CHAPTER 2.0 MOLE CONCEPT

CHAPTER 2:

DK014
SESSION 2022/2023
CHEMISTRY UNIT, KM52NS

2.1: Mole

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Learning Outcomes

a) Define mole in terms of :

mass of C-12
A.

b) molar mass
and molar volume of gas at room condition
and standard temperature pressure (STP).

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Introduction

The atoms, molecules or formula units
are the entities that react with one

another, so we would like to know the
numbers of them that we mix together.

But how can we possibly count entities that are so
small ?

To do this, chemists have devised a unit called the
MOLE to count chemical entities by weighing them.

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Mole

One mole is the amount of a substance that contains the same
number of particles that exists in exactly 12.00 g of carbon-12.

1 mol of carbon-12 has a mass of 12.00 g and contains 6.02 x 1023 atoms.

Avogadro s constant, NA = 6.02 x 1023 mol-1

Molar mass is the mass of a substance (in grams) that contains
6.02 x 1023 particles and has units of g mol-1.

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For any element, molar mass has the same value as atomic mass 57
or relative atomic mass.

Example:
Atomic mass of 12C = 12.00 amu
Relative atomic mass of 12C = 12.00
Molar mass of 12C atoms = 12.00 g mol-1

Relationship between mole, mass, Ar, Mr and amount of particles

Example:
Relative molecular mass, Mr of SO2 = 32 + 2(16) = 64
Molecular mass of SO2 = 64 amu
Mass of 1.0 mol of SO2 or 6.02 x1023 SO2 molecules = 64 g
Molar mass of SO2 = 64 g mol-1

Counting Atoms by Weighing Them
1 mole oxygen

6.02 x1023 of Molar mass of O2
O2 molecules = 32.00 g mol-1
= 32.00 g O2

1 Mole of 1 mole of Water
CaCO3
6.02 x1023 of Molar mass of H2O
6.02 x1023 formula H2O molecules = 18.00 g mol-1
units of CaCO3
= 18.00 g
= 100.09 g
1 mol of Copper
Molar mass of CaCO3
= 100.09 g mol-1 6.02 x1023 of Cu atom

= 63.55 g Molar mass of Cu

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= 63.55 g mol-1

Units for volume:
L @ dm-3

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Counting the number of atoms/molecules

Questions:
a) How many atoms are there in 0.8 mole of
copper?
b) How many moles of chlorine molecules in
8 × 1022 molecules of chlorine gas?

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Solution:

a)

0.8 mol = number of Cu atoms
6.02 ×1023 mol

Number of copper atoms = 0.8 × 6.02 × 1023
= 4.82 ×1023 atoms

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b) Moles of chlorine molecules

= 1.3 × 10-1 mol
or

1 mole of Cl2 6.02 x 1023 molecules of chlorine
? mole of Cl2 8 x 1022 molecules of chlorine

= x 1 mole of Cl2 of X 8 x 1022 molecules of chlorine
6.02 1023 molecules
chlorine

= 1.3 × 10-1 mol 62

Mole Concept

moles = Number of particles
moles NA
moles
= Mass m
Molar mass n=
n= V
= Gas volume
Molar volume Vm

Molar volume of gases, Vm = 22.4 dm3 mol-1 at STP 63
= 24 dm3 mol-1 at room temperature

Molar volume

Question:

How many liters 60 g of O2 will occupy at STP?

Solution: 60g
32gmol
Moles of O2 = 1 = 1.875 mol

1 mole of gas has a volume of 22.4 dm3 at STP 64
Thus, volume of O2 at STP

= 1.875 mol × 22.4 dm3 mol-1
= 42.0 dm3

2.2:
Empirical

and
Molecular Formulae

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Learning Outcomes:

Define the terms empirical and molecular formulae.
Determine empirical and molecular formulae
from mass composition and combustion data.

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Chemical Formula

Empirical formula Molecular formula

×n shows the exact number of atoms
of each element in the smallest unit
shows the simplest whole-number of a substance
ratio of the atoms in a substance

Determine from

Mass/percentage composition Combustion data

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By mass/percentage composition

Find the empirical formula:
1. Get the mass of each element by assuming a

certain overall mass for the sample (100 g is a
good mass to assume when working with percentages).
2. Convert the mass of each element to moles.
3. Find the ratio of the moles of each element.
4. Use the mole ratio to write the empirical formula.

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Example 1:
Determine the empirical formula of a compound contains 2.1% H, 65.3% O and 32.6% S.
Solution:

Assume mass of compound is 100g

Element H S O Empirical formula:
Mass(g) 2.1 32.6 65.3 H2SO4
Mole

Simplest ratio

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Example 2:

1.08 g of aluminium combines chemically with 0. 96 g of oxygen to form an oxide.
What is the empirical formula of the oxide? [Relative atomic mass: O, 16; Al, 27]

Solution:

Element Al O Empirical formula:
Mass(g) 1.08 Al2O3
0.96
Mole

Simplest ratio

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Example 3:

The composition of an organic compound is 38.71% C, 9.68% H and 51.61% O.
Its relative molecular mass is 62.
What are the empirical formula and molecular formula of the compound?

Solution: Empirical formula:
Assume mass of compound is 100g CH3O

Element C H O Molecular formula:
Mass(g) 38.71 9.68 51.61

Mole (Empirical formula)n= Mr
(CH3O)n = 62
(12+3+16)n = 62

n=2

Simplest ratio Molecular formula:

C2H6O2 71

BY COMBUSTION DATA

Combustion of 2.30g of an organic sample, X, yields 3.30 g CO2
and 1.80 g of H2O. Determined the empirical formula of X.

X contains C, H and O = 2.30 g

Find the mass of C and H of the organic
STEP 1 compound from the combustion data.

Find the composition by mass of C, H
STEP 2 and O.

Calculate the empirical formula by using
STEP 3 table.

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Example 1:

Combustion of 2.30 g of an organic sample, X, yields 3.30 g CO2 and 1.80 g of
H2O. Determined the empirical formula of X.

Solution:

Tips: mass of X containing C, H and O = 2.30 g
Find out the mass composition of C, H and O respectively in the sample !

Mass of C:
Molar mass of CO2 = 12 + (2 x 16) = 44 g/mol

1 mol CO2 1 mol C
44 g CO2 12 g C

3.30g CO2

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Mass of H:

Molar mass of H2O = (2 x 1) + 16 = 18 g/mol

1 mol H2 2 mol H

18 g H2 2gH

1.80 g H2

Mass of O:

Mass of C + mass of H + mass of O = 2.30 g
Mass of O = 2.30 0.9 0.2

= 1.20 g

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Element C H O
Mass (g) 0.90 0.20 1.20

moles

Simplest 1 x 3 =3 2.667 x 3 =8 1 x 3 =3
mole
ratio

Therefore, the empirical formula is = C3H8 O3

Keep in mind ! 75

Never round off values close to whole number in order to get a simple
ratio, but multiply the value by a factor until you get a whole number.

2.3:
Concentration

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Learning Outcomes:

Describe :
i) molarity
ii) percentage by mass
Perform calculations on molarity and

percentage by mass (include dilution,
use density as conversion factor)

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Concentration

is the amount of solute present in a given quantity of solvent in a solution.

Solution is a homogenous mixture formed when an amount
of solute dissolves completely in a solvent

Solute is the substance being dissolved and present in the
smaller amount in a solution.

Solvent is the substance doing the solving and present
in the larger amount in a solution.

Example:

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Solution Solvent Solute
Soft drink H2O Sugar, CO2

Air ( ) N2 O2, Ar, CH4

H2O N2 Sugar, CO2 O2, Ar, CH4

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CONCENTRATION Percentage by mass,
%w/w
MEASUREMENT:
Molarity,
M

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Definition Molarity
Symbol
Unit The number of mole of solute dissolved per unit volume.
M
mol L 1 @ mol dm 3 @ M

Formula

*Molarity depends on the volume of solution.

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Density

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Definition Percentage by mass
Symbol
Unit The percentage of the mass of solute per mass of solution.
Formula % w/w
%

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Example: 10% w/w NaOH
10 g of NaOH dissolved in 100 g of solution
10 g of NaOH dissolved in 90 g of solvent (water)
10 % of NaOH in 100 % of solution

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Example 1:
A solution is made by dissolving 2.355 g of sulfuric acid, H2SO4 in water. The total
volume of the solution is 50.0 mL. What is the molar concentration of sulfuric acid?
Solution:

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Example 2:
Concentrated hydrochloric acid solution is 36.5 % HCl by mass.
Its density is 1.18 g mL-1. Calculate the molarity of the solution.

Solution: nHCl

Molarity = moles of solute solution = msolution
volume of solution (L) Vsolution
1.18 g mL-1 =
Molarity = 1 mol Vsolution = 100 g
= Vsolution
84.75 10-3 L
100 g
= 11.8 mol L-1
1.18 g mL-1
84.75 mL 86

Example 3:

An aqueous solution contains 167 g CuSO4 in
820 mL of solution. The density of the solution is
1.195 g mL-1. Calculate the following:
a) percentage by mass of CuSO4

(17.0%)

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Solution:

msolution = Vsolution solution
= 820ml x 1.195g /ml
= 979.7g

%w/w of CuSO4

= (167g / 979.9g) x 100
= 17.0 %

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2.4 Chemical Equation

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

(a) Identify the oxidation number of an element in
a chemical formula.

(b) Write a balanced chemical equations by:
i. inspection method.
ii. ion-electron method for redox reaction.

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OXIDATION NUMBER

Definition :

A number equal to the magnitude of the charge an
atom would have if its shared electron are held
completely by the atom that attracts them more
strongly.

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Oxidation number of any atom can be
determined by applying the following rules :

1. The oxidation number of an atom or
molecule in its elementary form is zero.

Example : = 0 Cl2 = 0
Na = 0 O2 = 0
Cu = 0
Mg

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2. Monatomic ion, the oxidation number is
equal to the charge on the ion.

Example: = +1 Mg2+ = +2
Na+ = +3 S2- = -2
Al3+

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3. Fluorine and other halogens, the oxidation

number is:

-1 in its compound.
a positive number when combine with oxygen.

Example :

Oxidation number of F in NaF = -1
Oxidation number of Cl in HCl = -1
Oxidation number of Cl in Cl2O7 = +7

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4. Hydrogen oxidation number is:

+1 in its compound
-1 in metal hydrides

Example :

Oxidation number of H in HCl = +1
Oxidation number of H in NaH = -1
Oxidation number of H in MgH2 = -1

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5. Oxygen oxidation number is -2 in most of
its compound.

Example: = -2
Oxidation number of O in MgO = -2
Oxidation number of O in H2O

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Two exceptional cases of O :

- In peroxides, its oxidation number is -1

Example : = -1
Oxidation number of O in H2O2

- When combined with fluorine, =
possesses a

positive oxidation number.
Example :
Oxidation number of O in OF2

+2

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6. In neutral molecule, the sum of the oxidation
number of all atoms that made up the molecule
is equal to zero.

Example

The sum of oxidation number of H2O
= 2(+1) + 1(-2) = 0

The sum of oxidation number of HCl
= 1(+1) + 1(-1) = 0

The sum of oxidation number of KMnO4
= 1(+1) + 1(+7) + 4(-2) = 0

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7. The sum of oxidation number of the atoms in
polyatomic ions equal to the nett charge of
the ion.

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

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Example

Assign the oxidation number of Cr in
Cr2O72-.

Solution :
2 (ox. no of Cr) + 7 (ox. no of O) = -2
2 Cr + 7 (-2) = - 2
2 Cr = + 12
Cr = + 6

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mmm Can you help
me guys?

F = -1 Determine the oxidation number
for the underlined element.

IF7 IF7 = 0

1(I) + 7( F) = 0
1(I) + 7(-1) = 0

I = +7

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NaIO3

Na = +1 O = -2 I = ?

1(-1) + 1(I) + 3(-2) = 0
I = +5

K2Cr2O7

O = -2 K = +1

2(+1) + 2(Cr) + 7(-2) = 0

Cr = +6

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