Transcript Chapter 4 - ssc12chem

Chapter 4
Relative atomic mass and the
mole
A REMINDER
– atoms with same atomic
number but different mass numbers
 Isotopes – atoms with same number of
protons but different number of neutrons
 Ion – a charged atom
 Cation – a positively charged atom
 Anion – a negatively charged atom
 Isotopes
Relative Isotopic Mass
 Atoms
are extremely small with an equally
small mass.
 For example:

One atom of carbon has a mass of
approximately 2 x 10-23g
 Can
we work experimentally with
something this small?
Relative Isotopic Mass

Dalton (remember him) 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.
 However in 1961 it was decided by the
International Union of Physics and Chemistry
that the most common isotope of carbon, 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
The rule:
The relative isotopic mass (Ir) of an
isotope is the mass of an atom of that
isotope relative to the mass of an atom
of 12C taken as 12 units exactly.
There are no units for relative isotopic mass (Ir)
So what does this mean?
 All
isotopes of elements have relative
isotopic mass values relative to this
standard.
So what does this mean?

There are two different atoms of isotopes of
chlorine. These isotopes have different masses
because they have different numbers of
neutrons.
 Using 12C as the standard, the accurate relative
isotopic masses of the two chlorine isotopes are
34.969 (35Cl) and 36.966 (37Cl).
 Naturally occurring chlorine is made up of
75.80% of 35Cl and 24.20% 37Cl.
 What is the mass number of chlorine?

Table 4.2 Isotopic composition of some common elements.
Mass Spectrometer

Relative isotopic masses of elements can be
obtained using an instrument called a mass
spectrometer.
 This instrument separates the individual
isotopes in a sample of the elements and
determines the mass of each isotope, relative to
the 12C isotope, and the relative abundances of
each of the isotopes.
 This information presented graphically is called
a mass spectrum
Mass Spectrum
Figure 4.3 Mass spectrum of
atomic chlorine.

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.
How does a mass spectrometer
work?

Figure 4.4 The construction of
a mass spectrometer.
Relative Atomic Mass
 A naturally
occurring sample of an element
contains the same isotopes in the same
proportions, regardless of its source
 For most purposes it is sufficient to
imagine that hypothetical ‘average’ atoms
of the element are being used.
 This average is known as the relative
atomic mass (Ar) of an element.
Relative Atomic Mass
 The
relative atomic mass of an element
is, therefore, the weighted average of
the relative masses of the isotopes of
the element on the 12C scale.
 In
other words the relative atomic mass of
an element is the weighted average as it
takes into account the relative isotopic
mass and the abundance
Relative Atomic Mass
Worked example 4.1a
Calculate the average relative mass of one chlorine atom.
Solution:
If we had 100 atoms there would be 75.80 atoms of 35Cl
and 24.20 atoms of 37Cl.
Total relative mass of 100 atoms of chlorine:
=34.969 x 75.80 + 36.966 x 24.20
Average relative mass of one chlorine atom:
= 34.969 x 75.80 + 36.966 x 24.20
100
= 2650.65 + 894.58
100
The relative atomic mass of chlorine is 35.45
= 35.452
OR more simply Ar(Cl) = 35.45
You may have noticed …

The average mass of chlorine is closer to 34.969
that to 36.966. This reflects the greater
abundance of the lighter isotope.
 The relative atomic mass has no units, because
it is simply a measure of the mass of an element
compared with that of the carbon-12 isotope.
 In this exercise, the relative isotopic masses are
given to five significant figures, but the
abundances are only given to four significant
figures. The accuracy of the answer is therefore
limited to four significant figures.
Worked example 4.1c
7.9
1.1
1.0
The heights from the graph are:
 24Mg – 7.9 cm
 25Mg – 1.0 cm
 26Mg – 1.1 cm
So
– 7.9 cm
25Mg – 1.0 cm
26Mg – 1.1 cm
 24Mg


The percentage of the 24Mg isotope =
7.9
x100
7.9 + 1.0 + 1.1
= 79%
The percentage of the 25Mg isotope =
1.0
x100
7.9 + 1.0 + 1.1
= 10%
What is the percentage of the 26Mg isotope?
Can you work out the relative
atomic mass of these elements
Relative Molecular Mass (Mr)
 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, which is 12 exactly.
 The
relative molecular mass of a
compound is calculated by taking the sum
of the relative atomic masses of the
elements in the molecular formula
For example
 Consider


the following atmospheric gases:
Oxygen (O2): Mr (O2) = 2 x Ar (O)
= 2 x 16.0
= 32.0
Carbon dioxide (CO2): Mr (CO2)
= Ar (C) + 2 x Ar (O)
= 12.0 + 2 x 16.0
= 44.0
Relative Formula Mass
 For
non-molecular compounds, we use the
term relative formula mass.
 This is calculated the same way as relative
molecular mass.
 For example:

Find the relative formula mass of sodium chloride
Ar (Na) + Ar (Cl) = 23.0 + 35.5
= 58.5
Key Questions
 Try
key questions 1 - 4 on page 59 of your
text book.
The Mole

When chemists work with elements and
compounds, they cannot deal with quantities as
small as individual atoms, ions or molecules.

Just like you could not have used the paper that
your tore up it was just too small.

Chemists need a convenient quantity of atoms
or molecules – one that they can weigh out and
work with in a laboratory
The Mole
 When
at the supermarket you buy eggs by
the dozen not by the 15 or by the 10.
 It
is the same in chemistry.
 The
accepted convenient quantity for
chemists when working with chemicals is
the mole
The Mole
 A mole
is defined as the amount of
substance that contains the same number
of specified particles as there are atoms in
12 g of carbon-12.
 The mole has been defined in a very
convenient way:


1 atom of 12C has a relative atomic mass of
12 exactly
1 mol of atoms of 12C has a mass of 12 g
exactly
Referring to the mole
 It
is essential that when referring to a mole
of a substance that you specify the particle
you are talking about.
 One mole of oxygen could be several
things.

One mole of oxygen atoms
or

One mole of oxygen molecules
 They
are not the same thing
Molecules or atoms
 Amounts
of substances should also be
referred to by their formula.
 For instance:


1 mole of oxygen (O) atoms
2 mole of oxygen (O2) molecules
 The
specified particles may be atoms,
molecules, ions, depending on the
structure of the substance
Definitions
Atoms: the smallest part of an element that
can take part in a chemical reaction
Molecules: Two or more atoms held together
by covalent bonds
Ions: A charged atom or a charged group of
atoms
Particles

The number of particles in 1 mole is given the
symbol NA
 For example:




1 mole of hydrogen atoms contains NA hydrogen (H)
atoms.
1 mole of hydrogen molecules contains NA hydrogen
(H2) atoms.
1 mole of aluminium atoms contains NA aluminium
(Al) atoms.
1 mole of glucose molecules contains NA glucose
(C6H12O6) molecules.
NA
 Atoms
and molecules are so small, NA
needs to be a very big number if 1 mole of
a substance is to be an amount that is
convenient to work with.
 This
number is the number of particles in a
mole and is known as Avogadro’s constant
and has the symbol NA
For that reason
 Chemists
have selected exactly 12g of
carbon-12 atoms to define the mole.
 Note



that:
The symbol for amount of substance is n
The unit of measurement for amount of
substance is mol.
So n(glucose) = 2 mol or the amount of
glucose molecules in the sample is 2 moles.
Back to particles or NA

The number of carbon particles in exactly 12 g of
carbon-12 has been experimentally determined
to be 6.022045 x 1023

We shorten this number to 6.02 x 1023
NA = 6.02 x 1023 mol-1
 Therefore, 1 mol of particles contains
6.02 x 1023 particles

Worked example 4.2a
 Calculate
the number of O2 molecules in
2.5 mol of oxygen (O2).
 How
might we solve this?
Calculate the number of O2 molecules in 2.5 mol of
oxygen (O2).
 Solution:
By definition 1 mol of O2 molecules contains
6.02 x 1023 molecules.
Number of O2 molecules (N) in 2.5 mol of O2
= 2.5 x 6.02 x 1023
= 15.05 x 1023
= 1.5 x 1024
There are 1.5 x 1024 O2 molecules in 2.5 mol
of oxygen
The Rule
Number of particles (N) in a given amount of
substance
= amount of substance (n) x Avogadro’s
constant (6.02 x 1023)
OR
N = n x NA
Calculate the amount of copper atoms (mol)
represented by 3.01 x 1023 copper atoms
6.02 x 1023 copper atoms represents 1 mol of
copper atoms.
So, 3.01 x 1023 copper atoms represents:
3.01 x 1023
6.02 x 1023
= 0.500 mol of copper atoms
The relationship
 Amount
of substance (in mol) in a given
sample
= number of particles in sample
number of particles in 1 mol
OR
N
n= N
A
Your Turn
 Try
questions 5 a,c,e, 6 a,c and 7 b,d from
page 61 in your text book
Calculate the amount (in mol) of oxygen
atoms (O) in 5 mol of oxygen (O2) molecules
 Solution
One O2 molecule contains 2 O atoms
Therefore, 1 mol of O2 molecules contains 2
mol of O atoms
Therefore 5 mol of O2 molecules contains 10
mol of O atoms
More Questions
 Try
Question 8 from page 61 in your text
book
Molar Mass
 The
mass of 1 mol of a particular element
or compound is known as its molar mass.
 The
particles of different elements or
compounds, whether they are atoms, ions
or molecules will have different masses.
 The
masses of 1 mol samples of different
particles will obviously be different
Molar Mass

Consider an atom of 12C and one of 24Mg.
12C has a relative isotopic mass of 12
24Mg has a relative isotopic mass of 24

That is one 24Mg weighs approximately twice
that of 12C.

Since 1 mol of 12C weighs exactly 12 g 1 mol of
24Mg must weigh approximately twice as much
(24 g)
In general

The molar mass of an element is the relative
atomic mass of the element expressed in grams

The molar mass of a compound is the relative
molecular or relative formula mass of a
compound expressed in grams

The symbol for molar mass is M. The unit of
measurement of molar mass is g mol-1
Molar Mass
 M(CO2)
= 44 g mol-1 can be interpreted as:

The molar mass of carbon dioxide is 44 g mol-1

1 mole of carbon dioxide has a mass of 44 g

6.02 x 1023 molecules of carbon dioxide has a
mass of 44 g
Worked Example 4.3a
 Calculate:
a) the molar mass of table sugar, sucrose
(C12H22O11)
b) the mass of 2.5 moles of sucrose
Solution
a)
The molar mass of sucrose is its relative
molecular mass expressed in grams
So M(C12H22O11) = (12 x 12.0) + (22 x 1.0) +
(11 x 16.0)
= 342 g mol-1
b)
From part a, we know that 1 mole of sucrose
has a mass of 342 g.
So 2.5 moles of sucrose has a mass of
2.5 x 342 g, which is 855 g
The Rule
Amount of substance (mol)
Mass of a given
amount of substance (g)
m
n=
M
Molar Mass
(g mol -1)
Worked Example 4.3b
 Calculate
the mass of 0.35 mol of
magnesium nitrate (Mg(NO3)2.
 What
 How
is the formula we use?????
do we work out M
Solution

m=nxM
 (Mg(NO3)2
So:
m(Mg(NO3)2
= 24.3 + 2(14.0 + 3 x 16.0)
= 24.3 + 2 x 62.0
= 148.3 g mol-1
= n(Mg(NO3)2 x M(Mg(NO3)2
= 0.35 mol x 148.3 g mol-1
= 52 g
Why only two significant figures?
Your Turn
 Page
65
 Question 9 (every second one)
 Questions 10
Counting by Weighing
 If
we have a known mass of a substance
can we possibly know the number of
particles present?
 We
can by combining both the equations
we have learnt so far.
m = n x M
and
N = n x NA
 How
do we do this??
Worked Example 4.3c
 a)
Calculate the amount (in mol) of CO2
molecules present in 22 g of carbon
dioxide.

b) What is the number of molecules
present in this mass of CO2.
Solution
a)
m
Since n =
M
M(CO2) = 12.0 + 2 x 16.0
= 44.0 g mol-1
m(CO2)
n(CO2) =
M(CO2)
22 g
n(CO2) =
44.0 g mol-1
= 0.50 mol
Solution Part b)
b) Since N = n x NA, where N is the number of
specified particles:
N(CO2) = n(CO2) x NA
= 0.50 x 6.02 x 1023
= 3.0 x 1023
So, 22 g of carbon dioxide contains 3.0 x 1023 ??
Atoms, ions or molecules?
Molecules
Your Turn
 Page
65
 Question 11 (every second one)
 Question 12
 Question 13
Percentage Composition

The values for molar masses of elements in
compounds can be used to calculate the
percentage composition of a compound once its
formula is known.
 This type of calculation is important in chemistry.
 If for example, a company is producing
aluminium from alumina (Al2O3), the
management will want to know the mass of
aluminium that can be extracted given a quantity
of alumina.
The Rule
 Percentage
Composition
Mass of element in 1 mole of compound
% by mass
=
Mass of 1 mole of compound
of the element
x 100
Worked Example 4.4a
 Calculate
the percentage of aluminium in
alumina (Al2O3).
 Solution:
Step 1. Find the molar mass of Al2O3
M(Al2O3) = (2 x 27) + (3 x 16)
= 102 g mol-1
Solution

Step 2:
Find the percentage of aluminium in the alumina.
Since 1 mol Al2O3 contains 2 mol Al atoms:
Mass of aluminium in 1 mol (102 g) of Al2O3 = 2 x 27 g
= 54 g
Mass of Al in 1 mole of Al2O3
x 100
% of Al in Al2O3 =
Mass of 1 mole of Al2O3
54
=
x 100
102
= 52.9%
The company management, therefore, knows that aluminium
comprises about 53% by mass of any sample of alumina.
Your Turn
 Page
69 in text
 Question 14
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.
 You may have wondered how chemists
have determined the formulas of
compounds such as water, carbon dioxide,
nitrogen dioxide and other compounds.
Empirical Formulas
Compound
Empirical Formula
Whole number
ratio of elements
in compound
H:O
2:1
Water
H2O
Carbon dioxide
CO2
C:O
1:2
Calcium
Carbonate
CaCO3
Ca : C : O
1 : 1:3
Empirical Formulas

Empirical formulas are determined
experimentally, usually by determining the mass
of each element present in a given mass of
compound.
 To determine the empirical formula of a
compound, therefore, an experimentally
determined ratio by mass must be converted to
a ratio by numbers of atoms.
 This is done by calculating the amount (in mol)
of each element
Empirical Formulas – the process
Step 1: Measure the mass (m) of each
element in the compound
Step 2: Calculate the amount in mole (n)
of each element in the compound
Step 3: Calculate the simplest whole number ratio of moles
of each element in the compound
Step 4: Determine the empirical formula of the compound
Worked example 4.4b
 A compound
of carbon and oxygen is
found to contain 27.3% carbon and 72.7%
oxygen by mass. Calculate the empirical
formula of the compound.
Solution
Carbon
Oxygen
Step 1: Record the mass (m) of
each element. If masses are given
as a percentage assume that the
sample weighs 100g
27.3
72.7
Step 2: Calculate the amount in
moles (n) of each element
27.3
Step 3: Divide the number of
moles of each element by the
smallest value from step 2
2.27
Step 4: Obtain the simplest whole
number ratio
1
12.0
2.27
72.7
= 2.27
16.0
= 4.54
4.54
=1
2.27
=2
2
The empirical formula of this compound is therefore CO2
Your Turn
 Lets
do one together on the board
 Page 69 in text
 Question 15
Molecular Formula
 While
an empirical formula gives the
simplest whole number ratio of each
element in a compound, a molecular
formula gives the actual number of atoms
in one molecule of the compound.
 Note
that the empirical formula for a
compound can be the same or different
from its molecular formula
Molecular Formula
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.
Worked example 4.4e

A compound has the empirical formula CH. The molar
mass of this compound is 78 g mol-1. What is the
molecular formula of this compound?
 Solution
The molecule must contain a whole number of (CH) units.
The molar mass of a (CH) unit is 13 g mol-1.
If the compound has a molar mass of 78 g mol-1, then:
The number of CH units in a molecule
molar mass of compound
= molar mass of one unit
78 g mol-1
=
13 g mol-1
=6
The molecular formula of the compound is, therefore, 6 x CH
Ie C6H6
Your Turn
 Lets
do 4.4f together.
 Page 69 in text
 Question 16 and 17.
End of Outcome 1
 Yay,
you have just completed the theory
section from outcome 1.
 We still have a few practicals to finish but
your topic 3 questions are due.
 We will shortly be having a topic test on
topic 3. So start studying.