3.1 Formulae of Compounds

1
Chapter 3
Chemical Equations and Stoichiometry
3.1 Formulae of Compounds 3.2 Derivation of
Empirical Formulae 3.3 Derivation of Molecular
Formulae 3.4 Chemical Equations 3.5
Calculations Based on Equations 3.6 Simple
Titrations
2
Formulae of Compounds
3.1 Formulae of Compounds (SB p.54)
How can you describe the composition of compound
X?
1st way by chemical formula
C?H?
3
3.1 Formulae of Compounds (SB p.54)
How can you describe the composition of compound
X?
Compound X
2nd way by percentage by mass
Mass of carbon atoms inside . g
Mass of hydrogen atoms inside . g
4
3.1 Formulae of Compounds (SB p.54)
Check Point 3-1 Give the empirical
molecular and structural formula for the
following compounds
Answer
5
3.1 Formulae of Compounds (SB p.55)
The different types of formulae of some compounds
Compound Empirical formula Molecular formula Structural formula
Carbon dioxide CO2 CO2 O C O
Water H2O H2O
Methane CH4 CH4
Glucose CH2O C6H12O6
Sodium fluoride NaF Not applicable NaF-
6
3.2 Derivation of Empirical Formulae (SB p.56)
Solution The relative molecular mass of CO2
12.0 2 x 16.0 44.0 Mass of carbon in 2.93 g
of CO2 2.93 g x 12.0/44.0 0.80 g The relative
molecular mass of H2O 2 x 1.0 16.0
18.0 Mass of hydrogen in 1.80 g of H2O 1.80 g x
2.0/18.0 0.20 g
Example 3-1 A hydrogen was burnt completely in
excess oxygen. It was found that 1.00 g of the
hydrocarbon gives 2.93 g of carbon dioxide and
1.8 g of water. Find the empirical formula of the
hydrocarbon.? (R.a.m. H 1.0, C 12.0, O
16.0)
Answer
7
3.2 Derivation of Empirical Formulae (SB p.57)
Solution (contd) Let the empirical formula of
the hydrocarbon be CxHy. Mass of carbon in CxHy
Mass of carbon in CO2 Mass of hydrogen in CxHy
Mass of hydrogen in H2O The simplest whole
number ratio of x and y can be determined by the
following the steps in the below table.
8
3.2 Derivation of Empirical Formulae (SB p.57)
Carbon Hydrogen
Mass (g) 0.80 0.20
Number of moles (mol) 0.80/12.0 0.066 7 0.2/0.066 7 3
Relative number of moles 0.066 7/0.066 71 0.20/0.066 7 3
Simplest mole ratio 1 3
9
3.2 Derivation of Empirical Formulae (SB p.57)
Solution Mass of compound X 0.46g Mass of
carbon in compound X 0.88 g x 12.0/44.0 0.24
g Mass of hydrogen in compound X 0.54 g x
2.0/18.0 0..06g Mass of oxygen in compound X
0.46 g 0.24 g 0.06 g 0.16 g
Example 3-2 Compound X is known to contain
carbon, hydrogen and oxygen only. When it is
burnt completely in excess oxygen, carbon dioxide
and water are given out as the only products. It
is found that 0.46 g of compound X gives 0.88 g
of carbon dioxide and 0.54 g of water. Find the
empirical formula of compound X. (R.a.m. H
1.0, C 12.0, O 16.0)
Answer
10
3.2 Derivation of Empirical Formulae (SB p.57)
Solution (contd) Let the empirical formula of
compound X be CxHyOz. Therefore, the
empirical formula of compound X is C2H6O.
Carbon Hydrogen Oxygen
Mass (g) 0.24 0.06 0.16
Number of moles (mol) 0.02 0.06 0.01
Relative number of moles 2 6 1
Simplest mole ratio 2 6 1
11
3.1 Formulae of Compounds (SB p.58)
Sulphur Oxygen
Mass (g) 5 5
Number of moles (mol) 5/ 32.1 0.156 5/16.0 0.313
Relative Number of moles 0.156/0.1561 0.313/0.156 2
Simplest mole ratio 1 2
Answer
12
3.1 Formulae of Compounds (SB p.58)
M O
Mass (g) 19.85 25.61
Number of moles (mol) 19.85/31.0 0.64 25.61/16.0 1.6
Relative number of moles 0.64/0.64 1 1.6/0.64 2.5
Simplest mole ratio 2 5
13
3.1 Formulae of Compounds (SB p.58)
(c) Mass of Cu (22.940 - 21.430) g
1.51g Mass of O (23.321 - 22.940)
0.381 g The empirical
formula of the oxide is CuO.
Cu O
Mass (g) 1.51 0.381
Number of moles (mol) 1.51/63.5 0.0238 0.381/16.0 0.0238
Relative number of moles 0.0238/0.0238 1 0.0238/0.0238 1
Simplest mole ratio 1 1
14
Determination of Empirical Formula
3.2 Derivation of Empirical Formulae (SB p.58)
From Combustion by Mass
Composition by mass
Empirical formula
15
3.2 Derivation of Empirical Formulae (SB p.58)
Solution Let the empirical formula of the
hydrocarbon be CxHy, and the mass of the compound
be 100 g. Mass of carbon in the compound 75
g Mass of hydrogen in the compound(100 75) g
25 g Therefore, the empirical formula of
the hydrocarbon is CH4.
Example3-3 Compound A contains carbon and
hydrogen only. It is found that the compound
contains 75 carbon by mass. Determine its
empirical formula. (Relative atomic masses
C12, H1 )
Carbon Hydrogen
Mass (g) 75 25
Number of moles (mol) 75/12.0 6.25 25/1.0 25
Relative number of moles 6.25/6.25 1 25/6.25 4
Simplest mole ratio 1 4
Answer
16
3.2 Derivation of Empirical Formulae (SB p.59)
Solution Let the mass of phosphorus chloride be
100g. Then, Mass of phosphorus in the compound
22.55g Mass of chloride in the compound
77.45g Therefore, the empirical formula of
the phosphorus chloride is PCl3.
Example 3-4 The percentage by mass of phosphorus
and chlorine in a sample of a phosphorus chloride
are 22.55 and 77.45 respectively. Find the
empirical formula of the chloride. (R.a.m. P
31.0, Cl 35.5)
Phosphorus Chloride
Mass (g) 22.55 77.45
Number of mole (mol) 22.55/31.0 0.727 77.45/35.5 2.182
Relative number of moles 0.727/0.727 1 2.182/0.727 3
Simplest mole ratio 1 3
Answer
17
3.2 Derivation of Empirical Formulae (SB p.59)

• Check Point 3-3
• Find the empirical formula of vitamin C if it
  consists of 40.9 caarbon, 54.5 oxygen and 4.6
  hydrogen by mass. ( R.a.m. C 12.0, H 1.0, O
  16.0)
• Each 325 mg tablet of aspirin consists of 195.0
  mg carbon 14.6 mg hydrogen and 115.4mg oxygen.
  Determine the empirical formula of aspirin.
  (R.a.m. C 12.0, H 1.0, O 16.0)

Carbon Hydrogen Oxygen
Mass (g) 40.9 4.6 54.5
Number of moles (mol) 40.9/12.0 3.41 4.6/1.0 4.60 54.5/16.0 3.41
Relative number of moles 3.41/3.41 1 4.61/3.41 1.35 3.41/3.41 1
Simplest mole ratio 3 4 3
Answer
18
3.2 Derivation of Empirical Formulae (SB p.59)
Carbon Hydrogen Oxygen
Mass (g) 195.0 14.6 115.4
Number of moles (mol) 195.0/12.0 16.25 14.6/7.21 2.02 7.21/7.21 1
Relative number of moles 16.25/7.21 2.25 14.6/7.21 2.02 7.21/7.21 1
Simplest mole ratio 9 8 4
19
What is Molecular Formulae?
3.3 Derivation of Molecular Formulae (SB p.60)
Molecular formula
?
(Empirical formula)n
20
3.3 Derivation of Molecular Formulae (SB p.60)
Empirical formula
Molecular mass
Molecular formula
21
3.3 Derivation of Molecular Formulae (SB p.60)
Solution Let the empirical formula of the
hydrocarbon be CxHy. Mass of carbon in the
hydrocarbon 14.6g x 12.0/44.0 4.0g Mass of
hydrogen in the hydrocarbon 9.0g x 2.0/18.0
1.0g
Example 3-5 A hydrogen was burnt completely in
excess oxygen. It was found that 5.00 g of the
hydrocarbon gives 14.6 g of carbon dioxide and
9.0 g of water. Given that the relative molecular
mass of the hydrocarbon is 30.0, determine its
molecular formula. hydrocarbon.? (R.a.m. H
1.0, C 12.0, O 16.0)
Carbon Hydrogen
Mass (g) 4.0 1.0
Number of moles (mol) 4.0/12.0 0.333 1.0/1.0 1
Relative number of moles 0.333/0.333 1 1/0.333 3
Simplest mole ratio 1 3
Answer
22
3.3 Derivation of Molecular Formulae (SB p.60)
Solution (contd) Therefore, the empirical
formula of the hydrocarbon is CH3. The molecular
formula of the hydrocarbon is (CH3)n. Relative
molecular mass of (CH3)n 30.0 n x (12.0
1.0 x 3) 30.0
n 2 Therefore, the molecular formula of the
hydrocarbon is C2H6.
23
3.3 Derivation of Molecular Formulae (SB p.61)
Solution Let the empirical formula of the
hydrocarbon be CxHyOz. Mass of carbon in the
compound 44.44g Mass of hydrogen in the
compound 6.18g Mass of oxygen in the compound
49.38g
Example 3-6 Compound X is known to contain 44.44
carbon, 6.18 hydrogen and 49.38 oxygen by mass.
A typical analysis shows that it has a relative
molecular mass of 162.0. Find its molecular
formula(R.a.m. H 1.0, C 12.0, O 16.0)
Carbon Hydrogen Oxygen
Mass (g) 44.44 6.18 49.38
Number of moles (mol) 44.44/12.0 3.70 6.18/1.0 6.18 49.38/16.0 3.09
Relative number of moles 3.70/3.09 1.2 6.18/3.09 2 3.09/3.09 1
Simplest mole ratio 6 10 5
Answer
24
3.3 Derivation of Molecular Formulae (SB p.61)
Solution(contd) The empirical formula of
compound X is C6H10O5. The molecular formula of
compound X is (C6H10O5)n. Relative molecular mass
of (C6H10O5)n 162.0 n x (12.0 x 6 1.0
x 10 16.0 x 5) 162.0
n
1 Therefore, the molecular formula of compound is
C6H10O5.
25
3.3 Derivation of Molecular Formulae (SB p.61)
Water of Crystallization Derived from
Composition by Mass
Hydrated salt Anhydrous salt
CuSO4?5H2O Blue crystals Anhydrous CuSO4 White powder
Na2CO3?10H2O Colourless crystals Anhydrous Na2CO3 White powder
CoCl2 ?2H2O Pink crystals Anhydrous CoCl2 Blue crystals
26
3.3 Derivation of Molecular Formulae (SB p.61)
Example 3-7 The chemical formula of hydrated
copper(II) sulphate is known to be CuSO4.xH2O. It
is found that the percentage of water by mass in
the compound is 36. Find x.(R.a.m. H1.0,
O16.0, S32.1, Cu63.5)
Solution Let Relative molecular mass of
CuSO4?xH2O 63.5 32.1 16.0 x 4 (1.0x2
16.0)x 159.6 18x Relative molecular mass of
water of crystallization 18x 18x/(159.6
18x) 36/100 1800x
5745.6 648 x 1152x
5745.6 x 4.99 ?
5 Therefore, the chemical formula of hydrated
copper(II) sulphate is CuSO4 ?5H2O
Answer
27
3.3 Derivation of Molecular Formulae (SB p.63)

• Check Point 3-4
• Find Compound Z is the major component of a
  healthy drink. It contains 40.00 carbon, 6.67
  hydrogen and 53.33 oxygen.
• (i) Find the empirical formula of compound Z.
• (ii) If the relative molecular mass of compound Z
  is 180, finds its molecular formula.(R.a.m. C
  12.0, H 1.0, O 16.0)

Carbon Hydrogen Oxygen
Mass (g) 40.00 6.67 53.33
Number of moles (mol) 40.00/12.0 3.33 6.67/1.0 6.67 53.33/16.03.33
Relative number of moles 3.33/3.33 1 6.67/3.33 2 3.33/3.33 1
Simplest mole ratio 1 2 1
Answer
28
3.3 Derivation of Molecular Formulae (SB p.63)
29
3.3 Derivation of Molecular Formulae (SB p.63)
Check Point 3-4 (b) (NH4)2Sx contains 72.72
sulphur by mass is water. Find the value of x.
(R.a.m. H 1.0, N 14.0, O 16.0) (c) In
the compound MgSO4?nH2O, 51.22 by mass is water.
Find the value of n. (R.a.m. H 1.0, O
16.0, Mg 24.3, S 32.1)
(NH4) unit S
Mass (g) 27.28 72.72
Number of moles (mol) 27.28/18.0 1.52 72.72/32.1 2.27
Relative number of moles 1.52/1.52 1 2.27/1.52 1.49
Simplest mole ratio 2 3
Answer
30
3.3 Derivation of Molecular Formulae (SB p.63)
MgSO4 H2O
Mass (g) 48.78 51.22
Number of moles (mol) 48.78/120.4 0.405 51.22/18.0 2.846
Relative number of moles 0.405/0.405 1 2.846/0.4057
Simplest mole ratio 1 7
31
3.3 Derivation of Molecular Formulae (SB p.63)
Example 3-8 The chemical formula of ethanoic acid
is CH3COOH. Calculate the percentages by mass of
carbon, hydrogen and oxygen by mass respectively.
(R.a.m. C12.0, H1.0, O16.0 )
Solution Relative molecular mass of CH3COOH
12.0 x 2 1.0 x 4 16.0 x 2 60.0 by mass of
C 12.0 x 2/ 60.0 x 100 40.00 by mass of H
1.0 x 4 /60.0 x 100 6.67 by mass of O
16.0 x 2/60.0 x 100 53.33 The percentage by
mass of carbon, hydrogen and oxygen are 40.00,
6.67 and 53.33 respectively.
Answer
32
3.3 Derivation of Molecular Formulae (SB p.63)
Example 3-9 Calculate the mass of iron metal in a
sample of 20g of hydrated iron (II) sulphate,
FeSO4?7H2O. (R.a.m. Fe 55.8 , H1.0, O16.0
)
Solution Relative molecular mass of
FeSO47H2O 55.8 32.1 16.0 x 4
(1.0x216.0) x 7277.9 by mass of Fe
55.8/277.9 x 100 20.08 Mass of Fe 20g x
20.08 4.02g
Answer
33
3.3 Derivation of Molecular Formulae (SB p.63)

• Check Point 3-5
• (a) Calculate percentages by mass of potassium,
  chromium and oxygen in potassium chromate (VI),
  K2Cr2O7.(R.a.m. K 39.1 . Cr 52.0, O 16.0)
• (b) Find the mass of metal and water of
  crystallization in
• 100 g of Na2SO410H2O
• 70g of Fe2O38H2O.
• (R.a.m. H 1.0, O 16.0, Na 23, S 32.1,
  Fe 55.8)

Answer
34
3.3 Derivation of Molecular Formulae (SB p.63)
(b)( i) Molar mass of Na2SO410H2O 322.1 g
mol-1 Mass of Na 23.0 x 2 g mol-1/ 322.1 g
mol-1 x 100g 14.28 g Mass of
H2O 18.0 x 10 g mol-1/ 322.1 g mol-1 x 100g
14.28 g (ii)
Molar mass of Fe2O38H2O 303.6 g mol-1 Mass of
Fe 55.8 x 2 g mol-1/303.6g mol-1 x 70g
25.73 g Mass of H2O 18.0 x 8 g
mol-1/303.6g mol-1 x 70g
33.20 g
35
3.4 Chemical Equations (SB p.64)
Chemical Equations
a A b B ? c C d D
(can also be volume ratios for gases)
Stoichiometry relative no. of moles of
substances involved in a chemical reaction.
36
3.4 Chemical Equations (SB p.64)
Answer
37
3.5 Calculations Based on Equations (SB p.65)
Calculations Based on Equations
Calculations involving Reacting Masses
38
3.5 Calculations Based on Equations (SB p.65)
Example 3-10 Calculate the mass of copper formed
when 12.45g of copper(II) oxide is completely
reduced by hydrogen. (R.a.m. H1.0, O16.0, Cu
63.5 )
Answer
39
3.5 Calculations Based on Equations (SB p.65)
Solution Number of moles of CO2 formed 240cm3/
24000cm3 mol-1 0.01 mol From the equation, 2
moles of NaHCO3(s) will form 1 mole of
CO2(g). Number of moles of NaHCO3 required
0.01 x 2 0.02 mol Mass of NaHCO3 required
0.02 mol x(23.0 1.0 16.0 x 3) g mol-1 0.02
mol x 84.0g mol-1 1.68 g Therefore, the minimum
amount of sodium hydrogencarbonate required is
1.68g.
Answer
40
3.5 Calculations Based on Equations (SB p.66)
Calculations Based on Equations
Calculations involving Volumes of Gases
41
3.5 Calculations Based on Equations (SB p.66)
Example 3-12 Calculate the volume of carbon
dioxide formed when 20 cm3 of ethane and 70 cm3
of oxygen are exploded, assuming all volumes are
measured at room temperature and pressure.
Answer
42
3.5 Calculations Based on Equations (SB p.67)
Example 3-13 10 cm3 of a gaseous hydrocarbon was
mixed with 80cm3 of oxygen which was in excess.
The mixture was exploded and then cooled. The
volume left was 70cm3. Upon passing the resulting
gaseous mixture through concentrated sodium
hydroxide solution ( to absorb carbon dioxide),
the volume of the residual gas became 50 cm3.
Find the molecular formula of the hydrocarbon.
Answer
43
3.5 Calculations Based on Equations (SB p.68)

• Check Point 3-7
• Find the volume of hydrogen produced at R.T.P.
  when 2.43g of magnesium reacts with excess
  hydrochloric acid. (R.a.m. Mg 24.3 molar
  volume of gas at R.T.P. 24.0 dm3mol-1.
• Find the minimum mass of chlorine required to
  produced 100 g of phosphorus trichloride ( PCl3).
• 20 cm3 of a gaseous hydrocarbon and 150 cm3 of
  oxygen were exploded in a closed vessel. After
  cooling, 110 cm3 of gases remained. After passing
  through a solution of concentrated sodium
  hydroxide, the volume left was 50 cm3 . Determine
  the molecular of the hydrocarbon.
• Calculate the volume of carbon dioxide formed
  when 5cm3 of methane burns in excess
  oxygen, assuming all volumes are measured at room
  temperature and pressure.

Answer
44
3.5 Calculations Based on Equations (SB p.68)
(c)Volume of CxHy used 20 cm3 Volume of CO2
formed 60 cm3 Volume of O2 used 100
cm3 Volume of CxHy volume of CO2 1 x 20
60 x 3 Volume
of CxHy volume of O2 1 x y/4 20 100
x y/4 5
3 y/4 5
y 8
45
3.5 Calculations Based on Equations (SB p.68)
46
Simple Titrations
3.6 Simple Titrations (SB p.68)
Acid-Base Titrations
Acid-Base Titrationswith Indicators
Acid-Base Titrationswithout Indicators
(to be discussed in later chapters)
47
Finding the concentration of a solution
3.6 Simple Titrations (SB p.69)
Copper(II) sulphate

solute
Water
solvent
solution
48
Finding the concentration of a solution
3.6 Simple Titrations (SB p.69)
50 cm3
49
Finding the concentration of a solution
3.6 Simple Titrations (SB p.69)
50 cm3
50
Finding the concentration of a solution
3.6 Simple Titrations (SB p.69)
50 cm3
51
Finding the concentration of a solution
3.6 Simple Titrations (SB p.69)
50 cm3
52
Finding the concentration of a solution
3.6 Simple Titrations (SB p.69)
50 cm3
53
Finding the concentration of a solution
3.6 Simple Titrations (SB p.69)
50 cm3
54
Finding the concentration of a solution
3.6 Simple Titrations (SB p.69)
50 cm3
Solution A
55
Finding the concentration of a solution
3.6 Simple Titrations (SB p.69)
50 cm3
56
Finding the concentration of a solution
3.6 Simple Titrations (SB p.69)
50 cm3
57
Finding the concentration of a solution
3.6 Simple Titrations (SB p.69)
50 cm3
58
Finding the concentration of a solution
3.6 Simple Titrations (SB p.69)
50 cm3
59
Finding the concentration of a solution
3.6 Simple Titrations (SB p.69)
50 cm3
60
Finding the concentration of a solution
3.6 Simple Titrations (SB p.69)
50 cm3
61
Finding the concentration of a solution
3.6 Simple Titrations (SB p.69)
50 cm3
Solution B
62
Finding the concentration of a solution
3.6 Simple Titrations (SB p.69)
100 cm3
63
Finding the concentration of a solution
3.6 Simple Titrations (SB p.69)
100 cm3
64
Finding the concentration of a solution
3.6 Simple Titrations (SB p.69)
100 cm3
65
Finding the concentration of a solution
3.6 Simple Titrations (SB p.69)
100 cm3
66
Finding the concentration of a solution
3.6 Simple Titrations (SB p.69)
100 cm3
67
Finding the concentration of a solution
3.6 Simple Titrations (SB p.69)
100 cm3
68
Finding the concentration of a solution
3.6 Simple Titrations (SB p.69)
100 cm3
Solution C
69
Comment on the Concentrations of Solutions A, B
and C !
3.6 Simple Titrations (SB p.69)
Concentration of solution B is 2 times that of
the concentrations of solutions A B.
contain the same amount of solute (same
concentration)
Concentration is the amount of solute in a unit
volume of solution.
70
3.6 Simple Titrations (SB p.69)
Class Practice
Suppose the right-handed side figure shows the
number of solute particles in solution D. Draw
similar particle models for Solutions A, B and C.
71
3.6 Simple Titrations (SB p.69)
Class Practice Answers
72
Comment on the Concentrations of Solutions A, B
and C !
3.6 Simple Titrations (SB p.69)
no. of spoons
no. of moles
mass
Concentration is the amount of solute in a unit
volume of solution.
73
Molarity
3.6 Simple Titrations (SB p.69)
A way of expressing concentrations
Molarity is the number of moles of solute
dissolved in 1 dm3 (1000 cm3) of solution.
(M)
74
3.6 Simple Titrations (SB p.69)
In every 1 dm3 of the solution, 2 moles of HCl
is dissolved.
75
3.6 Simple Titrations (SB p.71)
Example 3-14 25.0cm3 of sodium hydroxide solution
was titrated against 0.067 M of sulphuric(VI)
acid using methyl orange as indicator. The
indicator changed colour from yellow to red when
22.5 cm3 of sulphuric(VI) acid had benn added.
Calculate the molarity of the sodium hydroxide
solution.
Answer
76
3.6 Simple Titrations (SB p.71)

• Example 3-15
• 2.52 g of a pure dibasic acid with formula
  mass of 126.0 was dissolved in water and made up
  to 250.0 cm3 in a volumetric flask 25.0 cm3 of
  this solution was found to neutralize 28.5 cm3 of
  sodium hydroxide solution.
• Calculate the molarity of the acid solution.
• If the dibasic acid is represented by H2X, write
  an equation for the reaction between the acid and
  sodium hydroxide.
• Calculate the molarity of the sodium hydroxide
  solution.

Answer
77
3.6 Simple Titrations (SB p.72)
Example 3-16 0.186g of sample of hydrate sodium
carbonate, NaCO2nH2O, was dissolved in 100 cm3
of distilled water in conical flask. 0.10 M by
hydrochloric acid was added from a burette, 2 cm3
at a time. The pH value of the solution was
measured by a pH meter. The result was recorded
and shown in the following figure. Calculate the
value of n in NaCO2nH2O.
Answer
78
3.6 Simple Titrations (SB p.73)
Answer
79
3.6 Simple Titrations (SB p.74)
80
3.6 Simple Titrations (SB p.74)
Solution (contd) (c) Neutralization is an
exothermic reaction. When more and more
sulphuric(VI) acid was added and reacted with
potassium hydroxide, the temperature rose. The
temperature rose to a maximum value at which the
equivalence point of the reaction was reached.
After that, any excess sulphuric (VI) acid added
cooled down the reacting solution, causing the
temperature to drop.
81
3.6 Simple Titrations (SB p.76)
Redox Titrations
Some Examples
Iodometric Titrations
I2(aq) 2S2O32-(aq) ? 2I-(aq) S4O62-(aq)
brown
colourless
82
3.6 Simple Titrations (SB p.76)
Redox Titrations
Some Examples
Iodometric Titrations
I2(aq) 2S2O32-(aq) ? 2I-(aq) S4O62-(aq)
brown
colourless
During titration brown ? yellow
83
3.6 Simple Titrations (SB p.76)
Redox Titrations
Some Examples
Iodometric Titrations
I2(aq) 2S2O32-(aq) ? 2I-(aq) S4O62-(aq)
brown
colourless
During titration brown ? yellow
84
3.6 Simple Titrations (SB p.76)
Redox Titrations
Some Examples
Iodometric Titrations
I2(aq) 2S2O32-(aq) ? 2I-(aq) S4O62-(aq)
brown
colourless
During titration brown ? yellow
End point blue black ? colourless(after
addition of starch indicator)
85
3.6 Simple Titrations (SB p.76)
Example 3-18 When excess potassium iodide
solution (KI) is added to 25.0 cm3 of acidified
potassium iodate solution (KIO3) of unknown
concentration, the solution turns brown. This
brown solution requires 22.0 cm3 of 0.05 M sodium
thiosulphate solution to react completely with
the iodine formed, using starch solution as
indicator. Find the molarity of the potassium
iodate solution.
Answer
86
3.6 Simple Titrations (SB p.76)
Redox Titrations
Some Examples
Titrations Involving Potassium Permanganate
MnO4-(aq) 8H(aq) 5Fe2(aq) ?Mn2(aq)
H2O(aq) 5Fe3(aq)
purple
colourless
87
3.6 Simple Titrations (SB p.76)
Redox Titrations
Some Examples
Titrations Involving Potassium Permanganate
MnO4-(aq) 8H(aq) 5Fe2(aq) ?
Mn2(aq) H2O(aq) 5Fe3(aq)
purple
colourless
During titration pale green ? yellow
88
3.6 Simple Titrations (SB p.76)
Redox Titrations
Some Examples
Titrations Involving Potassium Permanganate
MnO4-(aq) 8H(aq) 5Fe2(aq) ?
Mn2(aq) H2O(aq) 5Fe3(aq)
purple
colourless
During titration pale green ? yellow
End point yellow ? light purple
89
3.6 Simple Titrations (SB p.77)
Example 3-19 A piece of impure iron wire weighs
0.22g. When it is dissolved in hydrochloric acid,
it is oxidized to iron(II) ions. The solution
requires 36.5 cm3 of 0.02 M acidified potassium
manganate(VII) for complete reaction to form
iron(III) ions. What is the percentage purity of
the iron wire?
Answer
90
3.6 Simple Titrations (SB p.78)

• Check Point 3-8
• 5g of anhydrous sodium carbonate is added to 100
  cm3 of 2 M hydrochloric acid. What is the volume
  of gas evolved at room temperature and pressure?
• (R.a.m. C 12.0, O 16.0, Na 23.0 molar
  volume of gas at R.T.P. 24.0 dm3 mol-1)
• (b) 8.54g of impure hydrated iron(II) sulphate
  (formula mass of 392.14) was dissolved in water
  and made up to 250 cm3. 25cm3 of this solution
  required 20.76cm3 of 0.0203M acidified potassium
  manganate(VII) solution for complete reaction.
  Determine the percentage purity of the hydrated
  iron(II) sulphate.

Answer
91
3.6 Simple Titrations (SB p.78)
92
The END