CHAPTER 1 - MATTER

Solution:

c) mole of NH 3  8g 1
17.03 gmol

 0.4698 mol

mass of solvent  92 g

mole of solvent  92 g 1
18.02 gmol

 5.1054 mol

XNH3  0.4698
5.1054  0.4698

0.0843

100

Example 3:

A solution of hydrochloric acid contains
36% HCl by mass.
a) Calculate the mole fraction of HCl in
the solution.

ANSWER: 0.2173
b) Calculate the molality of HCl in the

solution.
ANSWER: 15.4069 m

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Solution:

a) %by mass of HCl  mass HCl x100%  36%
mass solution

assume that,

mass HCl  36 g

mass solution 100g

nHCl  36 g 1
36.46gmol

 0.9874 mol

nso l vent  64 g 1
18.02 gmol

 3.5516 mol

XHCl  0.9874
0.9874  3.5516
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 0.2175

Solution:

b) mass HCl  36 g
mass solution 100g
mass solvent  100  36
 64 g  0.064 kg

nHCl  36 g 1
36.46gmol

 0.9874 mol

molality HCl  0.9874
0.064 kg

15.42 molkg 1

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Example 4:

A 16 M commercial aqueous nitric acid,
HNO3 has a density of 1.42 g mL-1.
Calculate the percentage HNO3 by mass
in solution.
ANSWER: 71%

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Solution

Molarity of HNO 3  V of nHNO3 (L) 16 M
solution

assume that,

V of solution 1L 1000 mL

nHNO3 16 mol

mass HNO 3  63.02 gmol 1 x16 mol
1008 .32 g

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Solution

density of HNO 3  mass solution 1.42 gmL1
V solution

mass solution 1.42 x1000 1420 g

% by mass of HNO 3  mass solute x 100%
mass of solution

 1008.32 x100
1420

 71.01%

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Exercise

1. A solution contains 35% HBr by mass and has a
density of 1.30 g mL-1. Calculate the molality and
molarity of this solution. The molar mass for HBr is
80.91 g mol-1.
[ M=5.63 mol dm-3, m=6.66 mol kg-1]

2. A sulphuric acid solution contains 66% H2SO4 by
weight and has the density of 1.58 gmL-1. How
many moles of the acid are present in 1.00 L of the
solution. Molar mass for H2SO4 is 98.09 gmol-1.
[ 10.63 mol]

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3. Calculate the molality of a 5.86 M ethanol (C2H5OH)
solution which has a density of 0.927 gmL-1.
[8.88 mol kg-1]

4. The density of an aqueous solution containing 10.0%
of ethanol (C2H5OH) by mass is 0.984 g mL-1.

a) Calculate its molarity.
b) Calculate the molality of this solution.
c) What volume of the solution would contains

0.125 mole of ethanol?
[a) 2.14 M; b) 2.41 m; c)0.0587 L]

108

1.3 Stoichiometry

109

Learning outcomes…

At the end of today’s lesson, students should be
able to:

i. Write and balance chemical equation by inspection
method and redox equation by ion-electron method.

ii. define limiting reactant and percent yield

iii. perform stoichiometric calculations using mole
concept including limiting reactant and percent yield

110

Balancing Chemical Equation

• A chemical equation shows a chemical reaction
using symbols for the reactants and products.

• The formulae of the reactants are written on the left
side of the equation while the products are on the
right.

• Reactants are the starting substances in a chemical
reaction.

• Products are the substances formed as a result of a
chemical reaction.

111

Example: yB zC + wD

xA + Products

Reactants

• The total number of atoms of each element is the
same on both sides in a balanced equation.

• The numbers x, y, z and w (show the relative
number of molecules reacting) are called the
stoichiometric coefficients.

112

The methods to balance a chemical
equation are:

A) Inspection method
B) Ion-electron method

113

A) Inspection Method

a. Write down the unbalanced equation, with
the correct formulae of the reactants and
products.

b. Balance the metallic atoms, followed by
non-metallic atoms.

c. Balance the hydrogen and oxygen atoms.

d. Check to ensure that the total number of
atoms of each element is the same on
both sides of equation.

114

Example:
Balance the chemical equation by applying the
inspection method.

NH3 + CuO → Cu + N2 + H2O

Answer:
2NH3 + 3CuO → 3Cu + N2 + 3H2O

115

Exercise

Balance the chemical equation below by
applying inspection method.

a. Fe(OH)3 + H2SO4 → Fe2(SO4)3 + H2O
b. C6H6 + O2 → CO2 + H2O

c. N2H4 + H2O2 → HNO3 + H2O

116

Answer:

a. 2Fe(OH)3 + 3H2SO4 → Fe2(SO4)3 + 6H2O
b. C6H6 + 15/2O2 → 6CO2 + 3H2O

c. N2H4 + 4H2O2 → 2HNO3 + 5H2O

117

B) Ion-electron Method

• Mainly for redox reaction.
• Redox reaction is a reaction that involves

both reduction and oxidation.

118

Oxidation :
 A process of electron loss .
 When a species undergo oxidation,

- its oxidation number increase
- it loses one or more electron(s)
- act as a reducing agent
e.g

Mg Mg2+ + 2e

119

Reduction:
 A process of electron gain .
 When a species undergo reduction,

- its oxidation number decrease
- it gains one or more electron(s)
- act as an oxidation agent
e.g :

Br2 + 2e → 2Br-
Sn4+ + 2e → Sn2+

120

Rules for Assigning Oxidation Numbers

1. The oxidation number of an atom or a
molecule in its elementary form (or free
elements) is zero.

Example:
Na, Mg, O2, Br2, H2

121

2. For monoatomic ions, the oxidation
number is equal to the charge on the ion.

Example:

Ion Oxidation
number

Na+ +1

Al3+ +3

I- -1

S2- -2

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Exception:

a. Hydrogen bonded to metal, i.e metal hydrides
(e.g: NaH, MgH2) the oxidation number for
hydrogen, H is -1

1. b. Oxygen
- In peroxides, its oxidation number is -1
(e.g: H2O2 = -1).
- when combine with fluorine, posses a
positive oxidation number (e.g: in OF2 = +2)

c. Halogen bonded to oxygen (e.g: Cl2O7), the
oxidation number for halogen = +ve

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3. In a neutral compound, the total
oxidation number of all atoms that made
up the molecule is zero.

Example: 0
Oxidation number of H2O = 0
Oxidation number of HCl = 0
Oxidation number of KMnO4 =

124

4. For polyatomic ion, the total oxidation
number of all atoms that made up the ion is
equal to the net charge of the ion.

Example:

Oxidation number of MnO4- = -1
Oxidation number of Cr2O72- = -2
Oxidation number of NO3- = -1

125

Example

Assign the oxidation number of Cr in Cr2O72-
.

answer
2Cr  (7 x  2)   2

2Cr   12
Cr   6

126

Exercise

1. Assign the oxidation number of Mn in the
following chemical compounds.

i. MnO2 ii. MnO4-

2. Assign the oxidation number of Cl in the
following chemical compounds.

i. KClO3 ii. Cl2O72-

3. Assign the oxidation number of following:
i. Cr in K2Cr2O7
ii. U in UO22+
iii. C in C2O42-

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Answer ii. MnO4-

1. i. MnO2 Mn  4(2)   1
Mn   7
Mn  2(2)  0
Mn   4 ii. Cl2O72-

2. i. KClO3 2Cl  7(2)   2
2Cl   12
 1 Cl  3(2)  0 Cl  6
Cl   5

128

Answer

3. i. Cr in K2Cr2O7

2(1)  2Cr  (7 x  2)  0
2Cr   12
Cr   6

ii. U in UO22+

U  (2 x  2)   2
U 6

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Answer

3. iii. C in C2O42-

2C  (4 x  2)   2
2C   6
C3

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Balancing Redox Reaction using ion-
electron method

 Redox reaction may occur in acidic and basic
solutions.

 Follow the steps systematically so that
equations become easier to balance.

04/29/12 MATTER 131131

Steps for balancing redox equation

For reactions in acidic solution:

Example:
MnO4- + C2O42-  Mn2+ + CO2

1. Divide the equation into two half-reactions:
reduction reaction and oxidation reaction
oxidation: C2O42-  CO2
reduction: MnO4-  Mn2+

2. Balance atoms other than O and H in each half

reaction.

oxidation: C2O42-  2CO2 132
reduction: MnO -  Mn2+

3. Balance the oxygen atom by adding H2O
and the hydrogen atom by adding H+
oxidation: C2O42-  2CO2
reduction: MnO4- + 8H+  Mn2+ + 4H2O

4. Balance the charge by adding electrons
to the side with the greater overall charge
(more tve side).
oxidation: C2O42-  2CO2 + 2e
reduction: MnO4- + 8H+ + 5e  Mn2+ +
4H2O

133

5. Multiply each half-reaction with a coefficient to
equalize the number of electrons

oxidation: (C2O42-  2CO2 + 2e) x 5
reduction: ( MnO4- + 8H+ + 5e  Mn2+ + 4H2O)x2

6. Combine the two half-reactions and cancel out
species that appear on both sides of the equation

oxidation: 5C2O42-  10CO2 + 10e
reduction: 2 MnO4- + 16H+ + 10e  2Mn2+ + 8H2O
2MnO4- + 5C2O42- + 16H+ 2Mn2+ + 10CO2 + 8H2O

134

7. Check to make sure that there are same number
of atoms of each element and the same total
charge on both sides.

2MnO4- + 5C2O42- + 16H+ 2Mn2+ + 10CO2 + 8H2O

Total charge reactant Total charge product

= 2(-1) + 5(-2) +16(+1) = 2(+2) + 10(0) + 8(0)
= -2 – 10 + 16 = +4 + 0 + 0
= +4 = +4

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For reactions in basic solution

balance the equation as in acidic form, then
continue,

8. Add the same number of OH- as H+ on both

sides of the equation.

(OH- and H+ on the same side of the equation

can be combined to form H2O.)

2MnO4- + 5C2O42- + 16H+ 2Mn2+ + 8H2O +
10CO2

+ 16OH- +16OH-

2MnO4- + 5C2O42- + 8H2O  2Mn2+ + 10CO2 + 16OH-

136

Example 2

Balance the equation below as in acidic
medium

Cr2O72- + Cl-  Cr3+ + Cl2

137

Answer

6Cl  3Cl2  6e
6e 14H  Cr2O72  2Cr 3  7H2O
6Cl  14H  Cr2O72  2Cr 3  7H2O  3Cl2

138

Example 3

Balance the equation below as in basic
medium

CN- + MnO4-  CNO- + MnO2

139

Solution:

3H2O  3CN  3CNO   6e  6H

6e  8H  2 MnO   2MnO 2  4H2O
4

2OH   3CN  2H  2 MnO   2MnO  H2O  3CNO   2OH 
4
2

3CN  H2O  2 MnO   2MnO  3CNO   2OH 
4
2

140

Stoichiometry

• Stoichiometry is a quantitative study of
reactants and products in a chemical
reaction.

• The most important thing in stoichiometric
calculation is a balanced equation.

• The stoichiometric coefficient in a balanced
equation shows the ratio of amounts (in
moles) of reactants and products.

141

• Example:
2CO(g) + O2(g)  2CO2(g)

• It shows that 2 moles of CO reacts with 1 mole
of O2 to produce 2 moles of CO2.

• The relationship can be simplified using symbol
‘≡’ which means ‘stoichiometrically equivalent
to’.

2 moles of CO ≡ 1 mole of O2 ≡ 2 moles of CO2

142

• The calculation based on the chemical equation
can be solved by 4 simple steps:

1. write a balance equation.
2. calculate the number of mol (convert any

given quantities to mols)
3. use the equation (mol ratio) to determine

the mole of unknown.
4. convert mol calculated in 3. to the required

units.

143

Example 1

Excess barium nitrate was added to
aluminium sulphate and 6.260g barium
sulphate was recovered. Calculate the mass
of aluminium sulphate in the original solution.

144

Solution

1. write a balance equation.

3Ba(NO3)2 + Al2(SO4)3  3BaSO4 + 2Al(NO3)3

? (g) 6.260 g

2. calculate the number of mol

6.260 g
mol of BaSO4 = 233.36 g mol-

=2.682x10-2 mol

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3. use the equation (mol ratio) to determine the
mole of unknown.

from equation:
1 mol Al(SO4 )3  3 mol BaSO4

mol of Al2 (SO4 )3 = 2.682 x 10-2mol BaSO4x  1 mol Al2 (SO4 )3 
 3 mol BaSO4 
 

= 8.940 x 10-3mol

146

4. convert mol calculated in step 3 to the
required units.

mass of Al2 (SO4 )3 = 8.940 x 10-3mol x 342.14 g mol-1
=3.059 g

147

Example 2

Propane (C3H8) is a common fuel used for
cooking and home heating. What mass of
O2 is consumed in the combustion of 1.00
g of propane?

148

Solution

C3H8 + 5O2  3CO2 + 4H2O

1.00 g ? (g)

1.00 g
mol of C3H8 = 44.11 g mol-

=2.27x10-2 mol

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