The Structure of the Atom
Length: 2154 words (6.2 double-spaced pages)
Basic Atomic Particles :
Atoms are made up of the following particles :
Protons : Protons are positively charged particles with a mass of one
atomic mass unit. They are found in the nucleus at the centre of the
Neutrons : Neutrons are not charged and have a mass of one atomic mass
unit. They are found, with protons at the centre of the atom.
Electrons : Electrons are negatively charged particles with a mass of
1/1846th of an atomic mass unit. They are arranged in shells around
the central nucleus.
Useful definitions for atomic structure :
Atomic number (Z) : The number of protons in the nucleus of an atom
but is also equal to the number of electrons in the same atom (because
atoms don't carry an overall charge - the positives balance the
Mass Number, (A) (or atomic mass) : The number of neutrons and protons
in the nucleus of one atom of the element
The atomic number and mass number of a particular element can be found
from the periodic table. The atomic number is always displayed to the
bottom left of the symbol for the element. The mass number is always
displayed to the top left of the symbol for the element.
This symbol shows that sodium atoms have 11 electrons orbiting a
nucleus that contains 11 protons and (23 - 11=) 12 neutrons.
Sodium has an atomic number of 11 (and so each sodium atom contains 11
electrons and 11 protons).
The structure above represents the G.C.S.E. model of the atomic
structure of sodium. Electrons orbit a central nucleus in well defined
Simple Diagram of an Atom : The Sodium atom.
The first electron shell (closest to the nucleus) can contain at most
The second electron shell can contain at most 8 electrons.
The third electron shell can contain, at most, 18 electrons.
As a general rule electrons occupy the lowest available electron
As the number of negative charges (electrons) balances against the
number of positive charges (protons) the sodium atom has no overall
Q1. Define the following terms :
Atomic number, Mass No, Relative atomic mass, isotope.
Q2. Using a periodic table, give the correct symbols (including mass
No and atomic number for the following elements) : Hydrogen, Oxygen,
Carbon, Chlorine (35).
Q3. Using a periodic table, complete the following table.
No of Protons
No of neutrons
No of electrons
Q4. The mass of a single neutron is 1.65 x 10-24g and the mass of an
electron is 9.11 x 10-28g. From this data and the avogadro number
(6.02 x 1023) calculate the exact mass of one mole of :
(i) 1H atoms
(ii) 2H (deuterium) atoms
(iii) 3H (tritium) atoms
This question can be difficult. You might need to look at the working
out provided in the answers section for part (I) before completing
parts (ii) and (iii).
Q5. Explain why the exact RAM of carbon is 12.011 when the RAM's of
all other elements are found by comparison to 12C, which is given a
value of 12.000.
Locate the element chlorine on your copy of the periodic table. Notice
that it appears to have an atomic number (Z) of 17 and a mass number
(A) of 35.5.
How can this be? Surely the nucleus of the chlorine atom doesn't
contain 18.5 neutrons!!!
No! In fact there are two forms of chlorine atoms. They both contain
17electrons and 17 protons but one form (or isotope) of chlorine
contains 18 neutrons while the other contains 20.
Isotopes are atoms of the same element with the same atomic number (Z)
but different mass numbers (A).
Chemically the two chlorine isotopes, 37Cl and 35Cl are almost
identical as they differ only in their respective masses.
Q6 (HARD QUESTION). What physical properties might you expect to be
different for different isotopes of the same element?
The two forms of chlorine are not present in equal quantities.
In fact, in nature, only 25% of all chlorine is 37Cl, while 75% is 35Cl.
Therefore the average mass of all chlorine is :
Average mass = (37 x 25/100) + (35 x 75/100) = 35.5
The number 35.5 is the relative atomic mass of chlorine.
Relative Atomic Mass : The average relative mass of the atoms of an
element, taking into account natural abundance of isotopes, on a scale
where 12C has a mass of 12.000.
Q7. Lithium has two isotopes with atomic masses 6 and 7. 5.9% of all
Lithium is 7Li. Calculate the relative atomic mass of Li.
Q8. Indium (In) has two isotopes (of mass numbers 113 and 115) , 91%
of all Indium is 115In . Calculate the relative atomic mass of Indium.
Q9. The table below gives information on the various isomers of
Germanium. Use it to determine the Relative Atomic mass of Germanium.
Relative Abundance (%)
Q10. The table below gives information on the various isomers of
Palladium. Use it to determine the Relative Atomic mass of Palladium.
Relative Abundance (%)
The Mass Spectrometer.
How is one able to detect the presence of isotopes? The Mass
Spectrometer (shown below) allows us to discriminate between particles
based on their masses. The operation of the mass spectrometer is
A sample compound, for example a natural isotopic sample of atomic
chlorine, is placed into a mass spectrometer.
1) Gaseous (volatilised) atoms of the sample compound are passed
initially through a field of fast moving electrons, where they are
ionised according to the equation (ionisation) :
M(g) + e- è M+(g) + 2e-
2) The positively charged ions that result from ionisation are passed
through a series of slits which accelerate the particles
(acceleration) and focus them into a beam.
3) The beam of ions then passes through (and are deflected by) a
variable electromagnetic field (deflection). As the strength of the
magnetic field varies, a point is obtained at which each ion is
capable of passing through the spectrometer and colliding with a
sensitive screen that detects their presence (detection).
The extent of deflection in the magnetic field depends on the mass to
charge ratio (m/z) of the molecularion (As the molecularion has a
charge of 1+ the mass to charge ratio is equivalent to the relative
molecular mass of the sample compound)
How the mass spectrometer is able to differentiate between atoms of
One way of visualising the effect of a magnetic field on different
ions is to envisage firing differently sized steel ball bearings close
to a magnet.
Heavy ball bearings will only be slightly affected (if at all) by the
magnetic field while the path of the light ball bearings will be
strongly deflected by the magnetic field.
In the same way, at relatively light magnetic field strengths, ions of
small size are able to be deflected sufficiently to pass through to
the detector. However the magnetic field strength needs to be
increased considerably to deflect heavy ions enough so that they are
able to be detected.
During the mass spectrum of any sample, ionisation and acceleration
take place continually. The magnetic field starts at its lowest
setting before increasing gradually to its highest setting. The ions
are detected by the sensitive screen as they pass through the mass
spectrometer at the correct magnetic field strength. This very small
detection is photomultiplied before being sent to a suitable detector.
Increasing Magnetic Strength è
Notice that for the mass spectrum of atomic chlorine, there are two
One signal corresponds to a mass to charge ratio of 35 and is due to
the detection of 35Cl+ ions at the detector.
The other signal (at 37) is due to the presence of 37Cl atoms in the
original sample .
Notice that the peak due to 35Cl is three times higher (greater
relative intensity) than the peak of 37Cl. This reflects the fact that
there is three times as much 35Cl present in nature than there is 37Cl.
What does a mass spectrum look like?
Please note : Ionisation in a mass spectrometer is a precarious
business. Atoms are passed into a field of fast moving electrons. Most
of the time only one electron will be removed from the atoms at this
Cl + e- è Cl+ + 2e-
However, in some cases two electrons may be removed from the same
Cl + e- è Cl2+ + 3e-
This leaves us with a dipositive ion (2+) rather than a unipositive
one (1+) and the mass to charge (m/z) ratio of these ions will be half
of the mass to charge ratio of the (1+) ion.
How would this affect the mass spectrum of chlorine atoms?
We can see that on the true mass spectrum of a natural sample of
chlorine atoms, not only would there be the peaks due to 35Cl+ and 37Cl+
, but there would also appear much smaller peaks at exactly half of
the 1+ ion values.
These peaks would be due to the presence (in limited quantities) of
the 37Cl2+ ion and the 37Cl2+ ion.
Q11. A scientist wants to find the exact RAM of titanium. A sample of
titanium was placed into a mass spectrometer the following results
Mass to charge ratio (m/z)
a) Using a periodic table, give the correct symbol for the isotope
responsible for the peak at (m/z = 48). (1)
b) In addition to these peaks, a very small peak was found at mass to
charge ratio 24.
(i) Give the correct symbol for the ion causing this peak in the mass
(ii) Why was this peak not included in this data? (2)
c) Calculate the exact RMM of titanium (2)
Q12. A scientist wants to find the exact RAM of Zinc. A sample of zinc
was placed into a mass spectrometer the following results were
Mass to charge ratio (m/z)
Percentage of total peak area
a) Using a periodic table, give the correct symbol for the isotope
responsible for the peak at (m/z = 67). (1)
b) In addition to these peaks, very small peaks were found at mass to
charge ratio 32 and 33. Why were these peaks not included in this
data? Give the correct symbol for the ion causing this peak in the
mass spectrometer. (2)
c) Calculate the exact RMM of zinc.
Mass Spectrometry of molecules.
It is quite possible to place compounds (rather than elements) into a
If the information provided by the mass spectrum is used correctly we
can often determine the relative molecular mass of the molecules in
Molecules undergo the same four processes in a mass spectrometer that
atoms do : Ionisation, acceleration, deflection, detection
This means that molecules often become ionised (through the removal of
one electron) to form a molecularion.
M + e- è M+ + 2e-
The molecularion can then be detected. The peak produced would give a
mass to charge ratio equivalent to the relative molecular mass of the
However the conditions within a ,mass spectrometer are such that
molecules can be broken apart by passing through the field of fast
moving electrons :
M + e- è X + Y+ + 2e-
This random process is known as fragmentation and means that the mass
spectrum of molecules is often complicated by having any number of
peaks due to fragments in addition to the peak due to the
For example, the mass spectrum of ethanol, C2H5OH is shown above .
Notice that despite there being a number of peaks in the spectrum it
is relatively easy to locate the molecularion peak (and hence the
relative molecular mass of the compound).
The molecularion will always be the peak with the highest mass to
Attempt Questions on Mass Spectra Analysis
Molecules containing isotopic elements..
We have seen that there are two isotopic forms of chlorine, 35Cl and
37Cl, and they occur in the ratio 3:1 respectively in nature.
Because of this, molecules that contain chlorine (chloroalkanes - for
example) will form two molecularions that will always be in the ratio
Q1. Draw mass spectra showing the molecularions you would expect for
the following compounds :
Chloromethane (CH3Cl), Chloroethane(C2H5Cl) and Chloropropane (C3H7Cl).
Ans = All of these compounds should give mass spectra with a
characteristic region around the molecularion with a large peak at,
for example m/e = 50 (chloromethane) and other molecularion peak 2
mass units further (52) on but only 1/3rd of the size of the first
Therefore chloroethane would have a large molecularion peak with a
mass to charge ration of 64, with a smaller peak (one third of the
size) at 66.
peak with a mass to charge ration of 78, with a smaller peak (one
third of the size) at 80. Therefore chloropropane would have a large
Brominated Organic molecules.
Bromoalkanes can also often be identified by the shape of their
molecularion peaks. Bromine has two isotopes 79Br (50.5%) and 81Br
As a result mass spectra of compounds containing one bromine atom per
molecule consist of two molecularion peaks of near equal intensity two
mass units apart.
Q2. What would the shapes of the molecularion peaks be for
Bromomethane (CH3Br), Bromoethane (C2H5Br), Bromopropane (C3H7Br)?
Ans : Bromomethane would produce two molecularion peaks of near
identical relative intensity with mass units 94 and 96 (see above)
Bromoethane would produce two molecularion peaks of near identical
relative intensity with mass units 108 and 110
Bromopropane would produce two molecularion peaks of near identical
relative intensity with mass units 122 and 124 (see above)
How would you use this information?
The examiner may give you information that suggests the presence of a
halogen group within an organic molecule, you could use a mass
spectrum of the compound (if one is provided) to determine if chlorine
or bromine is present in the molecule, by looking for the presence of
multiple molecularion peaks and their relative shapes.
Q3. Draw the low resolution mass spectra you would expect for the
following compounds, clearly showing the molecularions that would
Question 3 (c) and (d) are particularly difficult but should prove a
useful puzzle for the mathematicians!!!
a) Should contain two peaks of equal size at 106 (containing 79Br) and
108 (containing 81Br)
b) Should contain a large molecularion peak at 62 (containing 35Cl)
and a smaller molecularion peak at 64 (containing 37Cl) that is one
third of the size of the other peak.
c) There should be three molecularion peaks in this spectrum, one at
186 (containing two 79Br atoms), one at 188 (containing one 79Br and
one 81Br - there are two possibilities as to how this might arise) and
one peak at 190 (containing two 81Br atoms). The heights of these peak
should be roughly in the ratio (1:2:1) reflecting the probability of
d) There should be three molecularion peaks in this spectrum, one at
98 (containing two 35Cl atoms), one at 100 (containing one 35Cl and
one 37Cl - there are two possibilities as to how this might arise) and
one peak at 102 (containing two 37Cl atoms). The heights of these peak
should be in the ratio (0.5625:0.375:0.0625) reflecting the
probability of each occuring.
(0.75 x 0.75 = 0.5625) : ((0.75 X 0.25) X2 = 0.375), (0.25 X 0.25 =