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
New Way Chemistry for Hong Kong A-Level Book 1
3.1
2
Formulae of
Compounds
New Way Chemistry for Hong Kong A-Level Book 1
3.1 Formulae of compounds (SB p.43)
Formulae of compounds
How can you describe the composition
of compound X?
1st way = by chemical formula
C? H ?
ratio of no. of
atoms
3
New Way Chemistry for Hong Kong A-Level Book 1
3.1 Formulae of compounds (SB p.43)
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
carbon atoms
hydrogen atoms
4
New Way Chemistry for Hong Kong A-Level Book 1
Check Point 3-1
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
5
NaF
Not
applicable
New Way Chemistry for Hong Kong A-Level Book 1
OH
Na+F-
3.2
6
Derivation of
Empirical
Formulae
New Way Chemistry for Hong Kong A-Level Book 1
3.2 Derivation of empirical formulae (SB p.45)
From combustion data
• During complete combustion, elements in a
compound are oxidized.
• e.g. carbon to carbon dioxide, hydrogen to
water, sulphur to sulphur dioxide
• From the masses of the products formed, the
number of moles of these atoms originally
present can be found
7
New Way Chemistry for Hong Kong A-Level Book 1
3.2 Derivation of empirical formulae (SB p.45)
The laboratory set-up used for determining the
empirical formula of a gaseous hydrocarbon
Example 3-2A
8
Example 3-2B
New Way Chemistry for Hong Kong A-Level Book 1
Check Point 3-2A
3.2 Derivation of Empirical Formulae (SB p.48)
From combustion by mass
Composition by mass
Empirical formula
Example 3-2C
9
Example 3-2D
New Way Chemistry for Hong Kong A-Level Book 1
Check Point 3-2B
3.3
10
Derivation of
Molecular
Formulae
New Way Chemistry for Hong Kong A-Level Book 1
3.3 Derivation of molecular formulae (SB p.49)
What is molecular formulae?
Molecular formula
?
= (Empirical formula)n
11
New Way Chemistry for Hong Kong A-Level Book 1
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
12
New Way Chemistry for Hong Kong A-Level Book 1
3.3 Derivation of Molecular Formulae (SB p.51)
Water of Crystallization Derived from
Composition by Mass
Example 3-3C
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)
Check Point 3-3A
13
New Way Chemistry for Hong Kong A-Level Book 1
3.3 Derivation of Molecular Formulae (SB p.52)
Find composition by mass from formula
Formula of a
compound
Example 3-3D
Example 3-3E
Check Point 3-3B
Composition by
mass
14
New Way Chemistry for Hong Kong A-Level Book 1
3.4
15
Chemical
Equations
New Way Chemistry for Hong Kong A-Level Book 1
3.4 Chemical equations (SB p.53)
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
16
New Way Chemistry for Hong Kong A-Level Book 1
Check Point 3-4
3.5
Calculations
Based on Chemical
Equations
17
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
Example 3-5A
18
Example 3-5B
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
Example 3-5C
Check Point 3-5
Example 3-5D
19
New Way Chemistry for Hong Kong A-Level Book 1
3.6
20
Simple
Titrations
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple titrations (SB p.58)
Simple titrations
Acid-base titrations
Acid-base titrations
with indicators
Acid-base titrations
without indicators
(to be discussed in
later chapters)
21
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple titrations (SB p.59)
Finding the concentration of a solution
Copper(II)
sulphate
solute
+
Water
solvent
Copper(II)
sulphate
solution
22
solution
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple titrations (SB p.59)
Finding the concentration of a solution
~50 cm3
23
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple titrations (SB p.59)
Finding the concentration of a solution
~50 cm3
24
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple titrations (SB p.59)
Finding the concentration of a solution
~50 cm3
25
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple titrations (SB p.59)
Finding the concentration of a solution
~50 cm3
26
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple titrations (SB p.59)
Finding the concentration of a solution
~50 cm3
27
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple titrations (SB p.59)
Finding the concentration of a solution
~50 cm3
28
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple titrations (SB p.59)
Finding the concentration of a solution
50 cm3
Solution A
29
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple titrations (SB p.59)
Finding the concentration of a solution
~50 cm3
30
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple titrations (SB p.59)
Finding the concentration of a solution
~50 cm3
31
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple titrations (SB p.59)
Finding the concentration of a solution
~50 cm3
32
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple titrations (SB p.59)
Finding the concentration of a solution
~50 cm3
33
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple titrations (SB p.59)
Finding the concentration of a solution
~50 cm3
34
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple titrations (SB p.59)
Finding the concentration of a solution
~50 cm3
35
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple titrations (SB p.59)
Finding the concentration of a solution
50 cm3
Solution B
36
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple titrations (SB p.59)
Finding the concentration of a solution
~100 cm3
37
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple titrations (SB p.59)
Finding the concentration of a solution
~100 cm3
38
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple titrations (SB p.59)
Finding the concentration of a solution
~100 cm3
39
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
40
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple titrations (SB p.59)
Finding the concentration of a solution
~100 cm3
41
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple titrations (SB p.59)
Finding the concentration of a solution
~100 cm3
42
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple titrations (SB p.59)
Finding the concentration of a solution
100 cm3
Solution C
43
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple titrations (SB p.59)
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.
44
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple titrations (SB p.59)
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.
45
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple titrations (SB p.59)
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 
volume of solution (in dm3 )
Unit: mol dm-3 (M)
46
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple titrations (SB p.59)
What does this
mean?
1 dm3
“In every 1 dm3 of the
solution, 2 moles of HCl
is dissolved.”
Example 3-6A
47
New Way Chemistry for Hong Kong A-Level Book 1
contains 2
moles of
HCl
Example 3-6B
3.6 Simple titrations (SB p.62)
Titration without an indicator
By change in pH value
Example 3-6C
48
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple titrations (SB p.62)
Titration without an indicator
By change in temperature
Example 3-6D
49
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple Titrations (SB p.65)
Redox titrations
1. Iodometric titration
in
conical
flask
in
burette
I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62-(aq)
brown
50
colourless
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple Titrations (SB p.65)
Redox titrations
1. Iodometric titration
Add starch
in conical flask
in burette
I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62-(aq)
brown
colourless
During titration : brown  yellow
51
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple Titrations (SB p.65)
Redox titrations
1. Iodometric titration
I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62-(aq)
brown
colourless
During titration : brown  yellow
52
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple Titrations (SB p.65)
Redox titrations
1. Iodometric titration
Example 3-6E
I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62-(aq)
brown
colourless
During titration : brown  yellow
End point : blue black  colourless
(after addition of starch indicator)
53
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple Titrations (SB p.66)
Redox titrations
2. Titrations involving potassium permanganate
In
burette
In conical
flask
MnO4-(aq) + 8H+(aq) + 5Fe2+(aq) Mn2+(aq) + H2O(aq) + 5Fe3+(aq)
purple
pale green
Example 3-6F
54
New Way Chemistry for Hong Kong A-Level Book 1
Check Point 3-6
The END
55
New Way Chemistry for Hong Kong A-Level Book 1
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
CH2
C3H6
H
H
H
H
C
C
C
H
H
(b) Nitric acid
(b) Nitric
(c) Ethanol
acid
(c) Ethanol
(d) Glucose
(d) Glucose
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
56
New Way Chemistry for Hong Kong A-Level Book 1
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
57
New Way Chemistry for Hong Kong A-Level Book 1
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
58
New Way Chemistry for Hong Kong A-Level Book 1
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.
59
New Way Chemistry for Hong Kong A-Level Book 1
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
60
New Way Chemistry for Hong Kong A-Level Book 1
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.
61
New Way Chemistry for Hong Kong A-Level Book 1
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.
62
New Way Chemistry for Hong Kong A-Level Book 1
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
63
New Way Chemistry for Hong Kong A-Level Book 1
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.
64
New Way Chemistry for Hong Kong A-Level Book 1
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
65
New Way Chemistry for Hong Kong A-Level Book 1
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.
66
New Way Chemistry for Hong Kong A-Level Book 1
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
67
New Way Chemistry for Hong Kong A-Level Book 1
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.
68
New Way Chemistry for Hong Kong A-Level Book 1
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
69
New Way Chemistry for Hong Kong A-Level Book 1
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.
70
New Way Chemistry for Hong Kong A-Level Book 1
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.
71
Answer
New Way Chemistry for Hong Kong A-Level Book 1
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.
72
New Way Chemistry for Hong Kong A-Level Book 1
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
73
New Way Chemistry for Hong Kong A-Level Book 1
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
74
1
New Way Chemistry for Hong Kong A-Level Book 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.
75
New Way Chemistry for Hong Kong A-Level Book 1
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
76
New Way Chemistry for Hong Kong A-Level Book 1
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
New Wayof
Chemistry
for HongX
Kong
The77empirical formula
compound
is A-Level
C6H10Book
O5 . 1
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.
78
New Way Chemistry for Hong Kong A-Level Book 1
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
79
New Way Chemistry for Hong Kong A-Level Book 1
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.
80
New Way Chemistry for Hong Kong A-Level Book 1
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
81
New Way Chemistry for Hong Kong A-Level Book 1
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.
82
New Way Chemistry for Hong Kong A-Level Book 1
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.
83
New Way Chemistry for Hong Kong A-Level Book 1
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.
84
New Way Chemistry for Hong Kong A-Level Book 1
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.
85
New Way Chemistry for Hong Kong A-Level Book 1
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
86
New Way Chemistry for Hong Kong A-Level Book 1
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.
87
New Way Chemistry for Hong Kong A-Level Book 1
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
88
New Way Chemistry for Hong Kong A-Level Book 1
3.3 Derivation of molecular formulae (SB p.53)
(a) Calculate the percentages by mass of potassium ,
chromium and oxygen in potassium chromate(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 %
89
New Way Chemistry for Hong Kong A-Level Book 1
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
90
New Way Chemistry for Hong Kong A-Level Book 1
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
91
New Way Chemistry for Hong Kong A-Level Book 1
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
92
New Way Chemistry for Hong Kong A-Level Book 1
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
93
New Way Chemistry for Hong Kong A-Level Book 1
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.
94
New Way Chemistry for Hong Kong A-Level Book 1
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)
95
New Way Chemistry for Hong Kong A-Level Book 1
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.
96
New Way Chemistry for Hong Kong A-Level Book 1
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
97
New Way Chemistry for Hong Kong A-Level Book 1
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.
98
New Way Chemistry for Hong Kong A-Level Book 1
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
99
New Way Chemistry for Hong Kong A-Level Book 1
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 :
100
New Way Chemistry for Hong Kong A-Level Book 1
+
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.
101
New Way Chemistry for Hong Kong A-Level Book 1
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 dm 3 mol -1
Volume of H2 = 2.4 dm3
102
New Way Chemistry for Hong Kong A-Level Book 1
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
(b) 2P(s) + 3Cl2(g)  2PCl3(l)
1
1
 No. of moles of Cl2 =  No. of moles of PCl3
2
3
1
100 g
Mass of Cl2
1


-1 =
2 (31.0  35.5  3) g mol -1
3 (35.5  2) g mol
Mass of Cl2 = 77.45 g
The minimum mass of chlorine required is 77.45 g.
103
New Way Chemistry for Hong Kong A-Level Book 1
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
104
New Way Chemistry for Hong Kong A-Level Book 1
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
105
New Way Chemistry for Hong Kong A-Level Book 1
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.
106
New Way Chemistry for Hong Kong A-Level Book 1
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.
107
New Way Chemistry for Hong Kong A-Level Book 1
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
108
New Way Chemistry for Hong Kong A-Level Book 1
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.
109
New Way Chemistry for Hong Kong A-Level Book 1
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 =
110
2.52 g
= 0.02 mol
-1
126.0 g mol
0.02 mol
= 0.08 M
3
3
250  10 dm
New Way Chemistry for Hong Kong A-Level Book 1
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)
111
New Way Chemistry for Hong Kong A-Level Book 1
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.
112
New Way Chemistry for Hong Kong A-Level Book 1
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
113
New Way Chemistry for Hong Kong A-Level Book 1
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.
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New Way Chemistry for Hong Kong A-Level Book 1
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:
115
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
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple titrations (SB p.63)
(a) Plot a graph of temperature against volume of sulphuric(VI)
acid added.
Answer
116
New Way Chemistry for Hong Kong A-Level Book 1
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.
117
New Way Chemistry for Hong Kong A-Level Book 1
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.
118
New Way Chemistry for Hong Kong A-Level Book 1
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
119
New Way Chemistry for Hong Kong A-Level Book 1
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)
1
From (1), Number of moles of IO3-(aq) =
 Number of moles of I2(aq)
1 3
From (2), Number of moles of I2(aq) =
 Number of moles of S2O32-(aq)
2
1
Number of moles of IO3-(aq) =
 Number of moles of S2O32-(aq)
6
1
Molarity of IO3-(aq)  25.0  10-3 dm3 =
 0.05 mol dm-3  22.0  10-3 dm3
6
Molarity of IO3-(aq) = 7.33  10-3 M
Therefore, the molarity of the acidified potassium iodate solution is 7.33 
10-3 M.
120
New Way Chemistry for Hong Kong A-Level Book 1
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
121
New Way Chemistry for Hong Kong A-Level Book 1
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 %.
122
New Way Chemistry for Hong Kong A-Level Book 1
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
123
New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple titrations (SB p.67)
(a) 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
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New Way Chemistry for Hong Kong A-Level Book 1
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
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New Way Chemistry for Hong Kong A-Level Book 1
3.6 Simple titrations (SB p.67)
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(b) MnO4-(aq) + 8H+(aq) + 5Fe2+(aq)  Mn2+(aq) + 4H2O(l) + 5Fe3+(aq)
No. of moles of MnO4- ions = 0.0203 M  20.76 dm 3
1000
-4
= 4.214  10 mol
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
% purity of FeSO4 = 8.26 g  100% = 96.72 %
8.54 g
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New Way Chemistry for Hong Kong A-Level Book 1