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Amount of Substance
Key terms in this chapter are:
Element
Compound
Mixture
Atom
Molecule
Ion
Relative Atomic Mass
Avogadro constant
Mole
Isotope
Relative Isotopic Mass
Relative Molecular Mass
Relative Formula Mass
Simple molecular structure
Giant ionic structure
Molecular formula
Empirical formula
Structural formula
Structural Isomer
Before starting, you must check up on elements, compounds and mixtures.
Write some brief notes to outline these ideas. The following link will help
you.
http://www.avogadro.co.uk/definitions/elemcompmix.htm
Relative Atomic Mass and the Avogadro constant
In this section you are going to learn:
•
•
•
•
•
The meaning of the term relative atomic mass for elements containing
atoms.
The meaning of the term relative isotopic mass for elements with isotopes.
That the relative atomic mass of an element when taken in grams is the
mass of 1 mole of that element.
About the Avogadro constant, its numerical value, the term 1 mol, and how
these relate to elements.
How to do simple calculations involving the mole.
We start the topic ‘Amount of Substance’ with a question…
Chemists don't deal with the masses of atoms and molecules measured in grams
because they would be dealing with extremely small numbers. Instead, the masses of
atoms and molecules are compared with that of the isotope carbon-12. 12C is assigned
a relative atomic mass of 12. As an example, consider an atom of magnesium. Two
atoms of 12C are required to be equal in mass to one atom of Mg therefore the relative
atomic mass of Mg is 24. So, using the scales below, one atom of magnesium is
balanced by two atoms of carbon-12. You should note that relative masses have no
units because in their calculation the units have cancelled.
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Q1.
Use a chemistry textbook or Periodic Table to look up the relative
atomic mass of silver. Now draw in the correct number of carbon
atoms needed to balance one atom of Ag.
Study the information in the table below. It refers to the atoms of elements.
Element
Ca
Na
Pb
Fe
Cl
H
He
Ag
Relative Atomic
Mass, Ar
40
23
207
56
35.5
1
4
108
40 g
23 g
207 g
56 g
35.5 g
1g
4g
108 g
Mass of
1 mol
In each of the above masses of elements, there is the same number of atoms. That is,
the relative atomic mass, Ar, of an element expressed in grams always contains the
same number of atoms.
This number can be determined experimentally. It is 6.023 x 1023. The number is a
constant, called the Avogadro constant, L (or NA). The mass of this amount of atoms
is called one mole (1 mol).
Amedeo Avogadro was born in Turin, Italy,
on 9th August 1776. It was in 1811 when he
published his famous hypothesis, now called
Avogadro's law. Avogadro died on the 9th
July 1856. The postage stamp pictured was to
remember Avogadro 100 years after his
death.
Avogadro's law is considered in a later part of this
course.
For an element with isotopes, the relative mass of the atoms of each of isotope of that
element can be referred to as the relative isotopic mass. If an element has no isotopes
then the term relative atomic mass is normally used. We will return to this idea in the
next section.
Note that chemists often use the term relative atomic mass when relative isotopic mass would be
strictly more correct.
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Relative Molecular Mass - Elements and compounds consisting of molecules
In this section you are going to learn:
•
•
•
•
•
That a molecule is a group of two or more covalently bonded atoms.
That some elements and some compounds consist of molecules.
That the relative molecular mass of an element or compound when taken in
grams is the mass of 1 mole of that substance.
About the units of the Avogadro constant.
How to do simple calculations involving the mole.
Study the information in the table below. It refers to the molecules of elements and
compounds.
Element or
compound
Relative
Molecular Mass,
Mr
Mass of
1 mol
CO2
SO2
N2
H2
Cl2
CH4
H2O
44
64
28
2
71
16
18
44 g
64 g
28 g
2g
71 g
16 g
18 g
NH3
17 g
A molecule is…
a group of two or more atoms that are bonded together by covalent bonding.
A molecule is an entity in itself. A dot-cross diagram representing methane,
CH4, is represented below.
The relative molecular mass, Mr, of an element or compound expressed in grams
always contains the same number of molecules. Again, this number is the Avogadro
Constant.
The Avogadro constant has units…
Since the Avogadro Constant is the number of particles (atoms, molecules, ions) in 1
mole it has units – they are mol-1 (particles mol-1, but the particles bit is omitted). The
Avogadro Constant is 6.023 x 1023 mol-1.
Relative molecular mass, Mr, can be calculated by adding together the relative atomic
masses of the constituent atoms, e.g. Mr (CO2) = 12 + 16 + 16 = 44.
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Q2.
The relative molecular mass of water is 18. Complete the diagram
below by balancing the number of water molecules against carbon
atoms.
Relative Formula Mass - Compounds consisting of ions (ionic compounds)
In this section you are going to learn:
•
•
•
•
That Sodium Chloride, NaCl, is an example of an ionic compound. It has a
giant ionic lattice structure.
The terms 'relative formula mass' and 'formula unit' as they apply to ionic
compounds.
What is meant by 1 mol of an ionic compound and how this relates to the
amounts in moles of the ions it contains.
How to do simple calculations involving the mole.
Giant ionic lattice structure (for NaCl) means that…
in the crystal structure of sodium chloride each Na+ ion is surrounded by six
Cl- ions and each Cl- ion by six Na+ ions. This arrangement of ions can in
theory go on and on indefinitely. This is why the structure is referred to as
'giant'. It is represented below.
You can read more about this at:
http://www.avogadro.co.uk/structure/chemstruc/ionic/g-ionic.htm
Also, the structure is held together by electrostatic attractions between the
oppositely charged ions. These are called ionic bonds. Sodium chloride
therefore is not molecular, it is ionic. Also, the formula NaCl is an empirical
formula (see later), and is sometimes called a formula unit.
It is best to use the term relative formula mass (rather than relative molecular mass)
for ionic compounds. For the formula unit Na+Cl- the relative formula mass is 23 +
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35.5 = 58.5. One mole of Na+Cl- is 58.5 g. This mass contains 1 mol (23 g) of Na+
ions and 1 mol (35.5 g) of Cl- ions, that is, two moles of ions in total.
Note that sometimes the term relative molecular mass is used to refer to ionic compounds.
Q3.
Work out the relative formula mass of FeSO4,7H2O(s).
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Q4.
What is the mass of 1 mole of iron(II) sulphate(VI)-7-water?
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Q5.
You have 2 moles of FeSO4,7H2O. How many moles of each type
of ion and molecule are present in this amount?
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Mole Calculations
In this section you are going to learn:
More about simple calculations involving the mole concept.
•
Think about the above ideas and you should be able to calculate the amount in moles
of a given mass of a substance, e.g. 35.5 g of Cl atoms is 1 mol. 35.5 g of Cl2
molecules is ½ mole of Cl2 molecules.
If you are given the amount in moles, then multiplying this by the mass of one mole
gives the amount in grams, e.g. 1½ mole of SO2 is 1½ x 64 = 96 g.
Q6.
Calculate the mass in grams of ¾ mole of sulphuric(VI) acid,
H2SO4.
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Q7.
You have 145 g of propanone, CH3COCH3, what amount of this is
there in moles?
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Balanced Chemical Equations
In this section you are going to learn:
That a balanced chemical equation gives information about the amounts of
reactants that react and of products that form.
How to do simple mole calculations involving chemical equations.
•
•
Balanced chemical equations give information about the molar amounts of substances
that react and the molar amounts of products that form.
For example,
2Mg(s)
+ O2(g) →
2MgO(s)
The above equation tells us that 2 moles of magnesium atoms react with 1 mole of
oxygen molecules to form 2 moles of magnesium oxide, an ionic compound. This
means that, for example, 12 g (½ mole) Mg will require 8 g (¼ mole) of O2 to react
completely to form 20 g (½ mole) MgO.
Q8.
If 12.80 g of magnesium metal is completely burned in excess
oxygen, what mass of magnesium oxide is obtained?
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Q9.
Some granulated zinc metal was added to an excess of a solution
of hydrochloric acid. The zinc reacted completely and 4 g of
hydrogen gas was formed. Calculate the mass of zinc in grams
added to the acid.
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Chemical Formulae
•
•
The meaning of empirical formula, molecular formula and structural formula
and how these are related.
About structural isomers and how to draw these for simple alkanes.
This section refers to compounds that are molecular.
With a little knowledge of organic chemistry you would know that C4H10 is the
formula for two alkanes. It tells us that in each of these there are four atoms of carbon
and ten atoms of hydrogen bonded together. This type of formula is called a
molecular formula. It gives the actual number of each type of atom in the molecule
but does not tell you the arrangement in which they are bonded.
The simplest whole-number ratio of C:H is 2:5. This gives the formula C2H5, and it is
called the empirical formula. (Remember that NaCl, for example, is an empirical
formula because of the nature of its giant ionic lattice structure.)
Q10. The molecular formula of glucose is C6H12O6. What is its empirical
formula?
...........................................................................................................
Q11. The molecular formula of sucrose is C12H22O11. What is its
empirical formula?
...........................................................................................................
So, the empirical formula of a covalent compound is the simplest whole-number ratio
of atoms in its molecules. Or, you could say ‘…the simplest whole-number ratio of
moles of atoms in 1 mole of the compound’. If you know the empirical formula of a
compound and also its relative molecular mass, then the molecular formula can be
found. This is illustrated below.
Q12. A hydrocarbon consists of 82.8% carbon and 17.2% hydrogen. Its
relative molecular mass is 58. Calculate the empirical formula and
the molecular formula of the compound.
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Page 8 of 14
A structural formula shows the arrangement in which the atoms are bonded. For
C4H10 these are shown below and are referred to as structural isomers:
Some additional questions for you to try...
1.
An atom of titanium has four times the mass of an atom of 12C.
What is its relative atomic mass?
...........................................................................................................
2.
An atom of helium has one-third the mass of an atom of 12C. What
is its relative atomic mass?
...........................................................................................................
3.
A molecule of water has 1½ times the mass of an atom of 12C.
What is its relative molecular mass?
...........................................................................................................
4.
What mass of 12C consists of 12.04 x 1023 atoms?
...........................................................................................................
5.
The relative atomic mass of lead is 207. What is the molar mass
(mass of 1 mole) of Pb?
...........................................................................................................
6.
What is the mass of 2 moles of molecules of ethanol, CH3CH2OH
(Ar (H) = 1; Ar (C) = 12; Ar (O) = 16)?
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7.
What mass of sulphur contains the same number of atoms as there
are present in 1.5 moles of magnesium (Ar (Mg) = 24; Ar (S) = 32)?
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8.
What mass of ammonia (NH3) contains the same number of
molecules as there are atoms present in 2 moles of sodium (Ar (H)
= 1; Ar (N) = 14; Ar (Na) = 23)?
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9.
How many moles of oxygen are there in 128 g of O2 (Ar (O) = 16)?
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10.
How many moles of sodium sulphate(VI), Na2SO4, are contained in
71 g (Ar (O) = 16; Ar (Na) = 23; Ar (S) = 32)?
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11.
How many moles of Al3+ are there in 2 moles Al2(SO4)3?
...........................................................................................................
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12.
In the following reaction, what mass (in grams) of SO2(g) is formed
when ½ mole of S reacts with oxygen (Ar (O) = 16; Ar (S) = 32;)?
S(s) + O2(g) → SO2(g)
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13.
In an experiment to find the formula of copper iodide, 0.17 g of Cu
reacts with 0.34 g of iodine. What is the formula of copper iodide
(Ar (Cu) = 63.55; Ar (I) = 126.90)?
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14.
In an experiment to find the formula of a lead oxide, 11.95 g of the
oxide, on reduction (removal of oxygen), gave 10.35 g of Pb. What
is the formula of the oxide (Ar (0) = 16; Ar (Pb) = 207)?
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15.
What is the percentage mass of S in SO2 (Ar (O) = 16; Ar (S) =
32)?
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16.
What is the percentage mass of water of crystallisation (H2O) in
FeSO4.7H2O(s) (Ar (H) = 1; Ar (O) = 16; Ar (S) = 32; Ar (Fe) = 56)?
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The Mass Spectrometer
In this section you are going to learn:
•
•
•
How the mass spectrometer works.
How relative isotopic mass and relative atomic mass are related, and how to
calculate a relative atomic mass from mass spectral data.
How the mass spectrometer is used in the analysis of molecular structure.
The existence of isotopes of elements was discovered using an instrument called a
mass spectrometer.
The mass spectrometer was invented by the English physicist Francis William Aston (18771945) when he was working in Cambridge with J. J. Thomson. Aston eventually discovered
many of the naturally occurring isotopes of non-radioactive elements. He was awarded the
Nobel Prize for Chemistry in 1922.
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Time-in-flight Mass Spectrometer
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The following description of a time-of-flight mass spectrometer is very simplified.
We assume all ions produced leave the source at the same time, with the same kinetic
energy.
A time-of-flight mass spectrometer identifies sample atoms (or molecules) by
measuring their flight time.
This diagram above shows the working principle of a linear time-of-flight mass
spectrometer.
It is necessary to put a charge on the atoms (or molecules) by bombarding them with
electrons emitted from a filament. Each atom (or molecule) loses one electron to form
an ion with a single positive charge.
M(g) + e- → M+(g) + 2e-
The sample ions (M+) are then accelerated by an electric field to give each of them the
same kinetic energy. They then enter into a part of the mass spectrometer, called the
flight path, which has no electric field. All of the ions travel the same distance
through this part of the instrument to reach the ion detector. The speed of an ion is
dependent upon its mass, with heavy ions having a lower velocity than light ones, and
so the time it takes for an ion to reach the detector is related to its mass.
The mass spectrometer measures the relative abundance (relative number) and relative
mass of each type of ion reaching the detector. This data is represented as a mass
spectrum. The mass of an atom or molecules is expressed relative to that of an atom
of the isotope carbon-12 (12C = 12).
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Where an element has two or more isotopes, the mass spectrometer measures the
relative isotopic masses and their relative numbers (relative abundances) and plots a
mass spectrum, as shown below.
The relative isotopic mass of a chlorine atom is either 35 or 75. However, the relative
atomic mass of chlorine is given as 35.5 in a chemical data book. This is because, for
an element with isotopes, the relative atomic mass of an element takes into account
the relative abundance and relative mass of each isotope present. In a typical sample
of naturally occurring chlorine, 100 atoms will consist of 75 of relative mass 35 and
25 of relative mass 37.
Using the mass spectrum on the left, the relative atomic mass of Cl is calculated as
follows:
Ar = (75/100 x 35) + (25/100 x 37) = 35.5
In the spectrum on the right, the most abundant isotope is assigned a relative
abundance of 100% and each of the other isotopes is assigned a relative abundance
relative to that.
Q13. Use the mass spectrum on the right above to calculate the relative
atomic mass, Ar, of chlorine.
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Q14. The mass spectrum of antimony shows it has two naturally
occurring isotopes at relative mass 121 (relative abundance
57.3%) and relative mass 123 (relative abundance 42.7%)
respectively. Draw a diagram to represent this mass spectrum and
calculate the relative atomic mass, Ar, of antimony.
Page 13 of 14
The mass spectrometer in the analysis of molecular structure
When the molecules of a substance enter the mass spectrometer to become ionised by
loss of an electron, the 'molecular ion' (M+) is formed. Some of these pass through the
instrument and remain intact, to produce a peak in the mass spectrum at the relative
molecular mass. Others are likely to fragment (break up) in various ways to produce
smaller positively charge ions, which produce corresponding peaks in the mass
spectrum. These numerous peaks can reveal how the atoms of the molecule were
bonded together, that is its molecular structure. There is more about these ideas in a
later module.
Some additional questions about mass spectrometry…
This chapter provides some additional questions to help you assess your learning.
When you have answered these and have checked your answers against the correct
ones, refer to the table that follows to find out what you may need to do to further
your progress.
1.
An atom of titanium has four times the mass of an atom of 12C.
What is its relative atomic mass?
...........................................................................................................
2.
An atom of helium has one-third the mass of an atom of 12C. What
is its relative atomic mass?
...........................................................................................................
3.
Using a mass spectrometer, it is possible to determine the number
of
A protons in an atom.
B energy levels in an atom.
C atoms in 1 mole of an element.
D isotopes of an element.
E neutrons in an atom.
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4.
Iron has four isotopes with relative numbers of atoms as follows:
54
Fe (5.8%), 56Fe (91.7%), 57Fe (2.2%), 58Fe (0.3%). Calculate the
relative atomic mass of iron.
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5.
The mass spectrum for naturally-occurring magnesium is shown
below. Singly charged Mg+ ions have produced the peaks. What is
the relative atomic mass of magnesium?
A 24.2
B 24.3
C 24.4
D 24.7
E 24.8
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6.
Which one of the following particles has seven protons, eight
neutrons and nine electrons?
A 78NB 714N2C 715ND 815N2E 715N2...........................................................................................................