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Powerpoint #2
The Mass Spectrometer
 The mass spectrometer can be used to measure the
mass of individual atoms.
 The mass of an individual hydrogen atom is
1.67 x 10-24g, and the mass of a carbon atom is
1.99 x 10-25g. The masses of all the elements are in the
range of 10-24 to 10-22g. Since these numbers are so
small, it makes sense to use relative masses instead.
 The relative atomic mass of an element, Ar, is the
average mass of an atom of the element taking into
account all its isotopes and their relative abundance,
compared to one atom of carbon-12
The Mass Spectrometer
 The results of the analysis of the mass spectrometer
are presented in the form of a mass spectrum.
 The horizontal axis shows the mass/charge ratio of the
different ions, and the relative abundance of the ions is
shown on the vertical axis.
 Ex. The mass spectrum of gallium shows that in a
sample of 100 atoms, 60 have a mass of 69, and 40
have a mass of 71. Calculate the relative atomic mass of
the element.
Total mass= (60 x 69) + (40 x 71) = 6980
Average mass= total mass/ number of atoms 6980/100=
69.80
The Mass Spectrometer
 Example: Deduce the relative atomic mass of the
element rubidium from the data given of 100 atoms.
70
60
50
40
30
20
10
0
77
23
85
86
87
88
The Mass Spectrometer
 Total mass = (85 x 77) + ( 87 x 23) = 8546
 Relative atomic mass = average mass of atoms
 Total mass/ number of atoms
 8546/100 = 85.46
 Boron exists in two isotopic forms. 10B and 11B. 10B is used as a
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control for nuclear reactors. Use your Periodic Table to find the
abundances of the two isotopes.
Solution: Consider a sample of 100 atoms.
Let x be 10B atoms. The remaining atoms are 11B.
Number of 11B atoms = 100 – x
Total mass = 10x + (100 – x)11
= 10x + 1100 – 11x
= 1100 – x
Average mass= total mass/ # of atoms
= 1100 – x/100
From the periodic table: the relative atomic mass of boron = 10.81
10.81 = 1100-x/100
1081 = 1100 – x
X = 1100 – 1081
X = 19.00
The abundance for 10B= 19.00% and 11B= 81.00%
Questions to be answered
1) Which ion would be deflected most in a mass
spectrometer?
A) 35Cl+
B) 37Cl+ C) 37Cl+2 D) (35Cl37Cl)+
2) What is the same for an atom of phosphorus-26 and
phosphorus-27?
A) atomic # and mass # B) # of protons and electrons
C) # of neutrons and electrons
D) # of protons and neutrons
3) Use the periodic table to find the percentage
abundance of neon-20, assuming that neon has only
one other isotope, neon-22.
Electron Arrangement
 Electromagnetic radiation comes in different forms of
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differing energy.
All electromagnetic waves travel at the same speed (c)
but can be distinguished by their different wavelengths
(λ).
The distance between two successive crests (or
troughs) is called the wavelength.
The frequency (f) is the number of waves which pass a
point in one second.
The wavelength and frequency are related by the
equation c=f λ where c= speed of light.
c= 3 x 108m/s
Electron Arrangement