicarbonate by multiplying mmol times FW
mg NaHCO3 = 4.345 mmol x (84.01 mg/mmol) = 365.01
% NaHCO3 = (365.01 x 10-3 g/0.4671 g) x 100 = 78.14%
68
We can use dimensional analysis to
calculate the mg NaHCO3 directly then get
the percentage as above.
? mg NaHCO3 = (0.1067 mmol HCl/ml) x
40.72 mL x (mmol NaHCO3/mmol HCl) x
(84.01 mg NaHCO3/ mmol NaHCO3) =
365.0 mg
69
A 0.4671 g sample containing Na2CO3 (FW =
106mg/mmol) was dissolved and titrated with
0.1067 M HCl requiring 40.72 mL. Find the
percentage of carbonate in the sample.
The equation should be the first thing to formulate
Na2CO3 +2 HCl g 2NaCl + H2CO3
mmol Na2CO3 = ½ mmol HCl
(mg Na2CO3/FW) = ½ x ( MHCl x VmL (HCl) )
70
(mg Na2CO3/FW) = ½ x ( MHCl x VmL (HCl) )
(mg Na2CO3/106) = ½ x 0.1067 x 40.72
mg Na2CO3 = 106 * ½ x 0.1067 x 40.72 = 230
% Na2CO3 = (230 x 10-3 g/0.4671 g ) x 100
= 49.3 %
71
How many mL of 0.25 M NaOH will react with
5.0 mL of 0.10 M H2SO4.
H2SO4 + 2 NaOH g Na2SO4 + 2 H2O
mmol NaOH = 2 mmol H2SO4
MNaOH x VmL(NaOH) = 2 {M(H2SO4) x VmL(H2SO4)}
0.25 x VmL = 2 x 0.10 x 5.0
VmL = 4.0 mL
72
We can also calculate the volume in one
step using dimensional analysis:
? mL NaOH = (mL NaOH/0.25 mmol NaOH) x (2
mmol NaOH/mmol H2SO4) x (0.10 mmol H2SO4
/ mL H2SO4) x 5.0 mL H2SO4 = 4.0 mL
73
A 0.1876 g of pure sodium carbonate (FW = 106
mg/mmol) was titrated with approximately 0.1
M HCl requiring 35.86 mL. Find the molarity
of HCl.
The first thing to do is to write the equation for
the reaction. You should remember that
carbonate reacts with two protons
Na2CO3 + 2 HCl g 2 NaCl + H2CO3
74
The second step is to relate the number of
mmol HCl to mmol carbonate, Where it is
clear from the equation that we have 2 mmol
HCl and 1 mmol carbonate. This is translated
to the following
mmol HCl = 2 mmol Na2CO3
Now let us substitute for mmol HCl by MHCl X
VmL, and substitute for mmol carbonate by
mg carbonate/FW carbonate. This gives
75
MHCl x 35.86 = 2 * 187.6 mg/ (106 mg/mmol)
MHCl = 0.09872 M
The same result can be obtained using
dimensional analysis in one single step:
? mmol HCl/mL = (187.6 mg Na2CO3 /35.86 mL
HCl) x ( mmol Na2CO3/106 mg Na2CO3) x (2
mmol HCl/mmol Na2CO3) = 0.09872 M
76
An acidified and reduced iron sample required
40.2 mL of 0.0206 M KMnO4. Find mg Fe (at
wt = 55.8) and mg Fe2O3 (FW = 159.7
mg/mmol).
The first step is to write the chemical equation
MnO4- + 5 Fe2+ + 8 H+ g Mn2+ + 5 Fe3+ + 4 H2O
mmol Fe = 5 mmol KMnO4
Now substitute for mmol Fe by mg Fe/at wt Fe
and substitute for mmol KMnO4 by molarity
of permanganate times volume, we then get:
77
[mg Fe/(55.8 mg/mmol)] = 5 x (0.0206
mmol/mL) x 40.2 mL
mg Fe = 231 mg
This can also be done in a single step as
follows:
? mg Fe = (0.0206 mmol KMnO4 /mL) x
40.2 mL x (5 mmol Fe/mmol KMnO4) x
(55.8 mg Fe/mmol Fe) = 231 mg
78
To calculate the mg Fe2O3 we set the following:
2Fe g Fe2O3
mmol Fe = 2 mmol Fe2O3
Substitute for mmol Fe in equation 1
2* [mg Fe2O3/ (159.7 mg/mmol)] = 5 x (0.0206 mmol/mL)
x 40.2 mL
mg Fe2O3 = 331 mg
The last step can also be done using dimensional
analysis as follows:
? mg Fe2O3 = (0.0206 mmol KMnO4 /mL) x 40.2 mL x (5
mmol Fe/mmol KMnO4) x (mmol Fe2O3 /2 mmol Fe) x (
159.7 mg Fe2O3/mmol Fe2O3) = 331 mg
79
A 1.00 g Al sample required 20.5 mL EDTA.
Find the % Al2O3 (FW = 101.96) in the sample
if 30.0 mL EDTA required 25.0 mL of 0.100 M
CaCl2.
We should know that EDTA reacts in a 1:1 ratio
with metal ions
mmol EDTA = mmol CaCl2
Molarity x VmL(EDTA) = Molarity x VmL (CaCl2)
MEDTA x 30.0 = 0.100 x 25.0
MEDTA = 0.0833 M
80
Al3+ + EDTA g Al-EDTA
mmol Al = mmol EDTA
However, 2Al g Al2O3
(1)
mmol Al = 2 mmol Al2O3 , substitute in equation (1)
2*mmol Al2O3 = mmol EDTA
Now we can solve the problem using relation (1)
2*[mg Al2O3/ ( 101.96 mg/mmol)] = 0.0833 x 20.5
? mg Al2O3 = 87.1 mg
81
% Al2O3 = (87.1 mg/1000 mg) x 100 = 8.71%
Another approach:
After calculation of the molarity of EDTA, one
can do the rest of the calculation in a single
step as follows:
? mg Al2O3 = (0.0833 mmol EDTA/mL) x 20.5
mL x(mmol Al/mmol EDTA) x (mmol
Al2O3/2mmol Al) x (101.96 mg Al2O3/mmol
Al2O3) = 87.1 mg
82
Lecture 12
Stoichiometric Calculations, Cont…..
Back Titrations
We have seen in previous sections that correct
solution of any problem involving reactions
between two substances requires setting up two
important relations:
1.
2.
3.
Writing a balanced chemical equation representing
stoichiometric relationships.
Formulating a relationship between the number of
mmol of substance A and mmol of substance B.
The last step in the calculation is to substitute for
the mmol A or B by either one of the following
according to given information:
mmol = M x VmL
mmol = mg/FW
84
Find the volume of 0.100 M KMnO4 that will
react with 50.0 mL of 0.200 M H2O2 according
to the following equation:
5H2O2 + 2KMnO4 + 6H+ g 2Mn2+ + 5O2 + 8H2O
Solution
We have the equation ready therefore the
following step is to formulate the relationship
between the number of moles of the two
reactants. We always start with the one we
want to calculate, that is
85
mmol KMnO4 = 2/5 mmol H2O2
It is clear that we should substitute M x
VmL for mmol in both substances as
this information is given to us.
0.100 x VmL = (2/5) x 0.200 x 50.0
VmL = 40.0 mL KMnO4
86
Find the volume of 0.100 M KMnO4 that will react
with 50.0 mL of 0.200 M MnSO4 according to the
following equation:
3MnSO4 + 2KMnO4 + 4OH- g 5MnO2 + 2H2O + 2K+ + 3SO42-
Again, we have the equation ready therefore we
turn to formulating a relationship for the
number of mmol for both substances.
mmol KMnO4 = 2/3 mmol MnSO4
87
mmol KMnO4 = 2/3 mmol MnSO4
It is clear that we should substitute M x
VmL for mmol in both substances as
this information is given to us.
0.100 x VmL = 2/3 x 0.200 x 50.0
VmL = 66.7 mL KMnO4
88
Find the volume of 0.100 M KMnO4 that will
react with 500 mg of H2C2O4 (FW =
90.0mg/mmol) according to the following
equation:
5 H2C2O4 + 2 KMnO4 + 6 H+ g 10 CO2 + 2 Mn2++ 8 H2O + 2 K+
Solution
We have the equation ready therefore we turn
to formulating a relationship for the number
of mmol for both substances.
mmol KMnO4 = 2/5 mmol H2C2O4
89
mmol KMnO4 = 2/5 mmol H2C2O4
It is clear that we should substitute M x
VmL for mmol permanganate while
substitute for mmol oxalic acid by
mg/FW. This gives:
0.100 x VmL = 2/5 x 500 mg / (90.0 mg/mmol)
VmL = 22.2 mL KMnO4
90
Back-Titrations
In this technique, an accurately known amount
of a reagent is added to analyte in such a
way that some excess of the added reagent
is left. This excess is then titrated to
determine its amount and thus:
mmol reagent taken = mmol reagent reacted
with analyte + mmol reagent titrated
Therefore, the analyte can be determined
since we know mmol reagent added and
mmol reagent titrated.
91
mmol reagent reacted = mmol reagent
taken – mmol reagent titrated
Finally, the number of mmol reagent
reacted can be related to the number of
mmol analyte from the stoichiometry of
the reaction between the two
substances.
92
Why Do We Use Back-Titrations?
Back-titrations are important especially in
some situations like:
1. When the titration reaction is slow. Addition
of an excess reagent will force the reaction
to proceed faster.
2. When the titration reaction lacks a good
indicator. We will see details of this later.
3. When the analyte is not very stable.
Addition of excess reagent will finish the
analyte instantly thus overcoming stability
problems.
94
Examples
A 2.63 g Cr(III) sample was dissolved and analyzed by
addition of 5.00 mL of 0.0103 M EDTA. The excess
EDTA required 1.32 mL of 0.0112 M Zn(II). Calculate
% CrCl3 (FW=158.4 mg/mmol) in the sample.
Solution
We should remember that EDTA reacts in a 1:1 ratio.
The first step in the calculation is to find the mmol
EDTA reacted from the relation:
mmol EDTA reacted = mmol EDTA taken – mmol EDTA
titrated
95
We substitute for both mmol EDTA taken and
mmol EDTA titrated by M x VmL for each. Also
since EDTA reacts in a 1:1 ratio, we can state the
following:
mmol EDTA reacted = mmol CrCl3.
mmol EDTA titrated = mmol Zn(II).
Therefore, we now have the following reformulated
relation:
mmol CrCl3 = mmol EDTA taken – mmol Zn(II)
Substitution gives:
96
mmol CrCl3 = 0.0103 x 5.00 – 0.0112 x 1.32
mmol CrCl3 = 0.0367 mmol
We can then find the number of mg CrCl3 by
multiplying mmol times FW.
mg CrCl3 = 0.0367 mmol x 158.4 mg/mmol =
5.81 mg
% CrCl3 = (5.81 mg/2.63x103 mg) x 100 =
0.221%
97
A 0.500 g sample containing sodium carbonate
(FW=106 mg/mmol) was dissolved and analyzed by
addition of 50.0 mL of 0.100 M HCl solution. The
excess HCl required 5.6 mL of 0.05 M NaOH solution.
Find the percentage of Na2CO3 n the sample.
First, write the chemical equations involved
Na2CO3 + 2 HCl g 2 NaCl + H2CO3
HCl + NaOH g NaCl + H2O
mmol HCl reacted = mmol HCl taken – mmol HCl titrated
98
mmol HCl reacted = 2 mmol Na2CO3
mmol HCl titrated = mmol NaOH
Now we can reformulate the above relation to
read
2 mmol Na2CO3 = mmol HCl taken – mmol
NaOH
mmol Na2CO3 = 1/2 ( 0.100 x 50.0 – 0.050 x 5.6)
= 2.36 mmol
? mg Na2CO3 = 2.36 mmol x 106 mg/mmol = 250
mg
% Na2CO3 = (250 mg/0.500 x103 mg) x 100 = 50.0%
99
Example
A 0.200 g sample containing MnO2 was dissolved and
analyzed by addition of 50.0 mL of 0.100 M Fe2+ to
drive the reaction
2 Fe2+ + MnO2 + 4 H+ g 2 Fe3+ + Mn2+ + 2 H2O
The excess Fe2+ required 15.0 mL of 0.0200 M KMnO4.
Find % Mn3O4 (FW= 228.8 mg/mmol) in the sample.
5 Fe2+ + MnO4- + 8 H+ g 5 Fe3+ + Mn2+ + 4 H2O
Solution
First, calculate mmol MnO2 then change it to mmol
Mn3O4 . We have the relation
100
mmol Fe2+ reacted = mmol Fe2+ taken – mmol
Fe2+ titrated
Now we should perform the following
substitutions:
1. mmol Fe2+ reacted = 2 mmol MnO2
2. mmol Fe2+ titrated = 5 mmol KMnO4
Now reformulate the above relation:
2 mmol MnO2 = mmol Fe2+ taken – 5 mmol
KMnO4
101
Substitution gives:
mmol MnO2 = 1/2( 0.100 x 50.0 – 5 x 0.0200 x 15.0) =
1.75 mmol
3 MnO2 = Mn3O4 + O2
mmol Mn3O4 = 1/3 mmol MnO2
mmol Mn3O4 = 1/3 x 1.75 mmol = 0.5833
?mg Mn3O4 = 0.5833 mmol x 228.8 mg/mmol = 133.5 mg
% Mn3O4 = (133.5 mg/200 mg) x 100 = 66.7%
102
Normality volumetric
Calculations
We have seen previously that solving volumetric
problems required setting up a relation between the
number of mmoles or reacting species. In case of
normality, calculation is easier as we always have
the number of milli equivalents (meq) of substance A
is equal to meq of substance B, regardless of the
stoichiometry in the chemical equation. Of course
this is because the number of moles involved in the
reaction is accounted for in the calculation of meqs.
Therefore, the first step in a calculation using
normalities is to write down the relation
meq A = meq B
103
The next step is to substitute for meq by one of the
following relations
meq = N x VmL
meq = mg/eq wt
meq = n x mmol
We should remember from previous lectures that:
eq wt = FW/n
N=nxM
Therefore, change from molarity or number of moles to
normality or number of equivalents using the above
relations.
104
Examples
A 0.4671 g sample containing sodium bicarbonate was
titrated with HCl requiring 40.72 mL. The acid was
standardized by titrating 0.1876 g of sodium
carbonate (FW = 106 mg/mmol) requiring 37.86 mL of
the acid. Find the percentage of NaHCO3 (FW=84.0
mg/mmol) in the sample.
This problem was solved previously using molarity.
The point here is to use normalities in order to
practice and learn how to use the meq concept. First,
let us find the normality of the acid from reaction
with carbonate (reacts with two protons as you
know).
105
Eq wt Na2CO3 = FW/2 = 53.0 and eq wt NaHCO3 = FW/1
= 84.0
meq HCl = meq Na2CO3
Now substitute for meq as mentioned above:
Normality x volume (mL) = wt (mg)/ eq wt
N x 37.86 = 187.6 mg/ (53 mg/meq)
NHCl= 0.0935 eq/L
106
Now we can find meq NaHCO3 where
meq NaHCO3 = meq HCl
mg NaHCO3 / eq wt = N x VmL
mg NaHCO3 /84.0 = 0.0935 x 40.72
mg NaHCO3 = 84.0 x 0.0935 x 40.72 mg
% NaHCO3 = (84.0 x 0.0935 x 40.72 mg/476.1
mg) x 100 = 67.2%
107
Lecture 13
Stoichiometric Calculations, cont….
Back to Normality Calculations
Use normalities to calculate how many mL of a
0.10 M H2SO4 will react with 20 mL of 0.25 M
NaOH.
Solution
We can first convert molarities to normalities:
N=nxM
N (H2SO4) = 2 x 0.10 = 0.20
N (NaOH) = 1 x 0.25 = 0.25
109
meq H2SO4 = meq NaOH
Substitute for meq as usual (either NVmL
or mg/eq wt)
0.20 x VmL = 0.25 x 20
VmL = 25 mL
110
Example
Find the normality of sodium carbonate (FW =
106) in a solution containing 0.212 g
carbonate in 100 mL solution if:
The carbonate is used as a monobasic base.
The carbonate is used as a dibasic base.
Solution
If carbonate is a monobasic base then eq wt =
FW/1 = 106/1 = 106 mg/meq
111
To find the normality of the solution we find the
weight per mL and then convert the weight
per mL to meq/mL. We have 212 mg/100 mL
which means 2.12 mg/mL. Now the point is
how many meq per 2.12 mg sodium
carbonate.
meq = mg/eq wt = 2.12/106 = 0.02 meq
Then the normality is 0.02 N
If carbonate is to be used as a dibasic salt then
the eq wt = FW/2 = 106/2 = 53 mg/meq.
112
To find the normality of the solution we find
the weight per mL and then convert the
weight per mL to meq/mL. We have 212
mg/100 mL which means 2.12 mg/mL.
Now the point is how many meq per 2.12
mg sodium carbonate.
meq = mg/eq wt = 2.12/53 = 0.04 meq
Then normality will be 0.04 N
113
How many mg of I2 (FW = 254 mg/mmol) should
you weigh to prepare 250 mL of 0.100 N
solution in the reaction:
I2 + 2e g 2 ISolution
To find the number of mg I2 to be weighed and
dissolved we get the meq required. We have:
meq = N x VmL
meq = 0.100 x 250 = 25.0 meq
mg I2 = meq x eq wt = meq x FW/2 = 25.0 x
254/2 = 3.18x103 mg
114
Find the normality of the solution containing
0.25 g/L H2C2O4 (FW = 90.0 mg/mmol).
Oxalic acid reacts as a diacidic
substance.
Solution
Since oxalic acid acts as a diacidic
substance then its eq wt = FW/2 =
90.0/2 = 45.0 mg/meq
115
We convert the mg acid per mL to meq acid
per mL We have 0.25 mg acid/mL
meq acid = 0.25 mg/45.0 mg/meq = 0.0056 meq
Therefore, the normality is 0.0056 meq/mL
116
The Titer Concept
In many situations where routine titrations are
carried out and to avoid wasting time in
performing calculations, one can calculate the
weight of analyte in mg equivalent to 1 mL of
titrant. The obtained value is called the titer of
the titrant. For example, an EDTA bottle is
labeled as having a titer of 2.345 mg CaCO3.
This means that each mL of EDTA consumed in
a titration of calcium carbonate corresponds to
2.345 mg CaCO3. If the titration required 6.75
mL EDTA then we have in solution 2.345 x 6.75
mg of calcium carbonate.
117
Example
What is the titer of a 5.442g/L K2Cr2O7 (FW = 294.2
mg/mmol) in terms of mg Fe2O3 (FW = 159.7
mg/mmol). The equation is:
6 Fe2+ + Cr2O72- + 14 H+ g 6 Fe3+ + 2 Cr3+ + 7 H2O
Solution
1 mL of K2Cr2O7 contains 5.442 mg K2Cr2O7 per mL.
Therefore let us find how many mg Fe2O3
corresponds to this value of K2Cr2O7.
118
mmol Fe2+ = 6*mmol K2Cr2O7
2Fe g Fe2O3
mmol Fe = 2 mmol Fe2O3
2*mmol Fe2O3 = 6*mmol K2Cr2O7
2*(mg Fe2O3/159.7) = 6*(5.442/294) * 1
mg Fe2O3 = 8.86 mg
Therefore, the titer of K2Cr2O7 in terms of
Fe2O3 is 8.86 mg Fe2O3 per mL K2Cr2O7
119