3
1
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 Chemical Equations
3.6
Simple Titrations
Stoichiometry (化學計量學)p.19
Deals with quantitative relationships
(a) among atoms, molecules and ions
RAM / RMM / Formula Mass
2
Stoichiometry (化學計量學)p.19
Deals with quantitative relationships
(b) among the constituent elements
of a compound
Empirical / Molecular Formulae
3
Stoichiometry (化學計量學)p.19
Deals with quantitative relationships
(c) among the substances
participating in chemical reaction
Calculations involving chemical
equation
4
3.1
5
Formulae of
Compounds
3.1 Formulae of compounds (SB p.43)
Empirical formula
Shows the simplest whole number
ratio of the atoms or ions present
E.g. Methane,
CH4
Sodium chloride, NaCl
6
3.1 Formulae of compounds (SB p.43)
Molecular formula
Shows the actual number of each
kind of atoms present in one
molecule
E.g.
CH4
methane
Ionic compounds do not have
molecular formulae
7
3.1 Formulae of compounds (SB p.43)
Structural formula
Shows the bonding order of atoms
in one molecule
H
E.g.
CH4
methane
H
C
H
8
H
3.1 Formulae of compounds (SB p.44)
Different types of formulae of some compounds
Compound
Empirical
formula
Molecular
formula
Structural
formula
Carbon
dioxide
Water
CO2
CO2
O = C =O
H2O
H2O
O
Methane
CH4
H
H
H
CH4
H
C
H
H
Glucose
CH2O
C6H12O6
OH
O H
H
H
OH
H
HO
OH
H
Sodium
fluoride
9
NaF
Not
applicable
OH
Na+F-
3.2
10
Derivation of
Empirical
Formulae
3.2 Derivation of empirical formulae (SB p.45)
From composition by mass
Example 1
Mg + N2
MgxNy
0.450 g excess
0.623 g
Mass of N used
= (0.623-0.450)g = 0.173 g
11
3.2 Derivation of empirical formulae (SB p.45)
From composition by mass
Example 1
Mg + N2
MgxNy
0.450 g excess
0.623 g
mMg
nMg
nN
12
MMg

mN
MN
0.450
1.5 3
24.31



0.173
1
2
14.01
Mg3N2
3.2 Derivation of empirical formulae (SB p.45)
Water of Crystallization Derived from
Composition by Mass
Example 2
Q.14
Example 3-3C
13
Check Point 3-3A
Q.14
CuSO4.xH2O
heat
10.0 g
CuSO4 + xH2O
6.4 g
(10.0–6.4) g
= 3.6 g
6.4 3.6
:
 0.04 : 0.2  1 : 5
CuSO4 : H2O =
160 18
x = 5
14
From combustion data
CxHyOz
Vitamin C
heat
CxHyOz
0.2000 g
15
+
O2 
excess
CO2(g)
0.2998 g
+
H2O(g)
0.0819 g
CxHyOz
+
O2 
CO2(g)
+
H2O(g)
Mass of C in sample = mass of C in CO2 formed
Mass of H in sample = mass of H in H2O formed
Mass of O in sample
= total mass of sample – mass of C – mass of H
16
12.0
mass of C  mass of CO2 
44.0
12.0
 0.2998 
 0.320 g
44.0
2.0
mass of H  mass of H2O 
18.0
2.0
 0.0819 
 0.0267 g
18.0
Mass of O in sample = (0.2000-0.0818-0.0091) g
= 0.109 g
17
Mass (g)
Number of
moles (mol)
Simplest
whole no.
ratio
C
H
O
0.0818
0.0091
0.109
0.0091
 9.1  10-3
1.0
0.109
 6.8  10 -3
16.0
4
3
0.0818
 6.8  10-3
12.0
3
C3H4O3
18
3.3
19
Derivation of
Molecular
Formulae
3.3 Derivation of molecular formulae (SB p.49)
What is molecular formulae?
Molecular formula
= (Empirical formula)n
20
3.3 Derivation of Molecular Formulae (SB p.49)
From empirical formula and known
relative molecular mass
Empirical formula
Molecular mass
Example 3-3A
Example 3-3B
Molecular formula
21
Q.15
CnH2n+2(l)
473 K, 1.00 atm
12.0 g
CnH2n+2(g)
3.28 dm3
3
1
1


12.0
g
(0.082
atm
dm
K
mol
)(473K)
mRT
M

1.00 atm3.28 dm3 
PV
= 142 g mol1
Relative molecular mass = 142
22
Q.15
CnH2n+2(l)
473 K, 1.00 atm
12.0 g
CnH2n+2(g)
3.28 dm3
RMM = 12n + 2n+2 = 142
 n = 10
The molecular formula is C10H22
23
Determination of Chemical Formulae
C, H, O are present
Qualitative
Analysis
Structural
formula
Ethanoic acid
CH3COOH
Quantitative
Analysis
Empirical
formula
CH2O
24
RMM
= 60.0
IR, NMR,
MS
Molecular
formula
C2H4O2
H
H
C
H
O
C
O
H
Structural formula : bond-line structure
25
Calculate the % by mass of the constituent elements of
soda alum.
Na2SO4·Al2(SO4)3·24H2O
2  23.0
 100%  5.02%
Na :
916.4
S:
4  32.1
 100%  14.0%
916.4
2  27.0
 100%  5.89%
Al :
916.4
26
Calculate the % by mass of the constituent elements of
soda alum.
Na2SO4·Al2(SO4)3·24H2O
40  16.0
 100%  69.8%
O:
916.4
H:
27
48  1.0
 100%  5.2%
916.4
A certain compound was known to have a formula which
could be represented as [PdCxHyNz](ClO4)2.
Analysis showed that the compound contained 30.15%
carbon and 5.06% hydrogen.
When converted to the corresponding thiocyanate,
[PdCxHyNz](SCN)2, the analysis was 40.46% carbon and
5.94% hydrogen.
Calculate the values of x, y and z.
(Relative atomic masses : C = 12.0, H = 1.0, N = 14.0,
O = 16.0, Cl = 35.5, S = 32.0, Pd = 106.0)
28
Let M be the formula mass of [PdCxHyNz](ClO4)2
Then, the formula mass of [PdCxHyNz](SCN)2
= M + 2(32.0+12.0+14.0) – 2(35.5+416.0)
= M – 83.0
% by mass of C in [PdCxHyNz](ClO4)2
12.0x

 100%  30.15%
M
% by mass of C in [PdCxHyNz](SCN)2
12.0(x  2)

 100%  40.46%
M  83.0
29
12.0x
 100%  30.15%
M
12.0(x  2)
 100%  40.46%
M  83.0
(1)
(2)
Solving by simultaneous equations,
x = 14, M = 557
% by mass of H in [PdCxHyNz](ClO4)2
1.0y

 100%  5.06%
557
y = 28
557 = 106.0 + 12.014 + 1.028 + 14.0z + 2(35.5+416.0)
z = 4
30
3.4
31
Chemical
Equations
3.4 Chemical equations (SB p.53)
Chemical equations
aA+bB cC+dD
a, b, c, d are stoichiometric coefficients
na : nb : nc : nd  a : b : c : d
32
3.5 Calculations Based on Equations (SB p.65)
Calculations based on equations
Calculations involving reacting masses
Example 3-5A
Example 3-5B
Check Point 3-4
33
3.4 Chemical equations (SB p.53)
Example (a) Excess oxygen
2Mg(s)
2.43 g
+ O2(g)  2MgO(s)
excess
?
2.43 g
nMg reacted 
 0.100 mol
1
24.3 g mol
nMg 2
 1
nMgO 2
 nMgO produced  0.100 mol
34
3.4 Chemical equations (SB p.53)
Example (a) Excess oxygen
2Mg(s)
2.43 g
+ O2(g)  2MgO(s)
excess
?
mMgO  0.100 mol  (24.3  16.0) g mol 1
= 4.03 g
35
3.4 Chemical equations (SB p.53)
Example(b) limiting reagent to be determined
2Mg(s)
2.43 g
+ O2(g)  2MgO(s)
1.28g
?
nMg  0.100mol
nO2
36
1.28 g

 0.040 mol
1
32.0 g mol
3.4 Chemical equations (SB p.53)
Example(b) limiting reagent to be determined
2Mg(s)
0.100 mol
nMg
nO2
+ O2(g)  2MgO(s)
0.040 mol
0.100 mol
2

 2.5 
0.040 mol
1
Mg is in excess,
O2 is the limiting reagent
37
?
3.4 Chemical equations (SB p.53)
Example(b) limiting reagent to be determined
2Mg(s)
+ O2(g)  2MgO(s)
0.040 mol
0.080 mol
mMgO  0.080 mol  (24.3  16.0) g mol
= 3.22 g
1
Q.16, 17
Check Point 3-4
38
Q.16
P4 (s)
4.00 g
+
5O2(g)
2P2O5(s)
6.00 g
4.00 g
nP4 
 0.0323 mol
1
31.0  4 g mol
nO2
39
6.00 g

 0.188 mol
1
32.0 g mol
limiting reactant
P4(s)
0.0323 mol
nO2
nP4
+
excess
5O2(g)
0.188 mol
0.188 mol

 5.82  5
0.0323 mol
O2 is in excess
40
2P2O5(s)
Q.16
P4(s)
+
5O2(g)
2P2O5(s)
20.0323 mol
0.0323 mol

mP2O5  2  0.0323 mol  142 g mol
= 9.17 g
41
1

Q.17
0.27 g
nAl reacted 
 0.01 mol
1
27 g mol
0.96 g
nCu produced 

0.015
mol
1
64 g mol
nAl : nCu  1 : 1.5 = 2 : 3
2Al(s) + 3Cu2+(aq)  2Al3+(aq) + 3Cu(s)
42
3.5 Calculations Based on Equations (SB p.66)
Calculations involving volumes of gases
1.Gases not at the same conditions
2.Gases at the same conditions
- Gay Lussac’s Law
43
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
2.8 dm3
25C
1.65 atm
35.0 dm3
31C
1.25 atm
? dm3
125C
2.50 atm
nCH4
PV
1.65 atm  2.8 dm3


 0.189 mol
3
1
1
RT 0.082 atm dm K mol  298 K
nO2
PV
1.25 atm  35.0 dm3


 1.75 mol
3
1
1
RT 0.082 atm dm K mol  304 K
nO2
nCH4
44
1.75 2


0.189 1
O2 is in excess and
CH4 is the limiting reactant
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
0.189 mol
VCO2 
nCO2 RT
0.189 mol
? dm3
125C
2.50 atm
P
0.189 mol  0.082 atm dm3 K 1 mol 1 (273  125) K

2.50 atm
= 2.47 dm3
45
Gay Lussac’s law :
When gases reacts, they do so in
volumes which bears a simple whole
number ratio to one another, and to
the volumes of gaseous products, all
volumes being measured under the
same conditions of temperature and
pressure.
Watch video
46
39.5 cm3
Ammonia
Fe as catalyst
20 cm3
H2(g) + CuO
nitrogen + hydrogen
10 cm3
heat
29.5 cm3
Cu(s) + H2O(l)
2NH3(g)  1N2(g) + 3H2(g)
VNH3 : VN2 : VH2  20 : 10 : 29.5  2 : 1 : 3
47
Gay Lussac’s law is an application
of the Avogadro’s law.
a A(g) + b B(g)  c C(g) + d D(g)
na : nb : nc : nd  a : b : c : d
At fixed T & P,
nV
Va : Vb : Vc : Vd  a : b : c : d
48
Gay Lussac’s law
a A(g) + b B(g)  c C(g) + d D(g)
At fixed T & P
Va : Vb : Vc : Vd  a : b : c : d
49
Determination of Molecular Formula
from Reacting Volumes of Gases
For CxHy
y
y

CxHy(g) +  x   O2(g)  xCO2(g) +
H2O(?)
4
2

VC x H y : VO2 : VCO2
50
y

 1:  x   : x
4

Determination of Molecular Formula
from Reacting Volumes of Gases
For CxHyOz
y
y z

CxHyOz(g) +  x    O2(g)  xCO2(g) + H2O(?)
2
4 2

VC x H y Oz : VO2 : VCO2
51
y z

 1:  x    : x
4 2

Q.18
y
y

CxHy(g) +  x   O2(g)  xCO2(g) +
H2O(l)
4
2

15 cm3
75 cm3
45 cm3
Volume of CO2 formed = volume of gas absorbed by NaOH
= (70 – 25) cm3 = 45 cm3
52
Before the reaction,
VO2  100cm3
After the reaction,
VO2  25cm3
 VO2used = 75 cm3
Q.18
y
y

CxHy(g) +  x   O2(g)  xCO2(g) +
H2O(l)
4
2

15 cm3
VCO2
VCxHy
VO2
75 cm3
45 cm3
x 45 cm
 
x 3
3
1 15 cm
3
y
4
x
75 cm


5
3
VCxHy
1
15 cm
y
(3  )  5  y  8
4
53
3
C3H8
Q.19
y
y z

CxHyOz(g) +  x    O2(g)  xCO2(g) + H2O(g)
2
4 2

50 cm3
VCO2
VCxHyOz
VH2O
VCxHyOz
VO2
VCxHyOz
54
100 cm3
x 100 cm3
 
x 2
3
1 50 cm
y
3
100
cm
2
y4
3
1
50 cm
x  4y  2z 100 cm3


2
3
1
50 cm
100 cm3
100 cm3
C2H4O2
4 z
(2   )  2  z  2
4 2
3.6
55
Simple
Titrations
3.6 Simple titrations (SB p.58)
Simple titrations
Simple titrations
Acid-base titrations
56
Redox titrations
3.6 Simple titrations (SB p.58)
Simple titrations
Acid-base titrations
Acid-base titrations
with indicators
Acid-base titrations
without indicators
Refer to Notes on ‘Detection of Equivalence
Point in Acid-base Titration
57
End point : The point at which the indicator undergoes
a sharp colour change in a titration
Equivalence point : The point at which the reaction is just
complete without excess of any reactant.
58
3.6 Simple titrations (SB p.62)
Titration without an indicator
By following the change
(a) in pH value (pH titration)
(b) in temperature (thermometric titration)
(c) in electrical conductivity
(conductometric titration)
in the course of the reaction
59
pH Titration Curves
Phenolphthalein
Equivalence point,
pH 7.00
Methyl orange
60
pH Titration Curves
Equivalence point,
pH 5.28
Methyl orange
61
pH Titration Curves
Equivalence point,
pH 8.72
Phenolphthalein
62
pH Titration Curves
Slow change in pH
around equivalence point
Equivalence point,
~ pH 7
Most indicators are
not suitable
63
Thermometric Titration
Temperature / K
Equivalence point
Volume of acid added / cm3
64
H+(aq) + OH(aq)  H2O(l) + heat
cooling by excess acid
Temperature / K
Equivalence point
Volume of acid added / cm3
65
Conductometric Titration – NaOH vs HCl
Electrical Conductivity
Equivalence point
Volume of acid added /cm3
66
Conductometric Titration – NaOH vs HCl
more mobile
+

+

Electrical Conductivity
H (aq) + Cl (aq) + Na (aq) + OH (aq)
less mobile
H2O(l) + Na+(aq) + Cl(aq)
Volume of acid added /cm3
67
Conductometric Titration – NaOH vs HCl
Electrical Conductivity
Beyond the equivalence point,
conductivity  sharply due to excess
H+ (most mobile) & Cl
Steeper
slope
Equivalence point
Volume of acid added /cm3
68
At the equivalence point
Electrical Conductivity
H+(aq) + Cl(aq) + Na+(aq) + OH(aq)
H2O(l) + Na+(aq) + Cl(aq)
conductivity > 0
Volume of acid added /cm3
69
Conductometric Titration – NH3 vs HCl
Electrical Conductivity
Equivalence point
Volume of acid added /cm3
70
Initial conductivity is low
NH3(aq) + H2O(l)
NH4+(aq) + OH(aq)
Electrical Conductivity
Equivalence point
Volume of acid added /cm3
71
H+(aq) + Cl(aq) + NH3(aq)
Electrical Conductivity
NH4+(aq) + Cl(aq)
less conducting
more
conducting
Volume of acid added /cm3
72
Electrical Conductivity
Beyond the equivalence point,
conductivity  sharply due to excess
H+ (most mobile) & Cl
steeper
slope
Equivalence point
Volume of acid added /cm3
73
Conductometric Titration – NaOH vs CH3COOH
Electrical Conductivity
Equivalence point
Volume of acid added /cm3
74
CH3COOH(aq) +
Na+(aq)
+
OH(aq)
more
conducting
Electrical Conductivity
less conducting
H2O(l) + Na+(aq) + CH3COO(aq)
Volume of acid added /cm3
75
Electrical Conductivity
Beyond the equivalence point,
conductivity  slowly because the
excess acid is weak
smaller slope
Equivalence point
Volume of acid added /cm3
76
Conductometric Titration – NH3 vs CH3COOH
Electrical Conductivity
Equivalence point
Volume of acid added /cm3
77
Initial conductivity is low
Electrical Conductivity
NH3(aq) + H2O(l)
NH4+(aq) + OH(aq)
Equivalence point
Volume of acid added /cm3
78
CH3COOH (aq) + NH3(aq)
Electrical Conductivity
CH3COO(aq) + NH4+(aq)
more
conducting
Volume of acid added /cm3
79
Electrical Conductivity
Beyond the equivalence point,
conductivity  slowly because the
excess acid is weak
smaller slope
Equivalence point
Volume of acid added /cm3
80
Conductometric Titration – Ba(OH)2 vs H2SO4
Electrical Conductivity
At equivalence point,
conductivity  0
because BaSO4 is insoluble in water
Equivalence point
Volume of acid added / cm3
81
3.6 Simple titrations (SB p.58)
Simple titrations
6B
Simple titrations
Acid-base titrations
82
Redox titrations
3.6 Simple Titrations (SB p.65)
Redox titrations
1. Iodometric titration
I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62-(aq)
brown
colourless
S2O3
2-(aq)
I2
brown
83
Very few
I2

yellow
During the
course of the
titration
3.6 Simple Titrations (SB p.65)
Redox titrations
1. Iodometric titration
I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62-(aq)
brown
colourless
Starch
Very few
I2
yellow
84
solution

Complex of
starch & I2
dark blue
During the
course of the
titration
3.6 Simple Titrations (SB p.65)
Redox titrations
Example 3-6E
1. Iodometric titration
I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62-(aq)
brown
colourless
At the end point
S2O32-(aq)
2I-(aq) +
Complex of
starch & I2
dark blue
85
S4O62-(aq)

colourless
3.6 Simple Titrations (SB p.66)
2. Titrations involving potassium manganate(VII)
MnO4-(aq) + 8H+(aq) + 5Fe2+(aq) Mn2+(aq) + H2O(aq) + 5Fe3+(aq)
purple
pale green
colourless
yellow
H+(aq)/MnO4(aq)
Fe2+ &
Fe2+
Fe3+
During the course of the titration
86
3.6 Simple Titrations (SB p.66)
2. Titrations involving potassium manganate(VII)
MnO4-(aq) + 8H+(aq) + 5Fe2+(aq) Mn2+(aq) + H2O(aq) + 5Fe3+(aq)
purple
pale green
colourless
yellow
H+(aq)/MnO4(aq)
Fe2+ &
Fe3+
At the end point
Example 3-6F
87
Fe3+ &
MnO4
Check Point 3-6
NaOH + HCl  NaCl + H2O
Na2CO3 + HCl  NaHCO3 + NaCl
Red in
phenolphthalein
colourless in phenolphthalein
Double indicator
titration
88
NaHCO3 + HCl  NaCl + H2O + CO2
yellow in
methyl orange
red in
methyl orange
Double indicator
titration
89
The END
90
3.1 Formulae of compounds (SB p.45)
Back
Give the
empirical,Empirical
molecular and
structural
formulae for the
Compound
Molecular
Structural
formula
formula
following compounds:
formula
(a) Propene
(a) Propene
(b) Nitric acid
(b) Nitric
(c) Ethanol
acid
(c) Ethanol
(d) Glucose
(d) Glucose
CH2
C3H6
H
H
H
H
C
C
C
H
HNO3
HNO3
O
H O
N
O
C2H6O
C6H12O6
C2H5OH
C6H12O6
H
H
H
C
C
H
H
Answer
OH
OH
O H
H
H
OH
H
H
OH
HO
91
H
OH
3.2 Derivation of empirical formulae (SB p.46)
A hydrocarbon was burnt completely in excess oxygen. It
was found that 1.00 g of the hydrocarbon gives 2.93 g of
carbon dioxide and 1.80 g of water. Find the empirical
formula of the hydrocarbon.
Answer
92
3.2 Derivation of empirical formulae (SB p.46)
The relative molecular mass of CO2 = 12.0 + 16.0  2 = 44.0
Mass of carbon in 2.93 g of CO2 = 2.93 g  12.0 = 0.80 g
44.0
The relative molecular mass of H2O = 1.0  2 + 16.0 = 18.0
2 .0
Mass of hydrogen in 1.80 g of H2O = 1.80 g 
= 0.20 g
18.0
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
93
3.2 Derivation of empirical formulae (SB p.46)
Back
Carbon
Hydrogen
Mass (g)
0.80
0.20
No. of moles
(mol)
0.80
 0.0667
12.0
0.20
 0.20
1.0
Relative no.
of moles
0.0667
1
0.0667
0.20
3
0.0667
Simplest
mole ratio
1
3
Therefore, the empirical formula of the hydrocarbon is CH3.
94
3.2 Derivation of empirical formulae (SB p.46)
Compound X is known to contain carbon, hdyrogen 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.
Answer
95
3.2 Derivation of empirical formulae (SB p.47)
Mass of compound X = 0.46 g
12.0
Mass of carbon in compound X = 0.88 g 
= 0.24 g
44.0
Mass of hydrogen in compound X = 0.54 g  2.0 = 0.06 g
18.0
Mass of oxygen in compound X = 0.46 g – 0.24 g – 0.06 g = 0.16 g
Let the empirical formula of compound X be CxHyOz.
96
3.2 Derivation of empirical formulae (SB p.47)
Back
Carbon
Hydrogen
Oxygen
Mass (g)
0.24
0.06
0.16
No. of moles
(mol)
0.80
 0.0667
12.0
0.06
 0.06
1.0
Relative no. of
moles
0.02
2
0.01
0.06
6
0.01
0.01
1
0.01
Simplest mole
ratio
2
6
1
0.16
 0.01
16.0
Therefore, the empirical formula of compound X is C2H6O.
97
3.2 Derivation of empirical formulae (SB p.47)
(a) 5 g of sulphur forms 10 g of an oxide on complete
combustion. What is the empirical formula of the oxide?
Answer
(a) Mass of sulphur = 5 g
Mass of oxygen = (10 – 5) g = 5 g
Sulphur
Oxygen
Mass (g)
5
5
No. of moles (mol)
5
 0.156
32.1
0.156
1
0.156
1
5
 0.313
16.0
0.313
2
0.156
2
Relative no. of
moles
Simplest mole
ratio
98
3.2 Derivation of empirical formulae (SB p.47)
(b) 19.85 g of element M combines with 25.61 g of oxygen to
form an oxide. If the relative atomic mass of M is 31.0,
find the empirical formula of the oxide.
Answer
(b)
M
O
Mass (g)
19.85
25.61
No. of moles (mol)
19.85
 0.64
31.0
0.64
1
0.64
2
Relative no. of
moles
Simplest mole
ratio
The empirical formula of the oxide is M2O5.
99
25.61
 1 .6
16.0
1.6
 2.5
0.64
5
3.2 Derivation of empirical formulae (SB p.47)
(c) Determine the empirical formula of copper(II) oxide using
the following results.
Experimental results:
Mass of test tube = 21.430 g
Mass of test tube + Mass of copper(II) oxide = 23.321 g
Mass of test tube + Mass of copper = 22.940 g
Answer
100
3.2 Derivation of empirical formulae (SB p.47)
Back
(c) Mass of Cu = (22.940 – 21.430) g = 1.51 g
Mass of O = (23.321 – 22.940) g = 0.381 g
Copper
Oxygen
Mass (g)
1.51
0.381
No. of moles
(mol)
1.51
 0.0238
63.5
0.381
 0.0238
16.0
Relative no.
of moles
0.0238
1
0.0238
0.0238
1
0.0238
Simplest
mole ratio
1
1
Therefore, the empirical formula of copper(II) oxide is CuO.
101
3.2 Derivation of empirical formulae (SB p.48)
Compound A contains carbon and hydrogen atoms only. It is
found that the compound contains 75 % carbon by mass.
Determine its empirical formula.
Answer
102
3.2 Derivation of empirical formulae (SB p.48)
Back
Let the empirical formula of compound A be CxHy, and the mass of the
compound be 100 g. Then, mass of carbon in the compound = 75 g
Mass of hydrogen in the compound = (100 – 75) g = 25 g
Carbon
Hydrogen
Mass (g)
75
25
No. of moles (mol)
75
 6.25
12.0
25
 25
1.0
Relative no. of
moles
6.25
1
6.25
25
4
6.25
Simplest mole ratio
1
Therefore, the empirical formula of compound A is CH4.
103
4
3.2 Derivation of empirical formulae (SB p.48)
The percentages by mass of phosphorus and chlorine in a
sample of phosphorus chloride are 22.55 % and 77.45 %
respectively. Find the empirical formula of the phosphorus
chloride.
Answer
104
3.2 Derivation of empirical formulae (SB p.48)
Back
Let the mass of phosphorus chloride be 100 g. Then,
Mass of phosphorus in the compound = 22.55 g
Mass of chlorine in the compound = 77.45 g
Mass (g)
Phosphorus
Chlorine
22.55
77.45
No. of moles (mol)
22.55
 0.727
31.0
77.45
 2.182
35.5
Relative no. of
moles
0.727
1
0.727
2.182
3
0.727
Simplest mole ratio
1
3
Therefore, the empirical formula of the phosphorus chloride is PCl3.
105
3.2 Derivation of empirical formulae (SB p.49)
(a) Find the empirical formula of vitamin C if it consists of
40.9 % carbon, 54.5 % oxygen and 4.6 % hydrogen by mass.
(a) Let the mass of vitamin C analyzed be 100 g.
Carbon
Hydrogen
Oxygen
Mass (g)
40.9
4.6
54.5
No. of moles
(mol)
40.9
 3.41
12.0
3.41
1
3.41
3
4.6
 4.60
1 .0
4 .6
 1.35
3.41
4
Relative no.
of moles
Simplest
mole ratio
The empirical formula of vitamin C is C3H4O3.
106
Answer
54.5
 3.41
16.0
3.41
1
3.41
3
3.2 Derivation of empirical formulae (SB p.49)
Back
(b) Each 325 mg tablet of aspirin consists of 195.0 mg carbon,
14.6 mg hydrogen and 115.4 mg oxygen. Determine the
empirical formula of aspirin.
Answer
(b) The masses of the elements are multiplied by 1000 first.
Mass (g)
Carbon
Hydrogen
Oxygen
195.0
14.6
115.4
No. of moles 195.0
 16.25
(mol)
12.0
Relative no. 16.25
 2.25
of moles
7.21
Simplest
9
mole ratio
14.6
 14.6
1.0
14.6
 2.02
7.21
8
The empirical formula of aspirin is C9H8O4.
107
115.4
 7.21
16.0
7.21
1
7.21
4
3.3 Derivation of molecular formulae (SB p.50)
A hydrocarbon was burnt completely in excess oxygen. It
was found that 5.0 g of the hydrocarbon gave 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.
Answer
108
3.3 Derivation of molecular formulae (SB p.50)
Let the empirical formula of the hydrocarbon be CxHy.
Mass of carbon in the hydrocarbon = 14.6 g  12.0 = 4.0 g
44.0
2 .0
Mass of hydrogen in the hydrocarbon = 9.0 g  18.0 = 1.0 g
Carbon
Hydrogen
Mass (g)
4.0
1.0
No. of moles (mol)
4.0
 0.333
12.0
1 .0
1
1 .0
Relative no. of
moles
0.333
1
0.333
1
3
0.333
Simplest mole ratio
109
1
3
3.3 Derivation of molecular formulae (SB p.50)
Back
Therefore, the empirical formula of the hydrocarbon is CH3.
Let the molecular formula of the hydrocarbon be (CH3)n.
Relative molecular mass of (CH3)n = 30.0
n  (12.0 + 1.0  3) = 30.0
n=2
Therefore, the molecular formula of the hydrocarbon is C2H6.
110
3.3 Derivation of molecular formulae (SB p.50)
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.
Answer
111
3.3 Derivation of molecular formulae (SB p.50)
Let the empirical formula of compound X is CxHyOz, and the mass of the
compound be 100 g. Then,
Mass of carbon in the compound = 44.44 g
Mass of hydrogen in the compound = 6.18 g
Mass of oxygen in the compound = 49.38 g
Carbon
Hydrogen
Oxygen
Mass (g)
44.44
6.18
49.38
No. of moles
(mol)
44.44
 3.70
12.0
3.70
 1 .2
3.09
6
6.18
 6.18
1.0
6.18
2
3.09
10
49.38
 3.09
16.0
3.09
1
3.09
5
Relative no. of
moles
Simplest mole
ratio
112 empirical formula of compound X is C6H10O5.
The
3.3 Derivation of molecular formulae (SB p.50)
Back
Let the molecular formula of compound X be (C6H10O5)n.
Relative molecular mass of (C6H10O5)n = 162.0
n  (12.0  6 + 1.0  10 + 16.0  5) = 162.0
n=1
Therefore, the molecular formula of compound X is C6H10O5.
113
3.3 Derivation of molecular formulae (SB p.51)
The chemical formula of hydrated copper(II) sulphate is
known to be CuSO4 · xH2O. It is found that the percentage of
water of crystallization by mass in the compound is 36 %.
Find the value of x.
Answer
114
3.3 Derivation of molecular formulae (SB p.51)
Relative formula mass of CuSO4 · xH2O
Back
= 63.5 + 32.1 + 16.0  4 + (1.0  2 + 16.0)x
= 159.6 + 18x
Relative molecular mass of water of crystallization = 18x
18 x
36

159.6  18 x 100
1800x = 5745.6 + 648x
1152x = 5745.6
x = 4.99
5
Therefore, the chemical formula of the hydrated copper(II) sulphate is
CuSO4 · 5H2O.
115
3.3 Derivation of molecular formulae (SB p.52)
(a) Compound Z is the major ingredient 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,
find its molecular formula.
Answer
116
3.3 Derivation of molecular formulae (SB p.52)
(a) (i) Let the mass of compound Z be 100 g.
Carbon
Hydrogen
Oxygen
Mass (g)
40.00
6.67
53.33
No. of moles
(mol)
40.00
 3.33
12.0
6.67
 6.67
1.0
53.33
 3.33
16.0
Relative no.
of moles
3.33
1
3.33
6.67
2
3.33
3.33
1
3.33
Simplest
mole ratio
1
2
1
Therefore, the empirical formula of compound Z is CH2O.
117
3.3 Derivation of molecular formulae (SB p.52)
(ii)Let the empirical formula of compound Z be (CH2O)n.
n  (12.0 + 1.0  2 + 16.0) = 180
30n = 180
n =6
Therefore, the molecular formula of compound Z is C6H12O6.
118
3.3 Derivation of molecular formulae (SB p.52)
(b) (NH4)2Sx contains 72.72 % sulphur by mass. Find the value
of x.
Answer
(b)
(NH4) unit
S
Mass (g)
27.28
72.72
No. of moles (mol)
27.28
 1.52
18.0
72.72
 2.27
32.1
Relative no. of
moles
Simplest mole ratio
1.52
1
1.52
2
2.27
 1.49
1.52
3
Since the chemical formula of (NH4)Sx is (NH4)2S3, the value of x
is 3.
119
3.3 Derivation of molecular formulae (SB p.52)
Back
(c) In the compound MgSO4 · nH2O, 51.22 % by mass is water.
Find the value of n.
Answer
(c)
MgSO4
H 2O
Mass (g)
48.78
51.22
No. of moles (mol)
48.78
 0.405
120.4
51.22
 2.846
18.0
Relative no. of
moles
Simplest mole ratio
0.405
1
0.405
1
2.846
7
0.405
7
Since the chemical formula of MgSO4 · nH2O is MgSO4 · 7H2O,
the value of n is 7.
120
3.3 Derivation of molecular formulae (SB p.52)
The chemical formula of ethanoic acid is CH3COOH.
Calculate the percentage of mass of carbon, hydrogen and
oxygen respectively.
Answer
121
3.3 Derivation of molecular formulae (SB p.52)
Relative molecular mass of CH3COOH
Back
= 12.0  2 + 1.0  4 + 16.0  2
= 60.0
12.0  2
 100%
60.0
= 40.00 %
% by mass of H = 1.0  4  100%
60.0
= 6.67 %
16.0  2
 100%
% by mass of O =
60.0
= 53.33 %
% by mass of C =
The percentage by mass of carbon, hydrogen and oxygen are 40.00 %,
6.67 % and 53.33 % respectively.
122
3.3 Derivation of molecular formulae (SB p.53)
Back
Calculate the mass of iron in a sample of 20 g of hydrated
iron(II) sulphate, FeSO4 · 7H2O.
Answer
Relative formula mass of FeSO4 · 7H2O
= 55.8 + 32.1 + 16.0  4 + (1.0  2 + 16.0)  7 = 277.9
55.8
 100%
% by mass of Fe =
277.9
= 20.08 %
Mass of Fe = 20 g  20.08 % = 4.02 g
123
3.3 Derivation of molecular formulae (SB p.53)
(a) Calculate the percentages by mass of potassium ,
chromium and oxygen in potassium dichromate(VI),
K2Cr2O7.
Answer
-1
(a) Molar mass of K2Cr2O7 = (39.1  2 + 52.0  2 + 16.0  7) g mol
= 294.2 g mol-1
(39.1 2)gmol 1
 100%
% by mass of K =
1
294.2gmol
= 26.58 %
(52.0  2)gmol 1
 100%
% by mass of Cr =
1
294.2gmol
= 35.35 %
(16.0  7)gmol 1
 100%
% by mass of O =
1
294.2gmol
= 38.07 %
124
3.3 Derivation of molecular formulae (SB p.53)
(b) Find the mass of metal and water of crystallization in
(i) 100 g of Na2SO4 · 10H2O
(ii) 70 g of Fe2O3 · 8H2O
Answer
125
3.3 Derivation of molecular formulae (SB p.53)
Back
(b) (i) Molar mass of Na2SO4 · 10H2O = 322.1 g mol-1
(23.0  2)gmol 1
Mass of Na =
 100g
1
322.1gmol
= 14.28 g
(18.0  10)gmol 1
Mass of H2O =
 100g
1
322.1gmol
= 55.88 g
(ii) Molar mass of Fe2O3 · 8H2O = 303.6 g mol-1
(55.8  2)gmol 1
Mass of Fe =
 70g
1
303.6gmol
= 25.73 g
(18.0  8)gmol 1
 70g
Mass of H2O =
303.6gmol 1
= 33.20 g
126
3.4 Chemical equations (SB p.54)
Back
Give the chemical equations for the following reactions:
• Zinc + steam  zinc oxide + hydrogen
(a) Zn(s) + H2O(g)  ZnO(s) + H2(g)
(b) Magnesium + silver nitrate  silver + magnesium
nitrate
(b) Mg(s) + 2AgNO3(aq)  2Ag(s) + Mg(NO3)2(aq)
(c) Butane + oxygen  carbon dioxide + water
(c) 2C4H10(g) + 13O2(g)  8CO2(g) + 10H2O(l)
Answer
127
3.5 Calculations based on chemical equations (SB p.55)
Calculate the mass of copper formed when 12.45 g of
copper(II) oxide is completely reduced by hydrogen.
Answer
128
3.5 Calculations based on chemical equations (SB p.55)
Back
CuO(s) + H2(g)  Cu(s) + H2O(l)
As the mole ratio of CuO : Cu is 1 : 1, the number of moles of Cu formed
is the same as the number of moles of CuO reduced.
12.45g
Number of moles of CuO reduced =
(63.5  16.0)gmol 1
= 0.157 mol
Number of moles of Cu formed = 0.157 mol
Mass of Cu
= 0.157 mol
63.5 g mol 1
Mass of Cu = 0.157 mol  63.5 g mol-1 = 9.97 g
Therefore, the mass of copper formed in the reaction is 9.97 g.
129
3.5 Calculations based on chemical equations (SB p.55)
Sodium hydrogencarbonate decomposes according to the
following chemial equation:
2NaHCO3(s)  Na2CO3(s) + CO2(g) + H2O(l)
In order to obtain 240 cm3 of CO2 at room temperature and
pressure, what is the minimum amount of sodium
hydrogencarbonate required?
(Molar volume of gas at R.T.P. = 24.0 dm3 mol-1)
130
Answer
3.5 Calculations based on chemical equations (SB p.55)
Back
Number of moles of CO2 required
240 cm3
=
= 0.01 mol
3
1
24000 cm mol
From the chemical equation, 2 moles of NaHCO3(s) give 1 mole of CO2(g).
Number of moles of NaHCO3 required = 0.01  2 = 0.02 mol
Mass of NaHCO3 required
= 0.02 mol  (23.0 + 1.0 + 12.0 + 16.0  3) g mol-1
= 0.02 mol  84.0 g mol-1
= 1.68 g
Therefore, the minimum amount of sodium hydrogencarbonate required is
1.68 g.
131
3.5 Calculations based on chemical equations (SB p.56)
Calculate the volume of carbon dioxide formed when 20 cm3
of ethane and 70 cm3 of oxygen are exploded, assuming all
volumes of gases are measured at room temperature and
pressure.
Answer
132
3.5 Calculations based on chemical equations (SB p.56)
Back
2C2H6(g) + 7O2(g)

4CO2(g)
+
6H2O(l)
2 mol
7 mol
:
4 mol
:
6 mol (from equation)
7 volumes
:
4 volumes :
- (by Avogadro’s law)
:
2 volumes :
It can be judged from the chemical equation that the mole ratio of CO2 :
C2H6 is 4 : 2, and the volume ratio of CO2 : C2H6 should also be 4 : 2
according to the Avogadro’s law.
Let x be the volume of CO2(g) formed.
Number of moles of CO2(g) formed : number of moles of C2H6(g) used
=4:2
Volume of CO2(g) : volume of C2H6(g) = 4 : 2
x : 20 cm3 = 4 : 2
x = 40 cm3
Therefore, the volume of CO2(g) formed is 40 cm3.
133
3.5 Calculations based on chemical equations (SB p.57)
10 cm3 of a gaseous hydrocarbon was mixed with 80 cm3 of
oxygen which was in excess. The mixture was exploded and
then cooled. The volume left was 70 cm3. 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
134
3.5 Calculations based on chemical equations (SB p.57)
Let the molecular formula of the hydrocarbon be CxHy.
Volume of hydrocarbon reacted = 10 cm3
Volume of O2(g) unreacted = 50 cm3 (the residual gas after reaction)
Volume of O2(g) reacted = (80 – 50) cm3 = 30 cm3
Volume of CO2(g) formed = (70 – 50) cm3 = 20 cm3
CxHy
+
O2

xCO2
1 mol
:
mol
:
x mol
volumes
:
x volumes
1 volume :
135
+
H2 O
3.5 Calculations based on chemical equations (SB p.57)
Back
Volume of CO2 (g) = x
Volume of C xHy (g) 1
20 cm3 = x
10 cm3
1
x=2
y
x

Volume of O 2 (g)
4
=
Volume of C xHy (g)
1y
3
x
30 cm
=
4
3
10 cm
1
y
x =3
4
y
2

As x = 2,
=3
4
y=4
Therefore, the molecular formula of the hydrocarbon is C2H4.
136
3.5 Calculations based on chemical equations (SB p.58)
(a) Find the volume of hydrogen produced at R.T.P. when
2.43 g of magnesium reacts with excess hydrochloric acid.
(Molar volume of gas at R.T.P. = 24.0 dm3 mol-1)
(a) Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g)
No. of moles of H2 = No. of moles of Mg
2.43 g
Volume of H2
=
24.3 g mol -1
24.0 dm3 mol -1
Volume of H2 = 2.4 dm3
137
Answer
3.5 Calculations based on chemical equations (SB p.58)
(b) Find the minimum mass of chlorine required to produce
100 g of phosphorus trichloride (PCl3).
Answer
3  35.5
mCl 
100 g  77.5 g
3  35.5  31.0
138
3.5 Calculations based on chemical equations (SB p.58)
(c) 20 cm3 of a gaseous hydrocarbon and 150 cm3 of oxygen
(which was in excess) were exploded in a closed vessel.
After cooling, 110 cm3 of gases remained. After passing
the resulting gaseous mixture through concentrated
sodium hydroxide solution, the volume of the residual
gas became 50 cm3. Determine the molecular formula of
the hydrocarbon.
Answer
139
3.5 Calculations based on chemical equations (SB p.58)
y
y
( x  ) O2(g)  xCO2(g) + H2O(l)
4
2
3
Volume of CxHy used = 20 cm
(c) CxHy(g) +
Volume of CO2 formed = (110 – 50) cm3 = 60 cm3
Volume of O2 used = (150 – 50) cm3 = 100 cm3
Volume of CxHy : Volume of CO2 = 1 : x
= 20 : 60
x=3
y
4
= 20 : 100
Volume of CxHy : Volume of O2 = 1 : x 
=5
140
3.5 Calculations based on chemical equations (SB p.58)
(c) As x = 3,
3
y
=5
4
y =2
4
y=8
The molecular formula of the hydrocarbon is C3H8.
141
3.5 Calculations based on chemical equations (SB p.58)
Back
(d) Calculate the volume of carbon dioxide formed when
5 cm3 of methane was burnt completely in excess oxygen,
assuming all volumes of gases are measured at room
temperature and pressure.
(d) CH4(g)
1 mol
2O2(g)
 CO2(g)
+ 2H2O(l)
:
2 mol
:
: 2 mol (from equation)
1 volume :
5 cm3
Answer
+
2 volumes :
1 mol
1 volume: - (from Avogadro’s law)
x cm3
It can be judged from the equation that the mole ratio of CO2 : CH4 is
1 : 1, the volume ratio of CO2 : CH4 should also be 1 : 1.
x
1
=
5 cm3 1
x = 5 cm3
The volume of CO2(g) formed is 5 cm3.
142
3.6 Simple titrations (SB p.61)
25.0 cm3 of sodium hydroxide solution was titrated against
0.067 M sulphuric(VI) acid using methyl orange as an
indicator. The indicator changed colour from yellow to red
when 22.5 cm3 of sulphuric(VI) acid had been added.
Calculate the molarity of the sodium hydroxide solution.
Answer
143
3.6 Simple titrations (SB p.61)
Back
2NaOH(aq) + H2SO4(aq)  Na2SO4(aq) + 2H2O(l)
Number of moles of NaOH(aq) 2
Number of moles of H2 SO 4 (aq) = 1
1
 Number of moles of NaOH(aq) = Number of moles of H2SO4(aq)
2
Number of moles of H2SO4(aq) = 0.067 mol dm-3  22.5  10-3 dm3
= 1.508  10-3 mol
Number of moles of NaOH(aq) = 2  1.508  10-3 mol
= 3.016  10-3 mol
3.016  10 -3 mol
Molarity of NaOH(aq) =
25.0  10 - 3 dm3
= 0.121 mol dm-3
Therefore, the molarity of the sodium hydroxide solution was 0.121 M.
144
3.6 Simple titrations (SB p.61)
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.
(a)Calculate the molarity of the acid solution.
(a) Number of moles of acid =
Molarity of acid solution =
145
2.52 g
= 0.02 mol
-1
126.0 g mol
0.02 mol
= 0.08 M
3
3
250  10 dm
Answer
3.6 Simple titrations (SB p.61)
(b) If the dibasic acid is represented by H2X, write an
equation for the reaction between the acid and sodium
hydroxide.
Answer
(b) H2X(aq) + 2NaOH(aq)  Na2X(aq) + 2H2O(l)
146
3.6 Simple titrations (SB p.61)
Back
(c) Calculate the molarity of the sodium hydroxide solution.
Answer
(c) Number of moles of H2X = 1  Number of moles of NaOH
2 3
-3
-3
0.08 mol dm  25.0  10 dm
1
=
 Molarity of NaOH  28.5  10-3 dm3
2
Molarity of NaOH = 0.14 M
Therefore, the molarity of the sodium hydroxide solution was 0.14 M.
147
3.6 Simple titrations (SB p.62)
0.186 g of a sample of hydrated sodium carbonate, Na2CO3 · nH2O,
was dissolved in 100 cm3 of distilled water in a conical flask.
0.10 M hydrochloric acid was added from a burette,
2 cm3 at a time. The pH value of the reaction mixture was
measured with a pH meter. The results were recorded and shown
in the following figure. Calculate the value of n in Na2CO3 · nH2O.
Answer
148
3.6 Simple titrations (SB p.63)
Back
There is a sudden drop in the pH value of the solution (from pH 8 to pH 3)
with the equivalence point at 30.0 cm3.
Na2CO3 · nH2O(s) + 2HCl(aq)  2NaCl(aq) + CO2(g) + (n + 1)H2O(l)
1
Number of moles of Na2CO3 · nH2O =
 Number of moles of HCl
2
0.186 g
-1
(23.0  2  12.0  16.0  3  18.0n) g mol1
=
 0.10 mol dm-3  30.0  10-3 dm3
2
106.0 + 18.0n = 124.0
n=1
Therefore, the chemical formula of the hydrated sodium carbonate is
Na2CO3 · H2O.
149
3.6 Simple titrations (SB p.63)
5 cm3 of 0.5 M sulphuric(VI) acid was added to 25.0 cm3 of
potassium hydroxide solution. The mixture was then stirred and
the highest temperature was recorded. The experiment was
repeated with different volumes of the sulphuric(VI) acid. The
laboratory set-up and the results were as follows:
150
Volume of H2SO4
added (cm3)
Temperature
(oC)
0
20.0
5
21.8
10
23.4
15
25.0
20
26.5
25
25.2
30
24.0
3.6 Simple titrations (SB p.63)
(a) Plot a graph of temperature against volume of sulphuric(VI)
acid added.
Answer
151
3.6 Simple titrations (SB p.63)
(b) Calculate the molarity of the potassium hydroxide solution.
(b) From the graph, it is found that the equivalence point of the
titration is
Answer
reached when 20 cm3 of H2SO4 is added.
Number of moles of H2SO4 = 0.5 mol dm-3  20  10-3 dm3
= 0.01 mol
2KOH(aq) + H2SO4(aq)  K2SO4(aq) + 2H2O(l)
2 mol
:
1 mol
From the equation,
mole ratio of KOH(aq) : H2SO4(aq) = 2 : 1
Number of moles of KOH(aq) = 2  0.01 mol = 0.02 mol
0.02 mol
Molarity of KOH(aq) =
= 0.8 M
3
3
25.0  10 dm
Therefore, the molarity of potassium hydroxide solution was 0.8 M.
152
3.6 Simple titrations (SB p.63)
Back
(c) Explain why the temperature rose to a maximum and then fell.
Answer
(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 would cool down the reacting
solution, causing the temperature to drop.
153
3.6 Simple titrations (SB p.66)
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 an
indicator. Find the molarity of the acidified potassium iodate
solution.
Answer
154
3.6 Simple titrations (SB p.66)
Back
IO3-(aq) + 5I-(aq) + 6H+(aq)  3I2(aq) + 3H2O(l) … … (1)
I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62-(aq) … … (2)
From (1), Number of moles of IO3-(aq) =
1
 Number of moles of I2(aq)
3
1
From (2), Number of moles of I2(aq) =
 Number of moles of S2O32-(aq)
2
1
 Number of moles of S2O32-(aq)
6
1
-3
3
Molarity of IO3 (aq)  25.0  10 dm =
 0.05 mol dm-3  22.0  10-3 dm3
6
Number of moles of IO3-(aq) =
Molarity of IO3-(aq) = 7.33  10-3 M
Therefore, the molarity of the acidified potassium iodate solution is
7.33  10-3 M.
155
3.6 Simple titrations (SB p.67)
A piece of impure iron wire weighs 0.22 g. 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)
solution for complete reaction to form iron(III) ions. What is the
percentage purity of the iron wire?
Answer
156
3.6 Simple titrations (SB p.67)
Back
MnO4-(aq) + 8H+(aq) + 5Fe2+(aq)  Mn2+(aq) + 4H2O(l) + 5Fe3+(aq)
Number of moles of MnO4-(aq) : Number of moles of Fe2+(aq) = 1 : 5
Number of moles of Fe2+(aq) = 5  Number of moles of MnO4-(aq)
= 5  0.02 mol dm-3  36.5  10-3 dm3
= 3.65  10-3 mol
Number of moles of Fe dissolved = Number of moles of Fe2+ formed
= 3.65  10-3 mol
Mass of Fe = 3.65  10-3 mol  55.8 g mol-1 = 0.204 g
0.204 g
Percentage purity of Fe =
 100 % = 92.73 %
0.22 g
Therefore, the percentage purity of the iron wire is 92.73 %.
157
3.6 Simple titrations (SB p.67)
(a) 5 g 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?
(Molar volume of gas at R.T.P. = 24.0 dm3 mol-1)
Answer
158
3.6 Simple titrations (SB p.67)
Na2CO3(s) + 2HCl(aq)  2NaCl(aq) + H2O(l) + CO2(g)
5g
No. of moles of Na2CO3 used =
(23.0  2  12.0  16.0  3) g mol -1
= 0.0472 mol
100
dm3
No. of moles of HCl used = 2 M 
1000
= 0.2 mol
Since HCl is in excess, Na2CO3 is the limiting agent.
No. of moles of CO2 produced = No. of moles of Na2CO3 used
= 0.0472 mol
Volume of CO2 produced = 0.0472 mol  24.0 dm3 mol-1
= 1.133 dm3
159
3.6 Simple titrations (SB p.67)
(b) 8.54 g of impure hydrated iron(II) sulphate (formula mass
of 392.14) was dissolved in water and made up to
250.0 cm3. 25.0 cm3 of this solution required 20.76 cm3 of
0.020 3 M acidified potassium manganate(VII) solution
for complete reaction. Determine the percentage purity of
the hydrated iron(II) sulphate.
Answer
160
3.6 Simple titrations (SB p.67)
Back
(b) MnO4-(aq) + 8H+(aq) + 5Fe2+(aq)  Mn2+(aq) + 4H2O(l) + 5Fe3+(aq)
20.76
dm 3
1000
-4
= 4.214  10 mol
No. of moles of MnO4- ions = 0.0203 M 
No. of moles of Fe2+ ions = 5  No. of moles of MnO4- ions
= 2.107  10-3 mol
No. of moles of Fe2+ ions in 25.0 cm3 solution = 2.107  10-3 mol
No. of moles of Fe2+ ions in 250.0 cm3 solution = 0.02107 mol
Molar mass of hydrated FeSO4 = 392.14 g mol-1
Mass of hydrated FeSO4 = 0.02107 mol  392.14 g mol-1 = 8.26 g
8.26 g
 100% = 96.72 %
% purity of FeSO4 =
8.54 g
161