Chapter 4: Relative Atomic Mass
and the Mole
Week 4, Lesson 1
Masses of Particles
Relative Isotopic Mass
• Chemists as early as John Dalton, two centuries ago,
used experimental data to determine the weight of
different atoms relative to one another.
• Dalton estimated relative atomic weights based on a
value of one unit for the hydrogen atom.
• In 1961, it was decided that the most common isotope
of 12C would be used as the reference standard.
• On this scale, the 12C isotope is given a relative mass of
exactly 12 units.
Relative Isotopic Mass cont…
“The relative isotopic mass (Ir) of an isotope is
the mass of an atom of the that isotope
relative to the mass of an atom of 12C taken as
12 units exactly.”
Isotopic Masses example…
• Chlorine has two isotopes.
• These have different masses as they have
different amounts of neutrons.
• Using the 12C isotope as a standard, the
relative isotopic masses of these two isotopes
are 34.969 (35Cl) and 36.966 (37Cl).
• Naturally occurring chlorine is made of 75.80%
of the lighter isotope and 24.20% of the
heavier isotope.
Isotopic Composition of Some
Common Elements
ELEMENT
ISOTOPES
RELATIVE ISOTOPIC
MASS
ABUNDANCE (%)
Hydrogen
1H
1.008
99.986
2H
2.014
0.014
3H
3.016
0.001
12C
12 exactly
98.888
13C
13.003
1.112
14C
14.003
Approx 10-10
16O
15.995
99.76
17O
16.999
0.04
18O
17.999
0.20
107Ag
106.9
51.8
108Ag
108.9
48.2
Carbon
Oxygen
Silver
Mass Spectrometer
• Relative isotopic masses of elements can be
obtained using an instrument called a mass
spectrometer.
• This separates the individual isotopes in a
sample of the element and determines the
mass of each isotope.
• The information is presented graphically and is
known as a mass spectrum.
Mass Spectrometer cont…
• In a mass spectrum showing the isotopes of
an element:
– The number of peaks indicates the number of
isotopes
– The position of each peak on the horizontal axis
indicates the relative isotopic mass
– The relative heights of the peaks correspond to
the relative abundance of the isotopes
Relative Atomic Mass
• A naturally occurring sample of an element
contains the same isotopes in the same
proportions regardless of its source.
• It is sufficient to imagine that the hypothetical
average atoms of the element are being used.
• This average is known as the relative atomic
mass.
• Its symbol is Ar.
Relative Atomic Mass cont…
• The relative atomic mass of an element is
actually a weighted average of the relative
isotopic masses because it takes into account
the relative amounts or abundances of each
isotope in natural samples of the element.
Average Relative Mass example…
1. Imagine taking 100 atoms from a sample of
chlorine of chlorine – there will be 75.80
atoms of 35Cl and 24.20 of 37Cl. Find the
relative atomic mass…
Equation to use: ((relative isotopic mass1 x %
abundance1) + (relative isotopic mass2 x % abundance2))
/100
OR
Ar = ∑(relative isotopic mass x %abundance) / 100

Average Atomic Mass cont…
• Imagine taking 100 atoms from a sample of
chlorine of chlorine – there will be 75.80
atoms of 35Cl and 24.20 of 37Cl. Find the
relative atomic mass…
Ar = 34.969 75.80 36.966 24.20
100
Ar =

2650.65  894.58
100
Ar = 35.452
Now you try…
• Find the relative atomic mass of the
following…
1.Hydrogen
2.Carbon
3.Oxygen
4.Silver
Equations with mass spectrums…
• A simplified mass spectrum of magnesium is
shown on page 58 (of textbook). From the
mass spectrum calculate the percentage
abundances of each of its three isotopes.
- The heights of the peaks for the graph are:
-
24Mg
= 7.9 cm
25Mg = 1.0 cm
26Mg = 1.1 cm
Equations with mass spectrums cont…
• So, the percentage of the 24Mg isotope
= 7.9 100%
= 79%
7.91.01.1
• The percentage of 25Mg isotope
=
= 10%
1.0
100

7.9 1.0 1.1
• The percentage of 26Mg isotope
1.1
=
11%
 =
100
7.9 1.0 1.1
Percentage Abundance example…
• Copper has two isotopes. 63Cu has a relative
isotopic mass of 62.95 and 65Cu has a relative
isotopic mass of 64.95. The relative atomic
mass of copper is 63.54. Calculate the
percentage abundance of the two isotopes.
1.Let x be the percentage abundance of 63Cu
2.So, 100-x is the percentage abundance of 65Cu
Percentage Abundance example…
3. Ar(Cu) = ∑(relative isotopic mass x %abundance)
100
So 63.54 = 62.95x  64.95(100 x)
100
6354 = 62.95x + 6495 – 64.95x

6354 = 6495 – 2x
2x = 6495 – 6354
2x = 141
x = 70.5
Relative Molecular Mass
• The relative molecular mass (Mr) of a
compound is the mass of one molecule of that
substance relative to the mass of a 12C atom.
• The Mr is calculated by taking the sum of the
relative atomic masses of the elements in the
molecular formula.
Relative Molecular Mass cont…
Calculate the Mr of CO2:
Mr(CO2) = Ar(C) + 2 x Ar (O)
= 12.0 + 2 x 16.0
= 44.0
For non-molecular compounds the term relative
formula mass is used. This is solved exactly the
same way.
Now you try…
Calculate the Mr of the following:
1.NaCl
2.H2O
3.H2SO4
Week 4, Lesson 2
The Mole
• A mole is defined as the amount of substance
that contains the same number of specified
particles as there are atoms in 12g of carbon12. The mole has been defined in a convenient
way:
– 1 atom of 12C has a relative atomic mass of 12
exactly
– 1 mol of atoms of 12C has a mass of 12g exactly.
The Mole cont…
• The number of particles in 1 mole is given by
the symbol NA.
• Because atoms and molecules are so small,
NA, needs to be a very large number if 1 mole
of a substance is to be an amount that is
convenient to work with.
• It is for this reason that chemists have
selected 12g of Carbon-12 atoms to define the
mole.
Symbols and Units
• The symbol for the amount of substance is n.
• The unit of measurement for the amount of
substance is mol.
• NA, also known as Avagadro’s Constant,
= 6.02 x 1023
• Therefore 1 mole of particles equals 6.02 x
1023.
• If given the amount of substance, we can
calculate the number of particles.
N = n x NA
• N = number of particles
• n = amount of substance
• NA = Avagadro’s Constant
• Find the number of O2 molecules in 2.5 mol of
oxygen.
– N = n x NA
– N = 2.5 x 6.02 x 1023
– N = 15.05 x 1023
– N = 1.5 x 1024
Rearranging the Equation
• You can also use this equation to find the
amount of substance, in mol, by rearranging
the equation.
N
n
NA
Molar Mass
• The mass of 1 mole of a particular element or
compound is known as the molar mass.
• Because the particles of different elements or
compounds have different mass, the masses
of 1 mol samples of different particles will be
different.
• For example, one dozen eggs.
Molar Mass cont…
• In general
– The molar mass of an element is the relative
atomic mass of the relative atomic mass
expressed in grams.
– The molar mass of a compound is the relative
molecular or relative formula mass of the
compound expressed in grams.
• The symbol for Molar Mass is M and its unit of
measurment is gmol-1.
Finding the Molar Mass
• Calculate the molar mass of table sugar,
sucrose (C12H22O11)
1. What are the mass numbers of the individual
elements?
C = 12.0
H=1
O = 16
2. Find the molar mass of each element.
C = 12 x 12
H = 22 x 1
O = 11 x 16
3. Add them together.
Mr = (12x12) + (22x1) + (11x16)
Mr = 342g mol-1
n = m / M or m = n x Mr
•
•
•
•
m = mass
n = amount of substance
M = Mr = molar mass
This means that is we have the mass of
substance we can find the amount in mol by
substituting the mass in for m and working
out the molar mass.
• Or, we can find the mass if the amount of
substance is known.
Week 4, Lesson 3
Formulas of Compounds
• Percentage Composition
– The values for molar masses of elements in
compounds can be used to calculate the
percentage composition of a compound once if
formula is known.
– For example if a company is producing aluminum
from alumina (Al2O3), the management would
want to know the mass of aluminium that can be
extracted from a given quantity of alumina.
Percentage Composition example
• Calculate the percentage of aluminium in alumina.
1. Find the molar mass of Al2O3.
M(Al2O3) = (2 x 27) + (3 x 16)
= 102gmol-1
2. Find the percentage of aluminium in alumina
Since 1 mol of alumina contains 2 mol of Al atoms:
Mass of aluminium in 1 mol (102g) of Al2O3 = 2x27
= 54g
Percentage of Al in Al2O3 = mass of element in 1 mol of compound
mass of 1 mole of compound
100
= 54/102 x 100
= 52.9%
x
Percentage Composition Formula
% by mass of the element =
mass of element in 1 mole of compound
Mass of 1 mole of compound
x100
Empirical Formulas
• The empirical formula of a compound is the
formula that gives the simplest whole number
ratio, by number of moles, of each element in
the compound.
• Empirical formulas are determined
experimentally usually by determining the
mass of each element present in a given mass
of compound.
Empirical Formulas Process
1. Measure the mass (m) of each element in the
compound.
2. Calculate the amount in mol (n) of each
element in the compound.
3. Calculate the simplest whole number ratio of
moles of each element in compound.
4. Empirical formula of compound.
Empirical Formula Example 1
• A compound of carbon and oxygen is found to be 27.3%
carbon and 72.7% oxygen by mass. Calculate the empirical
formula.
C
:
O
1. Mass
27.3
:
72.7
2. Moles
27.3
72.7
12
1. Find ratio
 2.27
2.27
1
2.27

1. Ratio
1
:
2. Empirical Formula is CO2


 4.54
16
4.54
2
2.27
2
Empirical Formula Example 2
• 9.0g of a compound of only carbon, hydrogen
and oxygen is found to contain 4.8g of oxygen
and 3.6g of carbon. Calculate the empirical
formula.
1.Find the mass of hydrogen:
9.0 = 4.8 + 3.6 + m(H)
9.0 = 8.4 + m(H)
M(H) = 0.6g
Empirical Formula Example 2
cont…
Carbon
Hydrogen
Oxygen
Step 1: mass (g)
3.6
0.6
4.8
Step 2: mol (n)
3.6
12.0 = 0.3
0.6
1.0
= 0.6
4.8
16.0 = 0.3
0.3
0.3
0.6
0.3
=2
0.3
0.3
Step 3: Find ratio
Step 4: Give
simplest ration
1
=1
2
Therefore, the empirical formula is CH2O.
1
= 1
Molecular Formulas
• While an empirical formula give the simplest whole
number ratio of an element in a compound, a
molecular formula gives the actual number of atoms
in one molecule of the compound.
• The empirical and molecular formula can be the
same or different.
• The molecular formula is always a whole number
multiple of the empirical formula.
• A molecular formula can be obtained from the
empirical formula if the molar mass of a compound is
known.
Molecular Formula Example 1
• A sample of a hydrocarbon was found to
contain 7.2g of carbon and 1.5g of hydrogen.
The molar mass of this compound was
determined to be 58g mol-1. What is the
molecular formula of the compound?
Example cont…
• Determine the empirical formula.
Carbon
Hydrogen
Step 1: m(g)
7.2
1.5
Step 2: n (mol)
7.2
12 = 0.6
1.5
1.0 = 1.5
Step 3: Divide by
smallest number
0.6
0.6 = 1
1.5
0.6 = 2.5
Step 4: Give
simplest whole
number ratio
2
5
• So the empirical formula is C2H5
Example cont…
• Determine the molecular formula.
– Empirical Formula is C2H5
– Molar Mass (C2H5) unit = 29g mol-1
– But molar mass of compound = 58g mol-1
– Number of C2H5 units in one molecule = 58
29
=2
Molecular formula = C2H5 x 2
= C4H10