# BSB 111 stoichiometry lecture notes 2020 eduhub ```How many litres can you get
from P50.00?
How much can you pay for 20 L?
Pay =
𝑷𝟗.𝟒𝟎
𝟏𝑳
∗ 𝟐𝐎𝐋 = 𝐏𝟏𝟖𝟖
How much steak can you get
from P400.00?
How much can you pay for 2 kg?
Amount of steak
𝟏 𝒌𝒈
=
𝑷𝟔𝟒.𝟗𝟓
∗ 𝑷𝟒𝟎𝟎. 𝟎𝟎
= 6.07 kg =6 kg
Reaction stoichiometry (outline)
Calculations in chemical reactions
❑ moles, mass and number of particles using molar mass
❑ Stoichiometric coefficients as moles of substance
❑ Calculating the moles of product from moles of reactant
❑ Mass calculations from chemical equations
The limits of a reaction
❑ Calculations for identifying the limiting reactant and excess reactant
❑ Theoretical, Actual and Percentage yields
The stoichiometry in solution
❑ Concentrations of solutions (Molarity)
❑ Prepare a solution of known concentration (sdt solution) starting with a solid
❑ Dilution (prepare a solution of known concentration from a concentrated one)
❑ Volumetric analysis-Titration (acid-base titration)
Skills to develop: Calculations in chemical reactions (Application of mole
concept)
❑ Do calculations of required amounts of reactants and products
o How much of the reactant is used in the chemical reaction.
o How much product will form from known amount of reactant.
❑ Calculate the theoretical yield of a product based on a limiting reactant.
❑ Calculate percentage yield from known amount of reactant
REACTION STOICHIOMETRY
What is chemical stoichiometry? quantitative relationships between amounts
of reactants used and amounts of products formed by a chemical reaction
Therefore we are interested in answering the following questions
How much (quantity) of the starting material is needed?
❑ To react with another starting material (reactant)
❑ How much product can be formed from a certain amount of reactant
2KI (aq)
+
Pb(NO3)2(aq)
PbI2 (s) +
2KNO3 (aq)
HCl (aq)
+
NaOH(aq)
NaCl (aq) +
H2O (l)
Fe2+ (aq)
+
MnO4-(aq)
Mn2+ (aq) +
Fe3+ (aq)
?kg
A
+
10kg
B
?kg
C
How much C can be produced
from 10 kg of A?
How much B is required to react
with 10 kg of A?
Stoichiometry rests upon the law of conservation of mass. In general, chemical
reactions involve rearrangement of atoms and valence electrons.
Therefore, chemical reactions neither create or destroy or transmute one
element into another hence the amount of each element must be the same
through out the overall reaction.
2NaHCO3 (s) ⎯⎯→ Na 2 CO3 (s) + H 2 O (l) + CO 2 ( g )
heat
(84.007 g/mol)
105.9888 g/mol
2* (84.007 g/mol)
=
168.014 g
=
18.01528 g/mol
44.01 g/mol
1 * 105.9888 g/mol + 1 *18.01528 g/mol + 1 * 44.01 g/mol
168.0141 g
mass on the left hand side = mass on the right hand side conservation of mass
Mole concept
The SI unit of an amount of a substance is called a mole, mol.
Familiar counting units we use every day are:
case = 24
dozen = 12
A mole is a counting unit for submicron particles (atoms, ions, electrons,
molecules, photons). Based on amount of carbon atoms in 12 g of carbon
(carbon 12 isotope) contains 6.02 * 1023 carbon atoms.
Therefore, 1 mole of anything contains 6.02 * 1023 of those things.
This number is called Avogrado’s number.
Therefore ; 1 mol of Na atoms = 6.02 * 1023 Na atoms
1 mol of Sucroce = 6.02 * 1023 molecules of sucrose
Molar mass (atomic mass for elements)
The molar mass of an atom is the mass of a sample of atoms that contain
one mole of atoms (6.02*1023 atoms).
Examples:
Element
Molar mass/ g/mol
interpretation
Mg
24.31
1 mol Mg weighs 24.31 g
Cu
63.55
1 mol Cu weighs 63.55 g
In other words,
6.02*1023 atoms Mg weigh 24.31 g
6.02*1023 atoms Cu weigh 63.55 g
Molar mass is a conversion factor between mass and moles.
mass
m
Number
of moles
n
M
Molar
mass
Examples
1. How many cans of coke are present in 3 cases?
m
2. How many sodium atoms are there in 5 moles of sodium sample?
3. What is mass of 5 moles sodium sample?
1 case = 24 coke cans
24 cans
no of cans in 3 cases =
* 3 cases = 72 coke cans
1 case
23
1 mole of Na has 6.022 * 10 Na atoms
6.022*1023 Na atoms
no of Na atoms =
*5 moles
1 mole
= 3.011*1024 Na atoms
mass = molar mass * moles
g
= 22.9897
*5 mols = 114.9485 g
mol
n
M
1
2
H
1.00794
17
He
4.00260
2
13
14
15
16
3
4
5
6
7
8
9
10
Li
6.941
Be
9.01218
B
10.81
C
12.011
N
14.0067
O
15.9994
F
18.9984
Ne
20.179
11
12
13
14
15
16
17
18
Na
22.9898
Mg
24.305
Al
26.9815
Si
28.0855
P
30.9738
S
32.06
Cl
35.453
Ar
39.948
33
34
35
36
Br
79.904
Kr
83.8
19
20
K
39.0983
Ca
40.08
3
4
5
6
21
22
23
24
Sc
44.9559
Ti
47.88
V
50.9415
Cr
51.996
7
8
25
Mn
54.9380
9
10
11
12
26
27
28
29
30
31
32
Fe
55.847
Co
58.9332
Ni
58.69
Cu
63.546
Zn
65.38
Ga
69.72
Ge
72.59
As
74.9216
Se
78.96
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
Rb
85.4678
Sr
87.62
Y
88.9059
Zr
91.22
Nb
92.9064
Mo
95.94
Tc
(98)
Ru
101.07
Rh
102.906
Pd
106.42
Ag
107.868
Cd
112.41
In
114.82
Sn
118.69
Sb
121.75
Te
127.6
I
126.9
Xe
131.29
55
56
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
Cs
132.905
Ba
137.33
Lu
174.967
Hf
178.49
Ta
180.948
W
183.85
Re
186.207
Os
190.2
Ir
192.22
Pt
195.08
Au
196.967
Bi
208.908
Po
(209)
At
(210)
Rn
(222)
87
88
103
104
105
106
107
108
109
Fr
(223)
Ra
110
Uun
111
Uuu
112
Uub
(269)
(272)
(269)
226.025
Lr
(260)
Rf
(261)
Db
(262)
Sg
(263)
Bh
(264)
Hs
(265)
Mt
(268)
Hg
200.59
Tl
204.383
Pb
207.2
114
Uuq
116
Uuh
118
Uuo
Molar mass of a compound/molecule
Molar mass the mass of one mole of a compound/molecule. It is equal to the sum
of molar masses of atoms in the compound.
What is the molar mass of ethanol (C2H5OH)? How many grams of C2H5OH are
there in 2.85 mol and how many molecules of C2H5OH are there in 2.85 mol?
molar mass of ethanol = 2 *C + 6*H + 1* O
= (2*12.011 + 6*1.00794 + 15.9994) gmol−1
= 46.06904 gmol−1
mass = molar mass * moles
g
= 46.06904
* 2.85 mols = 131.2968 g= 131.30 g ethanol
mol
1 mole of ethanol has 6.022 * 1023 ethanol molecules
6.022*1023 ethanol molecules
no of ethanol molecules =
* 2.85 moles
1 mole
= 1.716*1024 ethanol molecules
m
n
M
Flowchart for mole conversions
2A
B
mol B =
2 mol A : 1 mol B
Mole ratio
Moles A
Moles B
1 mol B
* 4 mol A = 2 mol B
2 mol A
4 mol
Mass A
1 mol =
Molar mass A
Coefficients
balanced
equation
1 mol = 6.02 * 1023
particles
Particles
A
1 mol =
Molar mass B
1 mol = 6.02 * 1023
particles
Particles
B
Mass B
Examples. 2Mg (s) + O2 (g)
2MgO (s)
How many moles are present in 122 g of Mg? How many atoms of Mg are present
in 122 g?
mass
122 g
no of moles =
=
= 5.075931 mol Mg
m
−1
molar mass
24.035 gmol
6.022*1023 Mg atoms
no of Mg atoms =
*5.075931 moles
1 mole
= 3.056*1024 Mg atoms
How many moles of O2 gas can react completely with 122 g of Mg?
mole ratio from balance equation: 2 mol Mg: 1 mol O 2
1 mol O 2
mol O 2 =
*5.075931 mol Mg
2 mol Mg
= 2.537965 mol O 2
Calculate the moles of MgO that can be formed from 122 g of Mg.
Calculate the mass (g) of MgO that can be formed from 122 g of Mg.
n
M
Examples. 2Mg (s) + O2 (g)
2MgO (s)
Calculate the moles of MgO that can be formed from 122 g of Mg.
Use mole ratio from balanced equation to relate two substances
mole ratio from balance equation: 2 mol Mg: 2 mol MgO; 1 : 1
mol MgO=
2 mol MgO
*5.075931 mol Mg
2 mol Mg
= 5.075931 mol MgO
Calculate the mass (g) of MgO that can be formed from 122 g of Mg.
Use the triangle to relate molar mass and moles
mass = molar mass * moles
g
= 40.3044
*5.075931 mols = 204.5824 g= 205 g MgO
mol
m
n
M
Mole ratios in chemical equations
A mole ratio is a relationship between the numbers of moles of any two
substances in a balanced equation. (gives a relationship between 2 chemicals
in a reaction)
Example:
FeCl3 (aq) +3NaOH (aq)
Fe(OH)3 (s) + 3NaCl (aq)
Relationships based on this balanced equation (mole ratios)
1 mol FeCl3 : 3 mol NaOH (reactant to reactant)
1 mol FeCl3 : 1 mol Fe(OH)3 (reactant to product)
1 mol FeCl3 : 3 mol NaCl (reactant to product)
3 mol NaOH : 1 mol Fe(OH)3 (reactant to product)
3 mol NaOH : 3 mol NaCl (reactant to product)
Why do reactions stop? Limiting and excess reactant
At the end of the lesson you should be able to:
o Do calculations to identify a limiting reagent/reactant
o Do calculations to identify the excess reactant
o Calculate the excess reactant/unreacted amount
o Use the limiting reactant to calculate the theoretical yield
Nothing will last for ever!!!!!!!!!!!!
Input
one
+
Input
two
output
+
leftover
Limiting reactant and excess reactant
The reactant that is completely consumed first is called the limiting reactant.
It limits OR gives the smallest amount of product that can be formed.
The reactant in abundance is the excess reactant, part of it remains unreacted
Example: NaCl (aq) + AgNO3(aq)
NaNO3 (aq) + AgCl (s)
Suppose 1.5 mol AgNO3 is mixed with 1 mol of NaCl, AgCl will form as precipitate
Input one
(NaCl)
1.0 mol
Input two
+ (AgNO3)
1.5 mol
Output (AgCl)
1.0 mol
Once 1 mol of NaCl is finished, 0.5 mol of AgNO3 will remain
un-reacted.
Therefore NaCl is the limiting reactant and AgNO3 is excess
reactant. The mass of AgCl formed depends on NaCl.
+
Leftover
0.5 mol AgNO3
Remaining excess amount = original amount excess reactant – reacted amount of excess reactant
Example: 2Mg (s) + O2 (g)
2MgO (s)
Which reactant will give the smallest amount of MgO (in moles) when 7.8 moles
of Mg and 4.7 moles of O2 are reacted?
Which is the limiting reactant? Which reactant is in excess, and how many moles
of it are left over?
Mole ratio: 2 mol Mg : 2 mol MgO OR 1 : 1
Remaining excess amount (O2)
2 mol MgO
=(original amount excess reactant
mol MgO =
* 7.8 mol Mg = 7.8 mol MgO – reacted amount of excess reactant)
2 mol Mg
= 4.7 mol O 2 - 3.9 mol O 2
Mole ratio: 1 mol O2 : 2 mol MgO OR 1 : 2
= 0.8 mol O 2
2 mol MgO
mol MgO =
* 4.7 mol O 2 = 9.4 mol MgO
1 mol O 2
Mg is limiting reactant becuase it gives smaller amount of MgO &amp; O2 excess
Mole of O2 reacting to consume all 7.8 mol Mg
Mole ratio: 2 mol Mg : 1 mol O OR 2 : 1
1 mol O 2
mol O 2 reacted=
* 7.8 mol Mg = 3.9 mol O 2
2 mol Mg
C3H8 (g) + 5O2 (g)
4H2O (l) + 3CO2 (g)
When 5.0 g of C3H8 and 10 g of O2 are reacted together, which is the limiting
reactant? Use CO2 as product for comparison. Which reactant is in excess and
how many moles of it are left over?
molar mass of C3H8 = 3*12.011 + 8*1.00794 = 44.09652 gmol−1
m
−1
molar mass of O 2 = 2*15.9994 = 31.9988 gmol
5.0 g
M
n
mols of C3H8 =
=
0.113388
mol
44.09652 gmol−1
10 g
mols of O 2 =
= 0.312512 mol
−1
31.9988 gmol
mols of CO2 produced from C3H8 Mol ratio: 1 mol C3H8 : 3 mol CO2
3 mol CO 2
mols of CO 2 =
*0.113388 mol C3H8 = 0.340163 mol CO 2
1 mol C3H8
mols of CO2 produced from O2
Mol ratio: 5 mol O2 : 3 mol CO2
3 mol CO 2
mols of CO 2 =
*0.312512 mol O 2 = 0.187507 mol CO 2
5 mol O 2
C3H8 (g) + 5O2 (g)
4H2O (l) + 3CO2 (g)
O2 is the limiting reactant because it produces smaller amount of CO2.
C3H8 is in excess.
mols of C3H8 to react with all 0.312512 mol O2
Mol ratio: 1 mol C3H8 : 5 mol O2
1 mol C3H8
mols of C3H8 reacted =
*0.312512 mol O 2 = 0.062502 mol C3 H8
5 mol O 2
Remaining excess amount (C3H8)
=(original amount excess reactant – reacted amount of excess reactant)
= (0.113388 – 0.062502 )
= 0.050885 mol C3H8
How much product? Theoretical yield and the actual yield
Theoretical yield: the calculated amount according to the chemical equation.
This is the maximum mass of the product we would expect if all the limiting
reactant is converted to the expected product.
Actual yield: the amount of product obtained experimentally in the lab.
The actual amount is often less than the theoretical yield because of side
reactions, product sticks to walls and loss of product during purification.
The percentage yield: ratio which tells us how efficient a chemical reaction is.
actual yield
Percentage yield =
∗100%
theoretical yield
Theoretical, actual and % yield
At the end of the lesson you should be able to:
o Use limiting reagent/reactant to calculate theoretical yield
o Calculate the % yield of a given reaction
Input
one
+
Input
two
product
Example: When 2.3 kg of CS2 reacts with excess Cl2, 3.6 kg of CCl4 is formed.
What is the theoretical yield and the % yield?
CS2 (l) + 2Cl2 (g)
CCl4 (l) + 2S (s)
3.6 kg of CCl4 = 3.6 * 103 g of CCl4
2.3 kg = 2.3*103 g of CS2
molar mass of CS2 = 12.011+(2*32.06) = 76.131 gmol−1
molar mass of CCl4 = 12.011+(4*35.453) = 153.823 gmol−1
2300 g
mols of CS2 =
= 30.21108 mol CS2
−1
76.131 gmol
mol of CCl4 produced from 30.21108 mol CS2
Mol ratio: 1 mol CCl4 : 1 mol CS2
Therefore; 30.21108 mol CCl4 will be produced from 30.21108 mol CS2
Theoretical mass = molar mass * moles
g
= 153.823
*30.21108 mols = 4647.16 g=4.6 kg
mol
What is the theoretical yield and the % yield?
Actual yield 3.6 kg of CCl4 = 3.6 * 103 g of CCl4
Theoretical mass = molar mass * moles
g
= 153.823
*30.21108 mols = 4647.16 g=4.6 kg
mol
Actual yield
*100%
Theoretical yield
3600 g
=
*100% =77.47%
4647.16 g
% yield of CCl4 =
solutions
Concentration of solutions
The concentration of a solution expresses the amount of solute
dissolved in a given solvent or solution.
Concentration may be expressed in many different ways.
Dilute and concentrated solution
A dilute solution contains a relatively small amount of dissolved solute and a
concentrated solution contains a relatively large amount of dissolved solute.
Different ways of expressing solution concentration
Mass percent solution (% m/m)
% mass = (mass of solute (g) / mass of solution (g)) * 100 %
mass of solution = mass of solute + mass of solvent
Mass per volume percent (% m/v)
% m/v = (mass of solute (g)/ volume of solution (mL)*100 %
Volume percent (% v/v)
Solutions that are mixtures of two liquids are expressed in volume percent with
respect to the solute.
% v/v = (volume of solute mL/ volume of solution mL)* 100 %
Mole fraction (x) = ( nx/nT)
Where nx is the moles of component x and nT is the total moles of the mixture.
example: for two component mixture of A and B.
Mole fraction of A (xA) = nA/(nA + nB)
Molarity
Molarity (M) = moles of a solute (mol)/ Volume of solution (L)
This means 1 mol/L or ( 1M) tells you that you have 6.02 * 1023 particles (ions,
molecules) in 1 L of the solution.
Adding 53.5 g of NaCl in 482 mL H2O results in 500 mL of a solution. What is
the molarity of this NaCl solution?
n
m
mol of solute
Molarity =
volume of solution (L)
M Vol(L)
mass
53.5 g
no of moles =
=
= 0.915469 mol NaCl
−1
molar mass 58.44 gmol
mol of solute
0.915469 mol
Molarity =
=
= 1.83 molL−1
volume of solution (L)
0.5 L
n
M
Preparation of solutions of precisely known concentrations (standard
solutions)
Learning outcomes
▪ At the end of this lesson you should be able to define a standard solution
▪ You should be able to prepare standard solutions from solids and make
relavent calculations
▪ You should be able to make solutions of known concentrations from
concentrated solutions and be able to solve dilution problems
Preparation of a solution of known concentration (molarity)
Molarity is wide spread because it allows calculation of moles of a solute.
No of moles = molarity (mol/L) * volume of solution (L)
n
M Vol(L)
This equation allows the preparation of a solution of known molarity (std soln).
The volume is determined by the volumetric flask used but it MUST be in L.
The mass is obtained from the molar mass of the solute using moles from the
equation above.
m
The mass weighed is then dissolved in distilled or deionised
water to make up the required concentration in a volumetric flask.
M
n
no of moles = molarity (mol/L) * Volume of solution (L)
mass of solute = mole of solute * molar mass of solute
weigh mass (sample) on balance
Procedure for making a standard solution
Examples:
1. How many grams of CuSO4 . 5H2O should be dissolved in a volume of 500.0 mL
to make 8.00 mM?
2. What is the concentration of 1.0 g of NaOH dissolved in 2000 mL?
3. Calculate the number of moles of 500 mL of 0.5 M sodium hydroxide solution.
1. How many grams of CuSO4 . 5H2O should be dissolved in a volume of 500.0 mL
to make 8.00 mM?
mol of solute
Molarity =
 mol of solute = molarity mol * volume of solution (L)
L
volume of solution (L)
mol of solute = molarity mol * volume of solution (L)
n
L
= 8.00 * 10−3 mol *0.5 L
M Vol(L)
L
= 0.004 mol
mass = molar mass * moles
m
(
(
(
)
)
)
g
= 249.68
*0.004 mols = 0.99872 g
mol
n
M
2. What is the concentration of 1.0 g of NaOH dissolved in 2000 mL?
n
m
mass
1.0 g
no of moles =
=
= 0.025002 mol
−1
molar mass 39.997 gmol
n
M Vol(L)
M
mol of solute
0.025002 mol
Molarity =
=
= 0.012501M
volume of solution (L)
2L
3. Calculate the number of moles of 500 mL of 0.5 M sodium hydroxide solution.
(
)
mol of solute
Molarity =
 mol of solute = molarity mol * volume of solution (L)
L
volume of solution (L)
(
mol of solute = molarity mol
L
) * volume of solution (L)
= 0.5 mol *0.5 L
L
= 0.25 mol
n
M Vol(L)
Dilution
Solutions that are used in the laboratory are often
purchased in a concentrated form. e.g. HCl 12 M.
Solutions of lower concentrations are prepared by
adding water in a process called DILUTION.
Conservation of moles
Only solvent is added to dilute a solution, the
number of moles of solute remains the same.
mol solute before dilution = mol solute after dilution
n
mol solute before dilution (6) = mol solute after dilution (6)
M Vol(L)
A volume of concentrated solution is transferred (pippete) to a fresh vessel and diluted to the desired final
volume.
Since only a solvent is added to dilute a solution, the number of moles of solute remains the same.
n
mol solute before dilution = mol solute after dilution
initial molarity * initial volume = final molarity * final volume
M1V1 = M2V2 EQUTATION FOR DILUTION ! M2 = M1V1/V2 conservation of moles
M
Vol(L)
V1 = Volume of stock solution needed to make the new solution
M1= Concentration of stock solution
V2 = Final volume of new solution
M2 = Final concentration of new solution
M2 = M1 * V1/V2
dilution factor
(10 ml/100 ml) =0.1
mol solute before dilution = mol solute after dilution
Example: A student is asked to make up a 2.00 L of 0.020 M of solution of NaBr
from a bottle of 1.0 M NaBr. What volume of 1.0 M NaBr is needed to prepare
this new solution?
mol solute before dilution = mol solute after dilution
Vol
M1V1 = M 2 V2
1.0 M
NaBr
M 2 * V2
V1 =
M1
?
2L
0.02 M
NaBr
0.02 M* 2.00 L
V1 =
1.0 M
= 0.04 L = 40 mL
Examples:
1. What mass of CuSO4 is needed to prepare a solution of 1M CuSO4 in a 250
2. What volume of the above solution is needed to make a dilute solution of 0.1
M in a 250 mL volumetric flask?
mol of solute = molarity mol * volume of solution (L)
L
Molar mass CuSO4 = 159.609 g/mol
= 1 mol *0.250 L
n
L
mass = molar mass * moles
= 0.250 mol
M Vol(L)
g
(
)
= 159.609
mol
= 39.9022 g
n
39.9022 g
*0.250 mols
m
M
2. What volume of the above solution is needed to make a dilute solution of 0.1
M in a 250 mL volumetric flask?
mol solute before dilution = mol solute after dilution
Vol
M1V1 = M 2 V2
1.0 M
CuSO4
M 2 * V2
V1 =
M1
?
250
mL
0.1 M
CuSO4
0.1 M* 0.250 L
V1 =
1.0 M
= 0.025 L = 25 mL
Titration (volumetric analysis)
At the end of this lesson,
• You are expected to perform calculations based of acid-base titration
(volumetric analysis)
• Define equivalence point
• Define end point
• Define analyte
• Define titrant
• Define an indicator
Solution stoichiometry (volumetric analysis)
For solutions, molarity (mol/L) and volume are used to find moles of solute.
no of moles = volume of solution (L) * molarity (mol/L) of solute
n
Volume of
Chem 1
Molarity
of Chem 1
Molarity of
Chem 2
Step 1
Moles of
Chem 1
Step 3
Step 2
Mole ratio
from balanced
equation
Volume of
of Chem 2
Moles of
Chem 2
How many moles of NaOH are required to neutralise 20.0 mL of 0.25 M H2SO4 ?
M Vol(L)
How many moles of NaOH are required to neutralize 20.0 mL of 0.25 M H2SO4 ?
2NaOH (aq) + H2SO4 (aq)
Na2SO4 (aq)
(
+
)
2H2O (l)
mol of solute
Molarity =
 mol of solute = molarity mol * volume of solution (L)
L
volume of solution (L)
(
mol of H 2SO 4 = molarity mol
L
) * volume of solution (L)
= 0.25 mol *0.020 L
L
= 0.005 mol
Mole ratio: 2 mol NaOH : 1 mol H2SO4
2 mol NaOH
mols of NaOH =
*0.005 mol H 2SO 4 = 0.01 mol NaOH
1 mol H 2SO 4
Titration
A quantitative experiment where one solution of known concentration
is used to find the unknown concentration of a second solution using volume
analysis. (Acid-base, redox, complexometric, precipitation titration)
A solution of known concentration is called a standard solution.
Finding the concentration of the second solution using a standard solution is
called standardizing.
Equivalence point: the point in titration at which you have added the titrant
that is exactly equivalent to the analyte according to the balanced equation.
An indicator signals the end point which coincides with the equivalent point.
End point: the point where the indicator changes colour.
Analyte: the substance whose concentration is being determined.
Titrant: a substance added to react with the analyte.
Indicator: compound that changes colour at end point of titration.
Tools needed for titration
Pipette fillers
Indicator e.g. methyl orange
Acid
base
pipette
Example: acid-base neutralization.
25.0 mL of 0.1 M NaOH solution required 23.5 mL of HCl for neutralisation.
What is the concentration of HCl?
Volume of
Chem 1
Molarity
of Chem 1
Molarity of
Chem 2
Step 1
Moles of
Chem 1
Step 3
Step 2
Mole ratio
from balanced
equation
Volume of
of Chem 2
Moles of
Chem 2
n
M Vol(L)
25.0 mL of 0.1 M NaOH solution required 23.5 mL of HCl for neutralization.
What is the concentration of HCl?
NaOH (aq) + HCl(aq)
NaCl(aq)
+
H2O(l)
(
)
mol of solute
Molarity =
 mol of solute = molarity mol * volume of solution (L)
L
volume of solution (L)
(
mol of solute NaOH = molarity mol
L
) * volume of solution (L)
= 0.1 mol *0.025 L
L
= 0.0025 mol
Mole ratio: 1 mol NaOH : 1 mol HCl
1 mol HCl
mols of HCl =
*0.0025 mol NaOH = 0.0025 mol HCl
1 mol NaOH
Molarity of HCl =
mol of solute
0.0025 mol
=
= 0.106 M=0.1 M
volume of solution (L)
0.0235 L
Redox titration- a volumetric analysis technique based on oxidation (e-loss) and
Reduction (e-gain) reaction
One solution is an oxidizing agent (e-receiver) while the other solution is a
reducing agent (e-donor)
Example of oxidizing agent permanganate ion (MnO4-)
Purple permanganate ion (MnO4- ) is reduced to colorless manganese ion (Mn2+)
Example :
A 0.0484 mol L-1 standard solution of potassium permanganate was titrated
against 25.00 mL of an iron (II) sulfate solution. The equivalence point, as
indicated by a faint pink colour, was reached when 15.50 mL of potassium
iron (II) sulfate solution.(N.B.The reaction takes place in acidic solution)
MnO4-(aq)
+
Fe2+(aq)
Mn2+(aq)
+
Fe3+(aq)
MnO4-(aq)
+
Fe2+(aq)
Mn2+(aq)
+
Balance the equation first by creating half reactions
Oxidation number of Mn
X + 4*(-2)=-1
X = 8-1=+7
MnO4-(aq) + 8H+ (aq) + 5e
Mn2+(aq) +
Fe2+(aq)
Fe3+(aq) + e
Fe3+(aq)
4H2O (l) reduction
oxidation
xply reduction rxn by 1 and oxidation rxn by 5
MnO4-(aq) + 8H+ (aq) + 5e
5Fe2+(aq)
Mn2+(aq) +
5Fe3+(aq) + 5e
4H2O (l) reduction
oxidation
Add the two equations above and get rid of electrons
MnO4-(aq) + 8H+ (aq) + 5Fe2+ (aq)
Mn2+(aq) + 4H2O (l) + 5Fe3+ (aq)
(
)
mol of solute
Molarity =
 mol of solute = molarity mol * volume of solution (L)
L
volume of solution (L)
mol of solute KMnO 4 = molarity mol * volume of solution (L)
L
= 0.0484 mol *0.01550 L
L
= 0.00075 mol
(
)
Mole ratio: 1 mol MnO4- : 5 mol Fe2+
mols of Fe 2+
5 mol Fe 2+
−
2+
=
*0.00075
mol
MnO
=
0.00375
mol
Fe
4
1 mol MnO −4
Molarity of Fe
2+
mol of solute
0.00375 mol
=
=
= 0.15 M
volume of solution (L)
0.025 L
Preparation of standard solutions (Summary)
Start from solid solute
n
mol of solute
Molarity =
volume of solution (L)
M Vol(L)
m
𝑚𝑎𝑠𝑠 = 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 ∗ 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠
n
M
Dilution of concentrated solutions
Vol
V1
M1
V2
M2
mol solute before dilution = mol solute after dilution
M1V1 = M 2 V2
Flowchart for mole conversions (Summary)
Mole concept
B
+
Limiting &amp; excess reactant 2A
Theoretical yield &amp; % yield
2 mol A : 1 mol B
Mole ratio
1 mol =
Mass A
Moles A
Moles B
Coefficients
balanced
equation
Molar mass A
1 mol = 6.02 * 1023
particles
Particles
A
C
1 mol =
Molar mass B
1 mol = 6.02 * 1023
particles
Particles
B
Mass B
Solutions/volumetric analysis
2A
Volume of
A (L)
Molarity
of A
B
+
2 mol A : 1 mol B
Step 1
C
Molarity of
B
Step 3
Volume of
B (L)
Step 2
Moles of A
Moles of B
Mole ratio
from balanced
equation
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