CHAPTER 2.0 MOLE CONCEPT

EXERCISE

Determine the oxidation number of

a) Fe atom in FeCl3
b) Cl atoms in CaCl2
c) Cl atom in HCl
d) N atom in NH4+

102

EXERCISE

1. Determine the oxidation number of Mn in the
following chemical compounds.
i. MnO2 ii. MnO4-

2. Determine the oxidation number of Cl in the
following chemical compounds.
i. KClO3 ii. Cl2O72-

3. Determine the oxidation number of the followings:
iiii.ii..CUrininUKOC2C2i2nr+2OC27O42-

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Balancing Chemical Equations

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

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

Example:

xA + yB zC + wD 104

Reactants Products

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 showing the relative
number of molecules reacting, are called the
stoichiometric coefficients.

105

Balancing a chemical equation by using
INSPECTION METHOD

1. Write down the unbalanced equation. Write the correct
formula for the reactants and products.

2. Balance the metallic element, followed by nonmetallic
atoms.

3. Balance the hydrogen and followed by oxygen atoms.

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

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Example 1

Balance the chemical equation below.
2O

H=1 2O
Cl = 1
Na = 1 H =1
O=1 Cl =1
Na =1
O=1

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Example 2

Balance the chemical equation below.

ClO2 + H2 3 + HClO2

2ClO2 + H2 3 + HClO2

H=2 H =2
Cl = 2 Cl =2
O=5 O=5

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EXERCISE

Balance the chemical equations below :

a. AgNO3 + Na2CrO4 2CrO4 + NaNO3

b. C6H6 + O2 2 + H2O

c. N2H4 + H2O2 3 + H2O

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Redox Reaction

Redox reaction is a reaction that involves both
reduction and oxidation.

Redox = reduction + oxidation

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OXIDATION

The substance loses one or more electrons.
Increase in oxidation number.
Acts as a reducing agent (reductant).

111

REDUCTION

The substance gains one or more electrons.
Decrease in oxidation number.
Acts as an oxidising agent (oxidant).

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EXAMPLE

Ox. no increase to +1

2Na(s) + Cl2

Ox.no decrease to -1

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Balancing Redox Reaction In Acidic Solution

Fe2+ + MnO4- 3+ + Mn2+

1. Divide the equation into two half equations, one

involving oxidation and the other reduction.

Oxidation: Fe2+ 3+

Reduction: MnO4- 2+

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2. Balance each half-reaction

a. balance the element other than

oxygen and hydrogen.

Oxidation : Fe2+ 3+

Reduction: MnO4- 2+

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b. Balance the oxygen atom by adding H2O
and hydrogen by adding H+.

Oxidation: Fe2+ Fe3+ Mn2+ + 4H2O
Reduction: MnO4- + 8H+

c. Balance the charge by adding electrons to
the side with the greater overall positive charge.

4H2O Oxidation: Fe2+ Fe3+ + 1e
Reduction: MnO4- + 8H+ + 5e Mn2+ +

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3. Multiply each half-equation by an integer, so
that number of electron lost in one half-equation
equals the number gained in the other.

Oxidation: 5 x (Fe2+ 3+ + 1e) 2+ +
5Fe2+ 5Fe3+ + 5e
8H+
Reduction: MnO4- +
4H2O

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4. Add the two half-equations and simplify where
possible by canceling species appearing on both sides of
the equation.
Oxidation: 5Fe2+ 3+ + 5e
Reduction: MnO4- + 8H+ 2+ +
4_H_2_O_________________________________________
5Fe2+ + MnO4- + 8H+ 3+ + Mn2+ + 4H2O

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

5Fe2+ + MnO4- + 8H+ 3+ + Mn2+ + 4H2O

Total charge reactant Total charge product
= 5(+2) + 1(-1) + 8(+1) = 5(+3) + 1(+2) + 4(0)
= (+10) + (-1) + (+8) = (+15) + (+2) + 0
= +17 = +17

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Example : In Acidic Solution

C2O42- + MnO4- + H+ 2 + Mn2+ + H2O

Solution:

1. Oxidation: C2O42- 2 Divide into two
Reduction: MnO4- 2+ half equation
Balanced atoms
2. Oxidation: C2O42- 2CO2 Balanced Charge
Reduction: MnO4- + 8H+ 2+ + 4H2O
2+ + 4H2O
3. Oxidation: C2O42- 2 + 2e
Reduction: MnO4- + 8H+ + 5e

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4. Oxidation: 5 x (C2O42- 2 + 2e) 5C2O42- 2 + 10e
Reduction: 2 x (MnO4- + 8H+ 2+ ++84HHB22OaOlan)ced
2MnO4- + 16H+ 2+ electrons

5. Oxidation: 5C2O42- 2 + 10e
Reduction:2MnO4- + 16H+ 2+ + 8H2O Cancel out
__________________________________________ electrons

5C2O42- + 2MnO4- + 16H+ 2 + 2Mn2+ + 8H2O

Final equation.

Note: The presence of H+ in final equation indicates acidic medium.

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Revision!!!
Five steps to balanced REDOX Equation.

1. Divide into two separate half equation
2. Balance the atoms other than O and H
Add HH+2Ototothtehesisdiedethtahtaht ahsaslelsesssHOataotmom
Add

3. Balance the charge on both side of the half
equation
(Add electrons to the side that has less
electrons)

4. Balance the electrons on both half equations to
equalise the number of electrons.

5. Combine the two half equations and cancel out the
electrons.

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Exercise

Balance the following redox equations in acidic medium

i. Cu + NO3 + H+ Cu2+ + NO2 + H2O
ii. MnO4- + H2SO3 Mn2+ + SO42- + H2O + H+
iii. Zn + SO42- + H+ Zn2+ + SO2 + H2O

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For Basic Solutions

Assume the solutions were acidic. Then
follow a simple three additional steps as
follows to the redox equations :

Additional steps

1. Add to both sides of the equation the same
number of OH- as there are H+ to eliminate
ALL H+

2. Combine H+ and OH- to form H2O.
3. Cancel any H2O that you can.

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Example Reduction

MnO4- + C2O42- MnO2 + CO32-

Solution Oxidation (basic solution)

Assume in acidic solution C2O42- CO32-

MnO4- MnO2 C2O42- 2CO32-
C2O42- + 2H2O 2CO32-
MnO4- + 4H+ MnO2 + 2H2O
2CO32- + 4H+
-1 + 4 = +3 0 C2O42- + 2H2O
-4 +4 = 0
MnO4- + 4H+ + 3e- MnO2 + 2H2O -2

C2O42- + 2H2O 2CO32- + 4H+ + 2e- 125

Red: 2(MnO4- + 4H+ + 3e- MnO2 + 2H2O)

Ox: 3(C2O42- + 2H2O 2CO32- + 4H+ + 2e-)

Red: 2MnO4- + 8H+ + 6e- 2 MnO2 + 4H2O

Ox: 3C2O42- +26H2O 6CO32- +412H+ + 6e-

2MnO4- + 3C2O42- + 2H2O 2 MnO2 + 6CO32- + 4H+

Check for net charge and atoms

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In basic solution

2MnO4- + 3C2O42- + 2H2O 2 MnO2 + 6CO32- + 4H+

2MnO4- + 3C2O42- + 2H2O + 4OH- 2 MnO2 + 6CO32- + 4H+ + 4OH-

2MnO4- + 3C2O42- + 2H2O + 4OH- 4H2O

2
2 MnO2 + 6CO32- +4H2O

2MnO4- + 3C2O42- + 4OH- 2 MnO2 + 6CO32- + 2H2O

127

Test yourself !

Balance the following redox equation :

1. Cr2O72- + Fe2+ + H+ Cr3++ Fe3+ + H2O
(basic solution)

2. CrO2- + ClO- CrO42- + Cl-
(basic solution)

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2.5 Stoichiometry

LEARNING OUTCOMES
At the end of this topic, students should be able to:

(a) Calculate the amount of reactant and product from
balanced chemical equation.
Calculation involve :
i. reacting masses and moles
ii. volume of gases at room condition & STP; and
iii. volume and concentration of solutions

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4.2 Stoichiometry

LEARNING OUTCOMES
At the end of this topic, students should be able to:

(b) Define :
i. limiting reactant
ii. Theoretical yield
iii. actual yield
iv. percentage yield

(c) Perform calculation involving limiting reactant and percentage
yield.

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STOICHIOMETRY

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

Balanced chemical equation is important in
stoichiometry calculation.

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Example

CaCO3 2 (aq) + CO2 (g) + H2O (l)

1 mole of CaCO3 reacts with 2 moles of HCl to
yield 1 mole of CaCl2, 1 mole of CO2 and 1 mole of
H2O.

Stoichiometry can be used for calculating the
amount of specific species in a reaction.

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Example 1

How many moles of hydrochloric acid, HCl are needed to
react with 0.5 mol of zinc, Zn?

Write reaction equation and balanced

Stoichio equivalence

Solve

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Example 2

How many moles of H2O will be formed when 0.25 mol of
C2H5OH burns in oxygen?

134

Exercise 1

A 16.50 mL volume of 0.1327 M KMnO4 solution is needed
to oxidise 20.00 mL of a FeSO4 solution in an acidic medium.
What is the concentration of the FeSO4 solution? The net ionic
equation is:

5Fe 2+ + MnO4- + 8H+ Mn 2+ + 5Fe 3+ + 4H2O

Answer : 0.5474 M 135

Exercise 2

How many millilitres of 0.112 M HCl will react with

exactly

21.2 mL of 0.150 M Na2CO3 according to the following
equation?

2HCl (aq) + Na2CO3(aq) 2NaCl (aq) + CO2 (g) + H2O (l)

Answer : 56.8 mL 136

Limiting Reactant & A
Percentage Yield N
A
How many cups of sundae can be made ? L
Which one limits the number of cups of sundae ? O
G
Y

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If this were a chemical reaction,

the syrup would be the limiting reactant or
limiting reagent.

A limiting reactant is the reactant that is
completely consumed in a chemical reaction
and limits the amount of products formed.

138

The reactants that do not limit the amount of
product, such as ice cream and cherry are in
excess.
An excess reactant is the reactant that is
present in a quantity greater than is required
to completely react with the limiting reactant.

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2 cups of sundae would be the theoretical yield.
Theoretical yield is the amount of product
that can be formed in a chemical reaction
based on the amount of limiting reactant.

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Suppose we accidentally spill the syrup and can
only come up with 1 cup of sundae.

1 cup of sundae would be the actual yield.

Actual yield is the amount of product
actually obtained from a reaction.

The actual yield is always equal to or less
than the theoretical yield because a small
amount of product is usually lost to other
reaction or does not form during a reaction.

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Percentage yield can be calculated as follows :

% yield = actual yield 100 %

theoretical yield

In the making of sundae, 100 % = 50 %
% yield = 1 sundae

2 sundae

we obtained only 50 % of our theoretical yield.

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Percentage Yield

The percentage yield is the ratio of the actual
yield (obtained from experiment) to the
theoretical yield (obtained from stoichiometry
equation) multiply by 100%.

Percentage yield = actual yield x 100%

theoretical yield

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Example 1

S + 3F2 6

If 4 mol of S react with 10 mol of F2 , which of the two reactants is the limiting
reagent?

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Example 2

C is prepared by reacting A and B :

In one process, 2 mol of A react with 9 mol of B.
a. Which is the limiting reactant?
b. Calculate the number of mole(s) of C?
c. How much of the excess reactant (in mol) is left at the end of the reaction?

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B

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147

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Example

During the extraction of iron, the iron of hematite (Fe2O3) was
reduced to iron metal by carbon monoxide according to the
equation,

Fe2O3 2

Theoretically 80.08 g iron will produce. If the mass of iron
extracted is 70 g, calculate the percentage yield.

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Solution

Mass of Fe will produced = 80.08 g = theoretical yield
Mass of Fe extracted = 70 g = actual yield

% yield = actual yield x 100%

theoretical yield

= 70 g x 100%

80.08 g

= 87.4%

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Exercise 1

In a certain experiment, 14.6 g of SbF3 was allowed to react with excess

CCl4. After the reaction was finished, 8.62 g of CCl2F2 was obtained.

3 CCl4 + 2 SbF3 3 CCl2F2 + 2 SbCl3

[ Ar Sb = 122, F = 19, C= 12, Cl = 35.5 ]

What was the theoretical yield of CCl2F2 in grams ?

b) What was the percentage yield of CCl2F2 ?

Answer : a) 11.6 g b) 74.31 % 151