Mole Cambridge chemistry IGCSEc04.pptxc06.pptx

PowerPoint Slides for Cambridge IGCSETM
Chemistry Student’s Book
© 2021 Marshall Cavendish Education Pte Ltd 1
THE MOLE
CHAPTER 6
The following content has not been through the Cambridge Assessment International Education endorsement process.

• What are some units we use to represent
different quantities of items?
• Why do we need to use units to represent
different quantities of items?
• Do you think we can count the number of atoms
in a substance like how we count the number of
rice grains in a sack of rice? Why?
Questions

In this section, you will learn the following:
• Describe relative atomic mass (Ar).
• Define relative molecular mass (Mr).
6.1 Relative Atomic Mass, Relative Molecular Mass
and Relative Formula Mass

What is relative atomic mass?
• The relative atomic mass (Ar) is the average mass
of the isotopes of an element compared to of the
mass of an atom of 12
C.
• Relative atomic mass is a ratio and therefore has no
unit.
6.1 Relative Atomic Mass, Relative Molecular Mass and Relative Formula Mass
Comparing the masses of different atoms relative to
one another, using Ar values

What is relative molecular mass?
• The relative molecular mass (Mr) of a molecular substance is the sum of the relative atomic masses of its constituent
elements.
• Relative molecular mass is a ratio and therefore has no unit.
Calculating the relative molecular mass of a molecular substance
The relative molecular mass of a molecular
substance is calculated by adding together the
relative atomic masses of all the atoms in its
chemical formula.

• Ionic compounds like sodium chloride do not exist as molecules.
• The relative molecular mass of an ionic compound is more accurately known as the relative formula mass.
• Like relative molecular mass, relative formula mass is given the symbol Mr and has no unit.
What is relative formula mass?
Calculating the relative formula mass of an ionic compound
The relative formula mass is calculated in exactly
the same way as relative molecular mass.

• State that the mole, mol, is the unit of amount of
substance and that one mole contains 6.02 × 1023
particles.
• Use the relationship, amount of substance (mol)
= in calculations.
6.2 The Mole and Molar Mass

The mole is the unit of amount of substance. The symbol for the mole is mol.
Number of particles in a mole:
• One mole of any substance contains 6.02 × 1023
particles.
• The particles could be atoms, molecules, ions and even subatomic particles such as electrons
and protons.
• The value 6.02 × 1023
is called the Avogadro constant or Avogadro number.
• Equal numbers of moles of all substances contain the same number of particles.
• Calculations involving the mole and the number of particles:
What is the mole?
6.2 The Mole and Molar Mass
Number of moles (amount of substance) =
number of particles
6.02 × 1023 ( Avogadro constant )

• The molar mass of an element is the
mass of one mole of atoms of the
element.
• The molar mass of an element is equal
to its relative atomic mass (Ar) in grams.
What is molar mass?
The molar mass of a molecular substance
is equal to its relative molecular mass (Mr)
in grams.
Molar mass of a molecular substance
Molar mass of an element
Molar masses of some elements or molecules
Element /
Substance
Formula Ar Mr
Molar mass
(g/mol)
Number of atoms in
one mole of element
/ substance
Neon Ne 20 20
6.02 x 1023
Carbon C 12 12
Sodium Na 23 23
Oxygen O2 2 x 16 = 32 32
Ammonia NH3
(1 x 14) + (3 x 1) =
17
17
Water H2O (2 x 1) + (1 x 16) =
18
18

The molar mass of an ionic compound is equal to its
relative formula mass (Mr) in grams.
Molar mass of an ionic compound
Molar masses of some ionic compounds
One mole of different substances have different masses

One mole of calcium bromide contains 18.06 × 1023
ions
• The number of ions in one mole of an
ionic compound is not equal to Avogadro
constant.
• It depends on the number of ions that
make up one formula unit.
The number of moles of a substance can be calculated using the following formula:
Number of moles =
mass of substance in g
molar mass of substance in g / mol
Calculations involving the mole and molar mass

6.3 Molar Volume of Gases
• Use the molar gas volume in calculations involving gases.

Avogadro’s Law
Volumes of oxygen gas and carbon dioxide gas at r.t.p.
• One mole of any gas occupies 24 dm3
(24 000 cm3
) at room
temperature and pressure (r.t.p.).
• This volume is called the molar volume of a gas.
Avogadro’s Law states that equal volumes of gases, under the
same conditions of temperature and pressure, contain the same
number of particles.
6.3 Molar Volume of Gases

How do we calculate the number of moles of a gas?
The number of moles of a gas can be determined in two ways.
This formula can be rearranged to give:
• Volume of gas in cm3
= number of moles × 24 000 cm3
• Volume of gas in dm3
= number of moles × 24 dm3

Do balloons of the same volume contain the same number
of particles?
According to Avogadro’s Law, the balloons contain the same number of gaseous
particles since they have the same volume.
Do balloons of the same mass contain the same
number of particles?
The equal masses of different gases do not
contain the same number of particles.
Calculating the number of moles of different gases with equal mass

6.4 Chemical Calculations
• Calculate stoichiometric reacting masses, volumes
of gases at r.t.p. and limiting reactants.

Stoichiometry is the relationship between the number of moles of reactants and the number of moles
of products involved in that chemical reaction. Consider the following chemical equation.
6.4 Chemical Calculations

Calculating masses of reactants and products
By using a balanced equation, we can calculate the mass of any reactant used up or the mass of
any product formed in a reaction.

Calculating volumes of gaseous reactants and products
Nitrogen dioxide is an air
pollutant that is released
from chemical plants.
Consider the following reaction:

What are limiting reactants?
• A limiting reactant is the reactant that is
completely used up in a reaction.
• It limits or determines the amount of
products formed in the reaction.
• The reactants that are not used up in a
reaction are said to be in excess.

IGCSE Chemistry Student Book
Results of experiment A
• In experiment A, the exact amounts of
reactants required are used.
• In experiments B and C, an excess of
one reactant is used.
Effect of limiting reactants
Consider the reaction below:

Results of experiment B

Results of experiment C
The amount of a product formed
in a reaction is always
determined by the amount of the
limiting reactant.

© 2021 Marshall Cavendish Education Pte Ltd
6.5 The Concentration of a Solution
27
• State that concentration can be measured in g/dm3
or
mol/dm3
.
• Calculate volumes and concentrations of solutions.
• Use experimental data from a titration to calculate the
moles of solute, or the concentration or volume of a
solution.

The concentration of a solution is given by the amount of a solute dissolved in a unit volume
of the solution.
We can express concentration as either:
• Concentration in g/dm3
(grams of solute per dm3
)
• Concentration in mol/dm3
(molar concentration)
6.5 The Concentration of a Solution

Concentration in g/dm3
The concentration of a solution can be determined using
the following formula:
By rearranging the formula,
Mass of solute in g = concentration in g/dm3
× volume
of solution in dm3
Concentration in mol/dm3
(molar concentration)
The concentration of a solution in mol/dm3
can be calculated by
using either of these formulae:
29

• Decrease the concentration of a solution by adding more solvent (see the figures below).
• Increase the concentration of a solution by increasing the amount of solute particles in the solution.
This measuring cylinder also contains one mole of NaCl
This measuring cylinder contains one mole of NaCl
How can we change the concentration of a solution?

What is volumetric analysis?
Chemists can check the concentration of substances by carrying
out volumetric analysis.
Titration in volumetric analysis
• To do volumetric analysis, we use a method called titration.
• A titration involves
a solution of unknown concentration (unknown);
a solution of known concentration (titrant).

Let’s Investigate 6A
• In a titration, we find out the volume of titrant
required to react with a fixed volume of the
unknown.
• From these results, we can calculate the
concentration of the unknown solution.
• Indicators such as methyl orange are used to
determine whether a titration has reached the
end-point
32

• Calculate empirical formulae and molecular formulae.
6.6 Empirical and Molecular Formulae

We can conduct experiments to find out the empirical formula of a compound.
The steps are as follows:
1. Find out the mass of the reactants taking part in the reaction.
2. Work out the relative numbers of moles (mole ratio) of the reactants used.
3. Find the formula of the compound using the mole ratio obtained.
How can we work out the empirical formula of a compound?

Let’s Investigate 6B
Sample results
Mass of crucible + lid = 26.52 g
Mass of crucible + lid + magnesium = 27.72 g
Mass of crucible + lid + magnesium oxide = 28.52 g
Experimental set-up to find the empirical formula of magnesium oxide
Finding the empirical formula of magnesium oxide
The empirical formula of magnesium oxide is thus MgO.
The table shows the calculations for finding the empirical
formula of magnesium oxide.
Calculations
Mass of magnesium = 27.72 − 26.52 = 1.20 g
Mass of magnesium oxide produced = 28.52 − 26.52 = 2.00 g
Mass of oxygen reacted = 2.00 − 1.20 = 0.80 g
35

How can we work out the molecular formula of a compound?
We can find the molecular formula of a substance if
we know two things about it:
1. The empirical formula
2. The relative molecular mass
They are related as follows:
• If empirical formula = AxBy,
• molecular formula = (AxBy)n, where n = 1, 2, 3, etc.
• To find n,
n =
relative molecular mass
relative mass from empirical formula

6.7 Percentage Yield, Percentage
Composition and Percentage Purity
• Calculate percentage yield, percentage composition by mass
and percentage purity.

6.7 Percentage Yield, Percentage Composition and Percentage Purity
How to find percentage yield?
• The theoretical yield of a reaction is the calculated amount of product that would be
obtained if the reaction were to be complete.
• The amount of pure product that is actually produced in the experiment is called the
actual yield.
• The percentage yield shows the relationship between actual yield and theoretical
yield.
Percentage yield =
actual yield
theoretical yield
× 100%

How to find percentage composition of a compound?
How to find percentage purity?
Percentage purity =
mass of pure substance in sample
mass of sample
× 100%

What have you learnt?
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Acknowledgements
• Slide 1: rice – ID 14378460 © Yap Kee Chan | Dreamstime.com
• Slide 2: video of coffee grains © kelvinreislavras | pixabay.com (https://pixabay.com/videos/coffee-grains-morning-drinks-31671/)
• Slide 17: balloons © serezniy | 123RF.com
• Slide 21: nitrogen dioxide from chemical plants © Teeravut Atthasak | 123RF.com

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