MCS lecture

Mole Concept & Stoichiometry

Outline of content

1. Introduction

2. Relative masses of atoms and molecules

3. Molecular and Empirical* Formulae

4. Mole Relationships including calculations

based on reacting masses and volumes (of
solutions and gases)

PAGE 1-1

2.1 Definitions

You should be able to:
•

Define relative atomic, isotopic, molecular
and formula masses

PAGE 1-3

The relative molecular mass, Mr, of a
substance

RELATIVE MOLECULAR MASS, Mr

PAGE 1-3

Multiple Choice

Does Relative Molecular Mass have units?

• Yes

• No

Multiple Choice

Are relative molecular mass and molar mass the same?

• Yes

• No

The relative molecular mass, Mr, of a
substance

RELATIVE MOLECULAR MASS, Mr

PAGE 1-3

The relative isotopic mass, Ar, of a particular
isotope

PAGE 2-3

RELATIVE ISOTOPIC MASS, Ar

Note:
Not average mass!

PAGE 1-3

Why average mass?

Atoms of the same element may not have the same
mass – exist as isotopes.

Note: The average mass is the weighted average of
the relative isotopic masses of the isotopes according
to their abundances.

PAGE 1-3

2.2 Calculate Ar from relative abundance of isotopes

You should be able to:
•

Calculate relative atomic mass, given
relative abundance of its isotopes

PAGE 1-4

O level Periodic Table

A level Periodic Table

3. Molecular and Empirical Formulae

You should be able to:
• Define empirical and molecular

formula

3. Molecular and Empirical Formulae

You should be able to:
• Calculate empirical* and molecular

formula, using composition by mass

Page 1-5

Worked Example 2:
Calculate the empirical formula of a compound that has the
Composition by mass: 12.8% carbon, 2.1% hydrogen and
85.1% bromine.

Information you need to solve this

problem:

Ar of C = 12.0
Ar of H = 1.0
Ar of Br = 79.9

Calculation: EF from Composition by mass

Page 1-5

Let the mass of a sample of compound be 100g.

C H Br

Mass / g 12.8 2.1 85.1

No. of mol 12.8 ÷ 12.0 2.1 ÷1.0 85.1 ÷ 79.9

= 1.067 = 2.1 =1.065

Simplest ratio 1 2 1

Worked Example 2:
Calculate the empirical formula of a compound that has the
Composition by mass: 12.8% carbon, 2.1% hydrogen and
85.1% bromine.

Solution:

Calculation: EF from Composition by mass

Page 1-5

C H O

% by mass 66.7 11.1 22.2

No. of mol

66.7 ÷ 12.0 11.1 ÷1.0 22.1 ÷ 16.0

= 5.558 = 11.1 =1.388

Simplest ratio 4 8 1

(a) Carbon is a major constituent of organic compounds, often

combined with the elements hydrogen, oxygen and
nitrogen. One such compound, B, contains C, 66.7%, H,
11.1%; O, 22.2% by mass. The relative molecular mass, Mr,
of B is 72. Calculate the empirical and the molecular
formula of B.

Solution:

∴ Empirical formula of compound is C4H8O

Lecture Practice 4

Page 1-7

Fill in the Blanks

Lecture Practice 4(a) What is the molecular formula for B?

4.1 & 4.2 Molar Mass & Molar Volume

You should be able to:
• Perform calculations using mole

concept, involving:
• Reacting masses
• Volumes of gases

Learning to count effectively

4 bowls of rice

5 356 grains of

rice

vs

The Mole and Avogadro Constant

Page 1-8

• Equal volumes of gases,under the

same temperature and pressure,
contain equal number of atoms or
molecules.

• One mole of any gas will have the

same volume under the same
temperature and pressure.

• This Law applies strictly to Gases only.

• Do not confuse it with Volume of solutions!!!!

Molar Volume (for gases only)

The Mole and Avogadro Constant

Page 1-9

Take a minute to recap your mole

concept.
1 mole of substance
No. of particles

Na

No. of Na atoms =
1 x 6.02 x 1023

S8

No. of S8 molecules =
1 x 6.02 x 1023

No. of S atoms =

2 x 6.02 x 1023

8 x 6.02 x 1023

1 x 6.02 x 1023

MgCl2
No. of Cl- ions =

No. of Mg2+ ions =

Can also start doing self-practice 3.
Page 1-8

•

Calculation: Molar Volume

Page 1-10

Fill in the Blanks

Lecture Practice 6. What is the volume occupied by 10 g of helium at r.t.p?

Important relationships involving “moles”:

No. of moles

of X

In aqueous solutions:

If X is a gas:

At r.t.p. (20°C and 1 atm)

At s.t.p. (0°C and 1 bar)

Page 1-11

REALITY
CHECK!

Page 1-17

5.3 Percentage Yield (actual yield)

4.3 Important relationships involving “moles”

You should be able to:
• Perform calculations using mole

concept, involving:
• Volumes and concentrations of

solutions

5.2 Limiting reagents

• In a reaction, 1 or more reagents may be present

in excess and not used up completely.

• The amount of product formed is determined by

the amount of reagent which is limited in the
reaction (hence it is completely used up).

• The reagent which is completely used up is

called the limiting reagent. The reaction stops
when the limiting reagent is consumed.

Use of mole concept in calculations

Page 1-15

Lecture Practice 8:
25.0 cm3 of 0.500 moldm-3 of hydrochloric acid, HCl, is reacted
with 0.500 g of solid calcium carbonate, CaCO3, under room
conditions. The equation for the reaction is as follows.

2HCl(aq) + CaCO3(s) → CaCl2(aq) + CO2(g) + H2O(l)

(i) Determine the limiting reagent.
(ii) Calculate the total volume of the gas formed under room
conditions.

Calculation: Limiting Reagent

Page 1-16

• In real life, when chemical reactions are carried out,

it is usually impossible to ensure that 100% of
expected products are formed.

• Why?

• All stoichiometric calculations we have done thus far

have had one assumption in common: 100% of
products assumed to be formed!

REALITY CHECK!

Percentage yield

Page 1-17

• The formation of 100% of products is known as the

theoretical yield. It is the maximum amount of
products expected to be obtained if there was no
loss in chemicals at all. This can be calculated using
stoichiometric ratios.

• The experimental yield is the actual amount of

products formed in real life, usually measured in
terms of mass.

REALITY CHECK!

Percentage yield

Page 1-17

•

Calculation: Limiting Reagent

Page 1-16

Worked Example 7:

0.2 mole of ammonia gas is mixed with 0.30 mole of oxygen
gas. What is the limiting reactant, and how much excess
reactant remains after the reaction has stopped?

4NH3(g) + 5O2(g) 🡢 4NO(g) + 6H2O(g)

Calculation: Limiting Reagent

Page 1-16

• The percentage yield can be calculated by:

REALITY CHECK!

Percentage yield (pg 1-17)

Percentage purity (pg 1-18)

• The percentage purity can be calculated by:

Page 1-17 & 18

0.5 g of magnesium was added to sulfuric acid. The mass of the salt
obtained at the end of the reaction is 1.58 g.
Calculate the percentage yield for the above reaction.
(Ar of Mg = 24.3, Ar of S = 32.1, Ar of O = 16.0)

Worked Example 9:

Solution:

Mg(s) + H2SO4(aq)→ MgSO4 (aq) + H2 (g)

Since Mg ≡ MgSO4
No. of moles of MgSO4 (expected to form) = 0.02058 mol

Theoretical massof MgSO4 formed = 0.02058 x (24.3 + 32.1 + 4(16.0))

= 2.477 g

Page 1-18

Calculation: Percentage Yield

When excess tin is refluxed (heated) with iodine in an organic
solvent, tin(IV) iodide is formed.

Sn + 2I2 → SnI4

When 3.18 g of iodine were reacted with tin, 1.95 g of SnI4
crystals were formed. Calculate the percentage yield of this
reaction. [Ar: Sn, 118.7; I, 126.9]

Lecture Practice 9:

Solution:

Page 1-18

Calculation: Percentage Yield

Worked Example 10:

Solution:

Calculation: Percentage Purity

Page 1-19

An alloy sample of 11.54g is made up of 0.0998 mol of pure

copper, Cu.

Calculate the percentage by mass of pure copper in the alloy.
[Ar: Cu, 63.5]

Lecture Practice 10:

Solution:

Calculation: Percentage Purity

Page 1-19

Chalk is made up of mostly pure calcium carbonate. In a reaction, 10 g of chalk

was reacted with an excess of dilute hydrochloric acid. 2.28 dm3 of carbon dioxide

gas was collected at room temperature and pressure (r.t.p.). The equation for the

reaction is as follows.

CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)

Calculate the percentage by mass of pure CaCO3 in chalk.

No. of moles of CO2 evolved =

= 0.0950 mol

Since CO2 ≡ CaCO3, no. of moles of CaCO3 = 0.0950 mol

Mass of CaCO3 = 0.0950 x 100.1 = 9.51 g

% by mass of CaCO3 in chalk =

x 100 % = 95.1 %

Lecture Practice 12:

CxHy(g) + (x + y/4)O2 (g) → xCO2(g) + (y/2)H2O(l)

Since mole ratio of CO2 : CxHy is

∴ molecular formula of the hydrocarbon is

Calculation: Combustion of Hydrocarbon

Page 1-23

Initial (cm3)

20

150

0

Final (cm3)

0

50

60

Total vol is
110 cm3

Vol used/
involved

(cm3)

20

100

60
(vol of CO2 = vol
reacted with KOH)

Ratio

1

5

3

Lecture Practice 12:

(b) 20 cm3 of a hydrocarbon A were mixed with 80 cm3 of
oxygen. After cooling the apparatus to room temperature and
pressure, the 60 cm3 of gas that remained was shaken with an
excess of aqueous potassium hydroxide, KOH, 20 cm3 of gas
remained. Deduce the molecular formula of A.

Calculation: Combustion of Hydrocarbon

Page 1-23