Chapter 3
Chemical Equations
and Stoichiometry
1
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
New Way Chemistry for Hong Kong A-Level Book 1
3.1
Formulae of Compounds (SB p.54)
Formulae of Compounds
How can you describe the composition of
compound X?
1st way = by chemical formula
C?H?
ratio of no. of atoms
2
New Way Chemistry for Hong Kong A-Level Book 1
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
carbon atoms
hydrogen atoms
3
Mass of hydrogen atoms inside
= …. g
New Way Chemistry for Hong Kong A-Level Book 1
3.1
Formulae of Compounds (SB p.54)
Check Point 3-1
Give
the empirical
molecular
and structural
formula
Compound
Empirical
Molecular
Structural
for the following
compounds
formula
formula
formula
(a) Propene
CH2
C3H6
H
H
H
H
C
C
C
H
H
(b) Nitric
acid
(c) Ethanol
(d) Glucose
HNO3
HNO3
O
H O
N
O
C2H6O
C6H12O6
C2H5OH
C6H12O6
H
H
H
C
C
H
H
OH
O H
H
H
OH
H
H
OH
HO
Answer
4
New Way Chemistry for Hong Kong A-Level Book 1
OH
OH
3.1
Formulae of Compounds (SB p.55)
The different types of formulae of some compounds
Compound
Carbon
dioxide
Water
Methane
Empirical
formula
CO2
Molecular
formula
CO2
Structural formula
H2O
H2O
O
CH4
CH4
O = C =O
H
H
H
H
C
H
H
Glucose
CH2O
C6H12O6
OH
O H
H
H
OH
H
HO
OH
H
Sodium
fluoride
5
NaF
Not applicable
New Way Chemistry for Hong Kong A-Level Book 1
OH
Na+F-
3.2 Derivation of Empirical Formulae (SB p.56)
Solution:
Example 3-1
The
relativewas
molecular
mass of in
CO
2
A hydrogen
burnt completely
excess
oxygen.
It was found that 1.00 g of the hydrocarbon gives
=
12.0
2 x 16.0
= 44.0
2.93
g of+ carbon
dioxide
and 1.8 g of water. Find the
empirical
formulainof2.93
the hydrocarbon.?
(R.a.m.* : H
Mass of carbon
g of CO2
= 1.0, C = 12.0, O = 16.0)
= 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
6
New Way Chemistry for Hong Kong A-Level Book 1
Answer
3.2 Derivation of Empirical Formulae (SB p.57)
Solution: (cont’d)
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.
7
New Way Chemistry for Hong Kong A-Level Book 1
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 0.066 7/0.066 7=1
number of
moles
Simplest
1
mole ratio
8
0.20/0.066 7
=3
New Way Chemistry for Hong Kong A-Level Book 1
3
3.2 Derivation of Empirical Formulae (SB p.57)
Solution:
Example
3-2
Mass of X
compound
X contain
= 0.46gcarbon, hydrogen and
Compound
is known to
oxygen only. When it is burnt completely in excess
Masscarbon
of carbon
in compound
X given out as the
oxygen,
dioxide
and water are
only
products.
It is found =that
0.46g g of compound X
= 0.88
g x 12.0/44.0
0.24
gives 0.88 g of carbon dioxide and 0.54 g of water. Find
theMass
empirical
formula ofincompound
X.X(R.a.m.* : H = 1.0,
of hydrogen
compound
C = 12.0, O = 16.0)
= 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
Answer
9
New Way Chemistry for Hong Kong A-Level Book 1
3.2 Derivation of Empirical Formulae (SB p.57)
Solution: (cont’d)
Let the empirical formula of compound X be CxHyOz.
Mass (g)
Number of moles
(mol)
Relative number
of moles
Simplest mole
ratio
Carbon
0.24
Hydrogen
0.06
Oxygen
0.16
0.02
0.06
0.01
2
6
1
2
6
1
Therefore, the empirical formula of compound X is
C2H6O.
10
New Way Chemistry for Hong Kong A-Level Book 1
3.1
Formulae of Compounds (SB p.58)
Check(a)Mass
Point of
3-2sulphur = 5 g
(a) 5 g of sulphur forms 10 g of an oxide on burning.What is the
empirical
formula
of the oxide?
Mass
of oxygen
= (10 – 5) g
(R.a.m. : O = 16.0, S = 32.1)
Sulphur
Oxygen
(b) 19.85
f of(g)
element M combines
with 25.61 g 5of oxygen to
Mass
5
form an oxide. If the relative atomic mass of M is 331.0,
find the empirical formula of the oxide.
Number of moles
5/ 32.1
5/16.0
(R.a.m.(mol)
: O =16.0)
= 0.156
= 0.313
(c) Relative
Determine
the empirical
formula of copper(II)
oxide
using
Number
0.156/0.156=1
0.313/0.156
=2
the following results.
of moles
Experimental
results:
Simplest mole
1
2
Mass of test
tube = 21.430 g
ratio
Mass of test tube + Mass of copper(II) oxide = 23.321g
The empirical formula of the sulphur oxide
Mass of test tube + Mass of copper = 22.940g
is SO2.
Answer
(R.a.m. : Cu = 63.5, O = 16.0)
11
New Way Chemistry for Hong Kong A-Level Book 1
3.1
Formulae of Compounds (SB p.58)
(b)
M
O
Mass (g)
19.85
25.61
Number of
moles (mol)
19.85/31.0
= 0.64
0.64/0.64
=1
25.61/16.0
= 1.6
1.6/0.64
= 2.5
2
5
Relative
number of
moles
Simplest mole
ratio
The empirical formula of the oxide is M2O5.
12
New Way Chemistry for Hong Kong A-Level Book 1
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
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
Simplest mole
ratio
0.0238/0.0238 =1
0.0238/0.0238
=1
1
1
The empirical formula of the oxide is CuO.
13
New Way Chemistry for Hong Kong A-Level Book 1
3.2 Derivation of Empirical Formulae (SB p.58)
Determination of Empirical Formula
From Combustion by Mass
Composition by mass
Empirical formula
14
New Way Chemistry for Hong Kong A-Level Book 1
3.2 Derivation of Empirical Formulae (SB p.58)
Solution:
Example3-3
Let
the empirical formula of the hydrocarbon be CxHy, and
the
mass of the
compound
be 100
g.
Compound
A contains
carbon
and hydrogen
only. It is
found
the compound
contains
75%
Mass
of that
carbon
in the compound
= 75
g carbon by mass.
Determine
its empirical
formula. (Relative
atomic
Mass
of hydrogen
in the compound=(100
–75)
g = 25 g
masses: C=12, H=1Carbon
)
Hydrogen
Mass (g)
75
25
Number of
75/12.0 = 6.25 25/1.0 = 25
moles (mol)
Relative number 6.25/6.25 = 1 25/6.25 = 4
of moles
Simplest mole
1
4
Answer
ratio
Therefore, the empirical formula of the hydrocarbon is CH4.
15
New Way Chemistry for Hong Kong A-Level Book 1
3.2 Derivation of Empirical Formulae (SB p.59)
Solution:
Example
Let the mass3-4
of phosphorus chloride be 100g. Then,
The
by mass
of phosphorus
and
chlorine in a
Masspercentage
of phosphorus
in the
compound =
22.55g
sample of a phosphorus chloride are 22.55% and 77.45%
Mass
of chloride
the
compound
= 77.45g
respectively.
Findinthe
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
0.727/0.727
=1
1
77.45/35.5
=2.182
2.182/0.727 =
3
3
Relative number of
moles
Simplest mole ratio
Therefore, the empirical formula of the phosphorus
Answer
chloride is PCl3.
16
New Way Chemistry for Hong Kong A-Level Book 1
3.2 Derivation of Empirical Formulae (SB p.59)
Check Point 3-3
(a) Let the mass of vitamin C analyzed be 100g.
(a) Find the empirical formula of vitamin C if it consists
Carbonoxygen
Hydrogen
of 40.9% caarbon, 54.5%
and 4.6%Oxygen
hydrogen
by mass. ( R.a.m.:
C = 12.0,
Mass (g)
40.9
4.6 H = 1.0,
54.5O =
16.0)
of moles
4.6/1.0 of 54.5/16.0
(b) Number
Each 325
mg tablet40.9/12.0
of aspirin consists
195.0 mg
(mol)
= 3.41 and 115.4mg
=4.60 oxygen.
=3.41
carbon
14.6 mg hydrogen
Determine
theofempirical
formula
of aspirin.
(R.a.m. :
Relative
number
3.41/3.41
4.61/3.41
3.41/3.41
C= 12.0,
16.0)
molesH = 1.0, O ==1
=1.35
=1
Simplest mole ratio
3
4
3
The empirical formula of vitamin C is C3H4O3.
Answer
17
New Way Chemistry for Hong Kong A-Level Book 1
3.2 Derivation of Empirical Formulae (SB p.59)
(b) In order to facilitate calculation, the masses of
the elements are multiplied by 1000 first.
Mass (g)
Number of
moles (mol)
Relative
number of
moles
Simplest mole
ratio
Carbon
195.0
Hydrogen
14.6
Oxygen
115.4
195.0/12.0
=16.25
16.25/7.21
= 2.25
14.6/7.21
=2.02
14.6/7.21
= 2.02
7.21/7.21
=1
7.21/7.21
=1
9
8
4
The empirical formula of aspirin is C9H8O4.
18
New Way Chemistry for Hong Kong A-Level Book 1
3.3 Derivation of Molecular Formulae (SB p.60)
What is Molecular Formulae?
Molecular formula
?
= (Empirical formula)n
19
New Way Chemistry for Hong Kong A-Level Book 1
3.3 Derivation of Molecular Formulae (SB p.60)
Empirical formula
Molecular mass
Molecular formula
20
New Way Chemistry for Hong Kong A-Level Book 1
3.3 Derivation of Molecular Formulae (SB p.60)
Solution: 3-5
Example
Let the empirical formula of the hydrocarbon be CxHy.
A hydrogen was burnt completely in excess oxygen. It
Mass
of carbon
in the
was
found
that 5.00
g ofhydrocarbon
the hydrocarbon gives 14.6 g of
carbon
and 9.0
g of water. Given that the relative
= 14.6gdioxide
x 12.0/44.0
= 4.0g
molecular
mass of the
hydrocarbon
is 30.0, determine its
Mass of hydrogen
in the
hydrocarbon
molecular formula. hydrocarbon.? (R.a.m.* : H = 1.0, C =
= 9.0g
12.0,
O x=2.0/18.0
16.0) = 1.0g
Mass (g)
Number of moles
(mol)
Relative number of
moles
Simplest mole ratio
21
Carbon
4.0
4.0/12.0
= 0.333
0.333/0.333 = 1
Hydrogen
1.0
1.0/1.0 = 1
1
3
New Way Chemistry for Hong Kong A-Level Book 1
1/0.333 = 3
Answer
3.3 Derivation of Molecular Formulae (SB p.60)
Solution: (cont’d)
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.
22
New Way Chemistry for Hong Kong A-Level Book 1
3.3 Derivation of Molecular Formulae (SB p.61)
Solution:
Example 3-6
Let the empirical formula of the hydrocarbon be CxHyOz.
Mass
of carbon
in the compound
44.44gcarbon, 6.18%
Compound
X is known
to contain =
44.44%
hydrogen
and 49.38%
oxygen
by mass.
A typical analysis
Mass of hydrogen
in the
compound
= 6.18g
shows that it has a relative molecular mass of 162.0. Find
Mass
of oxygen
in the compound
49.38g
its molecular
formula(R.a.m.*
: H == 1.0,
C = 12.0, O =
16.0)
Mass (g)
Number of
moles (mol)
Relative
number of
moles
Simplest mole
ratio
23
Carbon
44.44
44.44/12.0
= 3.70
3.70/3.09
= 1.2
Hydrogen
6.18
6.18/1.0
= 6.18
6.18/3.09
=2
Oxygen
49.38
49.38/16.0
= 3.09
3.09/3.09 =
1
6
10
5
New Way Chemistry for Hong Kong A-Level Book 1
Answer
3.3 Derivation of Molecular Formulae (SB p.61)
Solution(cont’d)
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.
24
New Way Chemistry for Hong Kong A-Level Book 1
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
25
New Way Chemistry for Hong Kong A-Level Book 1
3.3 Derivation of Molecular Formulae (SB p.61)
Example 3-7
Solution:
TheRelative
chemicalmolecular
formula of
hydrated
copper(II)
sulphate is
Let
mass
of CuSO
4xH2O
known to be CuSO4.xH2O. It is found that the percentage
=of63.5
+ 32.1
+ 16.0
x 4compound
+ (1.0x2 =is16.0)x
water
by mass
in the
36%. Find x.
=(R.a.m.
159.6 :+H=1.0,
18x
O=16.0, S=32.1, Cu=63.5)
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
Answer
Therefore, the chemical formula of hydrated copper(II)
sulphate is CuSO4 5H2O
26
New Way Chemistry for Hong Kong A-Level Book 1
3.3 Derivation of Molecular Formulae (SB p.63)
Check Point 3-4
(a) (i) Let the mass of compound Z be 100g.
(a) Find Compound Z isCarbon
the majorHydrogen
component Oxygen
of a
healthy
drink. It contains
40.00%6.67
carbon, 6.67%
Mass (g)
40.00
53.33
hydrogen and 53.33% oxygen.
(i) Find
the empirical
of compound
Number
of moles formula
40.00/12.0
6.67/1.0 Z.
53.33/16.0=
(mol)
= 3.33
= 6.67
3.33
(ii) If the relative molecular mass of compound Z is 180,
Relative
number
3.33/3.33
6.67/3.33
3.33/3.33
finds
its molecular
formula.(R.a.m.
: C= 12.0,
H=
of =
moles
=1
=2
=1
1.0, O
16.0)
Simplest mole
1
2
1
ratio
The empirical formula of compound Z is CH2O.
Answer
27
New Way Chemistry for Hong Kong A-Level Book 1
3.3 Derivation of Molecular Formulae (SB p.63)
(a) (ii) Let the molecular formula of
compound Z be (CH2O)n.
n x (12.0 = 1.0 x 2 = 16.0) = 180
30n = 180
n=6
The molecular formula of Z is C6H12O6.
28
New Way Chemistry for Hong Kong A-Level Book 1
3.3 Derivation of Molecular Formulae (SB p.63)
Check
Point 3-4
(b)
(NH4sulphur
) unit by mass
S is water.
(b) (NH4)2Sx contains 72.72%
(g) of x.
27.28
72.72
FindMass
the value
(R.a.m.:
H =of
1.0, N = 14.0,
O = 16.0)
Number
27.28/18.0
72.72/32.1
moles (mol)
1.52O, 51.22%
=2.27
(c) In the compound MgSO4=nH
by mass is
2
Relative
1.52/1.52
2.27/1.52
water.
Find the value of
n.
number of
=1
=1.49
moles
(R.a.m.:
H = 1.0, O = 16.0, Mg = 24.3, S = 32.1)
Simplest mole
2
3
ratio
Since the chemical formula of (NH4)2Sx is (NH4)2S3,
the value of x is 3.
Answer
29
New Way Chemistry for Hong Kong A-Level Book 1
3.3 Derivation of Molecular Formulae (SB p.63)
(c)
Mass (g)
Number of moles
(mol)
Relative number
of moles
Simplest mole
ratio
MgSO4
48.78
H2O
51.22
48.78/120.4
=0.405
0.405/0.405
=1
1
51.22/18.0
=2.846
2.846/0.405
=7
7
Since the chemical formula of MgSO4nH2O
is MgSO47H2O , the value of x is 7.
30
New Way Chemistry for Hong Kong A-Level Book 1
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
Solution:
and oxygen by mass respectively. (R.a.m. : C=12.0,
RelativeO=16.0
molecular
mass of CH3COOH
H=1.0,
)
= 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
31
New Way Chemistry for Hong Kong A-Level Book 1
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 , H=1.0, O=16.0 )
Solution:
Relative molecular mass of FeSO4·7H2O
= 55.8 + 32.1 + 16.0 x 4 + (1.0x2+16.0) x 7=277.9
% by mass of Fe = 55.8/277.9 x 100% = 20.08%
Mass of Fe = 20g x 20.08% = 4.02g
Answer
32
New Way Chemistry for Hong Kong A-Level Book 1
3.3 Derivation of Molecular Formulae (SB p.63)
Check Point 3-5
(a) Molar mass of K2Cr2O7
(a) Calculate percentages by mass of potassium,
chromium
and oxygen in potassium
chromate
= (39.1x2+52.0+16.0x7)
g mol-1 = 294.2
g mol-1(VI),
K2Cr2O7.(R.a.m. : K = 39.1 . Cr = 52.0, O = 16.0)
% by mass of K
(b) Find the mass of metal and water of crystallization in
= 39.1 x 2 g mol-1/294.2 g mol-1 x 100% = 26.58%
(i) 100 g of Na2SO4·10H2O;
% by mass of Cr
(ii) 70g of Fe2O3·8H2O.
-1/294.2g mol-1 x 100% =35.25%
=
52.0
x
2
g
mol
(R.a.m.: H = 1.0, O = 16.0, Na = 23, S = 32.1, Fe = 55.8)
% by mass of O
= 16.0 x 7 g mol-1/294.2g mol-1 x 100% = 38.07%
Answer
33
New Way Chemistry for Hong Kong A-Level Book 1
3.3 Derivation of Molecular Formulae (SB p.63)
(b)( i) Molar mass of Na2SO4·10H2O= 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 Fe2O3·8H2O= 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
34
New Way Chemistry for Hong Kong A-Level Book 1
3.4 Chemical Equations (SB p.64)
Chemical Equations
aA+bB cC+dD
mole
ratios
(can also be
volume ratios for
gases)
Stoichiometry
= relative no. of moles of substances involved
in a chemical reaction.
35
New Way Chemistry for Hong Kong A-Level Book 1
3.4 Chemical Equations (SB p.64)
Check Point 3-6
Give the chemical equations for the following reactions:
(a) Zinc + steam
zinc oxide + hydrogen
(a) Zn(s) + H2O(g)
ZnO(s) + H2(g)
(b) Magnesium + silver nitrate
(b) Mg(s) + 2 AgNO3(aq)
2Ag(s) + Mg(NO3)2(aq)
silver + magnesium nitrate
(c) 2C+4H
10H2O(l)
10(g) + 13O2(g)
2(g)++water
(c) Butane
oxygen
carbon 8CO
dioxide
Answer
36
New Way Chemistry for Hong Kong A-Level Book 1
3.5 Calculations Based on Equations (SB p.65)
Calculations Based on Equations
Calculations involving Reacting Masses
37
New Way Chemistry for Hong Kong A-Level Book 1
3.5 Calculations Based on Equations (SB p.65)
Solution:
Example 3-10
CuO(s) + H2(g)
Cu(s) + H2O(l)
Calculate the mass of copper formed when 12.45g of
As
the moleoxide
ratioisofcompletely
Cu : CuOreduced
is 1 : 1,bythe
number of
copper(II)
hydrogen.
moles
formed
is theCu
same
as) the number of
(R.a.m.of: Cu
H=1.0,
O=16.0,
= 63.5
moles of CuO reduced.
Number of moles of CuO reduced
= 12.45/ (63.5 + 13.0) g mol-1 = 0.157 mol
Number of mole of Cu formed = 0.157 mol
Mass of Cu / 63.5 g mol-1 = 0.157
Mass of Cu = 0.157 mol x 63.5 g mol-1= 9.97g
Therefore, the mass of copper formed in the reaction
is 9.97g.
Answer
38
New Way Chemistry for Hong Kong A-Level Book 1
3.5 Calculations Based on Equations (SB p.65)
Example 3-11
Solution:
Number
moles of CO2 formed
Sodium of
hydrogencarbonate
decomposes according to the
equation. 3 mol-1 = 0.01 mol
=following
240cm3/ 24000cm
2NaHCO
+ CO2(g)
+ H O(l)
From
the equation,
2 Na
moles
NaHCO
3(s)
2COof
3(s)
3(s) will2form 1
mole of CO2(g).
In order to obtain 240 cm3 of CO2 at room temperature
Number
of moles
ofisNaHCO
3 required
and pressure,
what
the minimum
amount of sodium
=hydrogencarbonate
0.01 x 2 = 0.02 molrequired?
Mass
of NaHCO
(R.a.m.
: H = 1.0,
C =12.0, O = 16.0, Na = 23.0;
3 required
-1
= 0.02
mol
x(23.0
+
1.0
+
16.0
x
3)
g
mol
molar volume of gas at R.T.P. = 24.0 dm3mol-1)
= 0.02 mol x 84.0g mol-1
= 1.68 g
Answer
Therefore, the minimum amount of sodium
hydrogencarbonate required is 1.68g.
39
New Way Chemistry for Hong Kong A-Level Book 1
3.5 Calculations Based on Equations (SB p.66)
Calculations Based on Equations
Calculations involving Volumes of Gases
40
New Way Chemistry for Hong Kong A-Level Book 1
3.5 Calculations Based on Equations (SB p.66)
Solution:
Example 3-12
Number
of the
moles
of COof
2 formed
Calculate
volume
carbon dioxide formed when
3 + ethane
C
7O2(g) and 704CO
+ 6H2O(l)
202Hcm
cm23(g)
of oxygen
are exploded,
6(g) of
assuming
volumes :are
measured
room
2 mol
: all
7 mol
4 mol
: 6 at
mol
temperature and pressure.
(from equation)
2 volumes: 7 volumes : 4 volumes : (by Avogadro’s law)
It can be judged from the equation that the mole ratio of CO2 :
C2H6 is 4 :2, and the volume ratio of CO2 : C2H6 should also be
4:2.
Let x be the volume of CO2(g) formed
x /20cm3 = 4/2
x = 40 cm3
Answer
Therefore, the volume of CO2 formed is 40 cm3.
41
New Way Chemistry for Hong Kong A-Level Book 1
3.5 Calculations Based on Equations (SB p.67)
Solution:
Example
3-13
Let the molecular formula of the hydrocarbon be CxHy.
3
10Volume
cm3 ofofa hydrogen
gaseous hydrocarbon
reacted = 10 was
cm3 mixed with 80cm
ofVolume
oxygenofwhich
was in excess. The mixture was
O2(g) unreacted = 50 cm3
exploded and then cooled. The volume left was 70cm3.
Volume
of O2(g)
= 30
cm3 mixture through
Upon
passing
the reacted
resulting
gaseous
concentrated
sodium
hydroxide
solution
( to absorb
Volume of CO
= 20 cm
2(g) formed
3
carbon
of the
residual
gas
became
CxHy3 dioxide),
+ (x +the
y/4)volume
O2
CO
+
y/2
H
O
2
50 cm . Find the molecular formula 2of the hydrocarbon.
1 volume : (x + y/4) volumes : x volumes
Volume of CO2 (g)/ volume of CxHy(g)
= 20 cm3/ 10cm3 = 2
X =2
Volume of O2(g) / volume of CxHy(g)
=(x + y/4) / 1= 30/ 10
Answer
(x + y/4)= 3
Y=2
Molecular formula is C2H4
42
New Way Chemistry for Hong Kong A-Level Book 1
3.5 Calculations Based on Equations (SB p.68)
Check Point 3-7
(a) Find
the volume
ofof
hydrogen
produced
at R.T.P.
when
(a)
No.
of
moles
H
=
No.
of
moles
of
Mg
2
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
3 mol
-1 = 2.43
-1
3
-1
Volume
of
H
/
24.0
dm
g
/
24.3
g
mol
2
dm mol .
3 chlorine required to produced
(b) Find
the minimum
mass
Volume
of H2 = 2.4
dmof
100 g of phosphorus trichloride ( PCl3).
3 of oxygen
(b)cm3
1/3 of
x no.
of moles
of Cl2 = 1/2x
no. cm
of moles
of PCl3
(c) 20
a gaseous
hydrocarbon
and 150
were exploded in a closed vessel. After cooling, 110 cm3 of
-1
gases1/3
remained.
a solution
of
x mass After
of Cl2passing
/ (35.5 through
x 2) g mol
concentrated sodium hydroxide, the volume left was 50
3 . Determine the molecular of the hydrocarbon.
cm
= 1/2 x 100g / (31.0 + 35.5 + 3 ) g mol-1
(d) Calculate the volume of carbon dioxide formed when
3 of methane burns in excess oxygen, assuming all
5cm
Mass
of Cl2 = 77.45g
volumes are measured at room temperature and pressure.
Answer
43
New Way Chemistry for Hong Kong A-Level Book 1
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
44
New Way Chemistry for Hong Kong A-Level Book 1
3.5 Calculations Based on Equations (SB p.68)
(d)Volume of CxHy used = 20 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 / 5 = 1/1
x=5
The volume of carbon dioxide gas is 5 cm3.
45
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple Titrations (SB p.68)
Simple Titrations
Acid-Base Titrations
Acid-Base Titrations
with Indicators
Acid-Base Titrations
without Indicators
(to be discussed
in later chapters)
46
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple Titrations (SB p.69)
Finding the concentration of a solution
Copper(II)
sulphate
solute
+
Water
solvent
Copper(II)
sulphate
solution
47
solution
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple Titrations (SB p.69)
Finding the concentration of a solution
~50 cm3
48
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple Titrations (SB p.69)
Finding the concentration of a solution
~50 cm3
49
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple Titrations (SB p.69)
Finding the concentration of a solution
~50 cm3
50
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple Titrations (SB p.69)
Finding the concentration of a solution
~50 cm3
51
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple Titrations (SB p.69)
Finding the concentration of a solution
~50 cm3
52
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple Titrations (SB p.69)
Finding the concentration of a solution
~50 cm3
53
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple Titrations (SB p.69)
Finding the concentration of a solution
50 cm3
Solution A
54
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple Titrations (SB p.69)
Finding the concentration of a solution
~50 cm3
55
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple Titrations (SB p.69)
Finding the concentration of a solution
~50 cm3
56
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple Titrations (SB p.69)
Finding the concentration of a solution
~50 cm3
57
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple Titrations (SB p.69)
Finding the concentration of a solution
~50 cm3
58
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple Titrations (SB p.69)
Finding the concentration of a solution
~50 cm3
59
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple Titrations (SB p.69)
Finding the concentration of a solution
~50 cm3
60
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple Titrations (SB p.69)
Finding the concentration of a solution
50 cm3
Solution B
61
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple Titrations (SB p.69)
Finding the concentration of a solution
~100 cm3
62
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple Titrations (SB p.69)
Finding the concentration of a solution
~100 cm3
63
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple Titrations (SB p.69)
Finding the concentration of a solution
~100 cm3
64
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple Titrations (SB p.69)
Finding the concentration of a solution
~100 cm3
65
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple Titrations (SB p.69)
Finding the concentration of a solution
~100 cm3
66
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple Titrations (SB p.69)
Finding the concentration of a solution
~100 cm3
67
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple Titrations (SB p.69)
Finding the concentration of a solution
100 cm3
Solution C
68
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple Titrations (SB p.69)
Comment on the Concentrations of Solutions
A, B and C !
Concentration of solution B is 2 times that of
the concentrations of solutions A & B.
2 x the amount of solute
contain the same amount of solute
(same concentration)
Concentration is the amount of solute in a unit volume of solution.
69
New Way Chemistry for Hong Kong A-Level Book 1
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.
70
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple Titrations (SB p.69)
Class Practice Answers
71
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple Titrations (SB p.69)
Comment on the Concentrations of Solutions
A, B and C !
no. of
spoons
mass
no. of
moles
Concentration is the amount of solute in a unit volume of solution.
72
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple Titrations (SB p.69)
Molarity
A way of expressing concentrations
Molarity is the number of moles of solute dissolved
in 1 dm3 (1000 cm3) of solution.
number of moles of solute
Molarity 
3
volume of solution (in dm )
Unit: moles/dm3 (M)
73
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple Titrations (SB p.69)
What does this mean?
1 dm3
contains 2
moles of
HCl
“In every 1 dm3 of the solution, 2 moles of HCl
is dissolved.”
74
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple Titrations (SB p.71)
Solution:
Example
3-14
Number
of moles
of NaOH(aq)
2
= 1 was titrated
25.0cmof3 of
sodium
hydroxide
solution
Number
moles
of H
SO
(aq)
2
4
against 0.067 M of sulphuric(VI) acid using methyl
½ xorange
Number
of molesThe
of NaOH(aq)
as indicator.
indicator changed colour from
3 of sulphuric(VI) acid had
yellow
to
red
when
22.5
cm
= Number of moles of H2SO4 (aq)
benn added. Calculate
the molarity
of the sodium-3
-3
-3
3
= 0.067
mol dm
x 22.5 x 10 dm = 1.508 x 10 mol
hydroxide
solution.
Number of moles of NaOH(aq) = 2 x 1.508 x 10-3 mol
= 3.016 x 10-3 mol
Molarity of NaOH(aq)
= 3.012 x 10-3 mol / 25.0 x 10-3 mol
= 0.1221 mol dm-3
Answer
The molarity of NaOH is 0.121M
75
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple Titrations (SB p.71)
Example 3-15
Solution:
2.52 g of a pure dibasic acid with formula mass of 126.0
-1
3
(a)Number
of
moles
of
acid
=
2.52
g/
126.0
g
mol
was dissolved in water and made up to 250.0 cm in a
volumetric flask 25.0 cm3 of=this
solution
was found to
0.02
mol
neutralize 28.5 cm3 of sodium hydroxide solution.
Molarity of acid solution = 0.02 mol / 250 x 10-3 dm3
(a) Calculate the molarity of the acid solution.
= 0.08M
(b) If
the
dibasic
acid is representedNa
by H
an
2X, write
(b)
H
X(aq)
+
2NaOH(aq)
X(aq)
+
2NaOH(l)
2
2 acid and sodium
equation
for the reaction between the
(c) hydroxide.
Number of moles of H2X
=½ x number
of moles
ofsodium
NaOH hydroxide solution.
(c) Calculate
the molarity
of the
Molarity of NaOH = 0.14M
Answer
76
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple Titrations (SB p.72)
Example 3-16
Solution:
0.186g of sample
of hydrate
(a)Number
of moles
of acid = 2.52 g / 126.0 g mol-1
sodium carbonate, NaCO2·nH2O,
There
is a sudden
was dissolved
in 100drop
cm3 in
of the pH value of the solution
3.
(from
pH
3
to
pH
8)
with
the
end
point
at
30.0
cm
distilled water in conical flask.
0.10
M by hydrochloric acid was
Na
2CO3·nH2O(s) + 2 HCl(aq)
added from a burette, 2 cm3 at a
2NaCl(aq)
+ CO2(g) + (n+1)H2O(l)
time. The pH value of
the
solution of
was
measured
a pH
Number
moles
of Naby
2CO3·nH2O
meter. The result was
recorded-3
-3
3
=and
½ xshown
0.1 mol
dm
x
30
x
10
dm
in the following
figure.
106.0 + 18.0n = 124.0
Calculate the valuenof= n1in
NaCO2·nH2O.
Answer
The formula
is Na2CO3·H2O
77
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple Titrations (SB p.73)
Solution:
Example
3-17
(a)
3 of 0.5M sulphuric(VI) acid was added
5 cm
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:
(a) Plot the graph of temperature against
volume of sulphuric(VI) acid added.
(b) Calculate the molarity of the potassium
hydroxide solution.
(c) Explain why the temperature rose to a
maximum and the fell.
78
New Way Chemistry for Hong Kong A-Level Book 1
Answer
3.6 Simple Titrations (SB p.74)
Solution: (cont’d)
(b) From the graph, it is found that the end point of the
titration is reached when 20 cm3 of H2SO4 is added.
Number of moles of = 0.5 mol dm-3 x 20/1000 dm3
= 0.01 mol
2KOH(aq) + H2SO4(aq)
K2SO4(aq) + 2H2O(l)
2 mol
1 mol
Mole of KOH(aq) : H2SO4 = 2 : 1
Number of moles of KOH(aq)
= 2 x 0.01 mol = 0.02 mol
Molarity of KOH(aq)
= 0.02 mol / (25 x 10-3 dm3) = 0.8M
79
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple Titrations (SB p.74)
Solution: (cont’d)
(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.
80
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple Titrations (SB p.76)
Redox Titrations
Some Examples
in
Iodometric Titrations
conical
flask
in
burette
I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62-(aq)
brown
81
colourless
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple Titrations (SB p.76)
Redox Titrations
Some Examples
Add starch
Iodometric Titrations
in conical flask
in burette
I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62-(aq)
brown
colourless
During titration : brown  yellow
82
New Way Chemistry for Hong Kong A-Level Book 1
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
New Way Chemistry for Hong Kong A-Level Book 1
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)
84
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple Titrations (SB p.76)
Solution: 3-18
Example
IO3-(aq) + 5I- + 6H+(aq)
3H2O(l) …(1)
2(aq) + (KI)
When excess potassium iodide3Isolution
is added to
I2(aq)cm
+32S
O3 2-(aq) potassium
2I-(aq)
+ S4solution
O6 2-(aq)(KIO
…(2)
25.0
of2acidified
iodate
3) of
3-(aq)
unknown
the solution
turns brown. This
From (1),concentration,
Number of moles
of IO
brown solution requires 22.0 cm3 of 0.05 M sodium
=1/3 x number
of moles
of I2(aq)
thiosulphate
solution
to react
completely with the iodine
formed,
starch
indicator. Find the
From(2),using
Number
of solution
moles ofasI2(aq)
molarity
of the potassium
2-(aq)
=1/2 x number
of moles ofiodate
S2O3solution.
Number of moles of IO3-(aq)
= 1/6 x number of moles of S2O3 2-(aq)
Molority of IO3-(aq) x 25.0/1000 dm3
=1/6 x 0.05 mol dm-3 x 22.0/1000 dm3
Answer
Molarity of IO3-(aq) = 7.33 x 10-3 M
85
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple Titrations (SB p.76)
Redox Titrations
In burette
Some Examples
In conical
flask
Titrations Involving Potassium Permanganate
MnO4-(aq) + 8H+(aq) + 5Fe2+(aq) Mn2+(aq) + H2O(aq) + 5Fe3+(aq)
purple
86
colourless
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple Titrations (SB p.76)
Redox Titrations
In burette
Some Examples
In conical
flask
Titrations Involving Potassium Permanganate
MnO4-(aq) + 8H+(aq) + 5Fe2+(aq) 
Mn2+(aq) + H2O(aq) + 5Fe3+(aq)
purple
colourless
During titration : pale green  yellow
87
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple Titrations (SB p.76)
Redox Titrations
In burette
Some Examples
In conical
flask
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
88
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple Titrations (SB p.77)
Solution:
Example 3-19
MnO4-(aq) + 8H+(aq) + 5Fe2+(aq)
A piece of impure iron wire
weighs 0.22g. When 3+
it is
2+
Mnacid,
(aq)it+is4H
5Fe
(aq)
2O(l) + to
dissolved in hydrochloric
oxidized
iron(II)
ions. The
36.5 cm3 of 0.02 M acidified
Number
ofsolution
moles ofrequires
Fe2+(aq)
manganate(VII)
for 4-complete
reaction to form
=potassium
5 x number
of moles of MnO
(aq)
iron(III) ions. What is the percentage purity of the iron
=wire?
5 x 0.02 mol dm-3 x 36.5 x 10-3 dm3
= 3.65 x 10-3 mol
Number of moles of Fe dissolved
= number of mole of Fe 2+ formed = 3.65 x 10-3 mol
Mass of Fe = 3.65 x 10-3 mol x 55.8 g mol-1= 0.204g
Percentage purity of Fe
Answer
= 0.204g/0.22g x 100% = 92.73%
89
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple Titrations (SB p.78)
Check
Point 3-8
(a) Na CO (s) + 2 HCl (aq)
2
3
3
(a)5g of anhydrous sodium 2NaCl(aq)
carbonate +isHadded
to
100
cm
(g)
2O(l) + CO
of 2 M hydrochloric acid. What is the volume
of2gas
evolved
at moles
roomoftemperature
No. of
Na2CO3 usedand pressure?
-1 = 0.0472 mol
+16.0 Na
x 3)g
(R.a.m.=: 5g
C /=(23.0x
12.0,2O+12
= 16.0,
= mol
23.0;
molar volume of
3 mol-1)
gas at R.T.P.
=
24.0
dm
No. of moles of HCl used
(b) 8.54g
of ximpure
hydrated
iron(II)
= 2M
100/ 1000
dm3 = 0.2
mol sulphate (formula
mass of 392.14) was dissolved in water and made up to
3. 25cm
Since
HCl is3inofexcess,
Na2CO3 is
the limiting
agent. 3 of
250 cm
this solution
required
20.76cm
0.0203M
acidified
potassium
manganate(VII) solution
No.
of
moles
of
CO
produced
for complete reaction.2 Determine the percentage purity
of the
hydrated
sulphate.
=no.
of molesiron(II)
of Na2CO
3 used = 0.0472 mol
Volume of CO2 produced
= 0.0472 mol x 24.0 dm3 mol-1= 1.133 dm3
90
New Way Chemistry for Hong Kong A-Level Book 1
Answer
3.6 Simple Titrations (SB p.78)
(b)No. of moles of MnO4= 0.0203M x 20.76/1000 dm3 = 4.214 x 10-4 mol
No. of moles of Fe2+
= 5 x no. of moles of MnO4- = 2.107 x 10-3 mol
No. of mole of Fe2+ in 25.0 cm3 solution = 2.107 x 10-3 mol
No. of mole of Fe2+ in 250.0 cm3 solution = 0.021 07 mol
Molar mass of hydrated FeSO4 = 392.14 g mol –1
Mass of hydrated FeSO4
= 0.021 07 mol x 392.14 g mol –1= 8.26g
5 purity of FeSO4 = 8.26g/8.54g x 100% = 96.72%
91
New Way Chemistry for Hong Kong A-Level Book 1
The END
92
New Way Chemistry for Hong Kong A-Level Book 1