The Mass Spectrometer
Topic 2.2
Review of Topic 2.1
Mass Number (A)
protons + neutrons
A
Z
X
n+/n-
Atomic Number (Z)
number of protons
The symbol (X) for a
given element
Charge (n)
Atoms have no charge,
so this is left blank.
Ions are atoms that
have gained or lost
electrons, the charge
is indicated here.
Mass Spectrometer (Topic 2.2)
• has many applications, but one of the simplest
is to determine the natural abundances of the
isotopes of a particular element
– the relative atomic mass can be calculated from
the data from the mass spectrometer
Mass spectrometer video (2:26)
http://www.youtube.com/watch?v=_L4U6ImYSj0
• a positively charged particle is deflected along
a circular path that is proportional to its
mass/charge ratio
– m/z
• mass is m
• charge is z
• occurs in a vacuum
• the machine can be adjusted in order to look
at certain particles
• Five parts (a simple diagram with these parts is required)
– vaporization
• a substance is first converted to a vapor/gaseous
state
– ionization
• the sample (atoms or molecules) are bombarded
with a stream of electrons
– the collisions knock one or more electrons off to make
positive ions
– normally one electron is knocked off leaving a 1+
charge
–acceleration
• sample is accelerated through a magnetic
field
–deflection
• ions are deflected with a magnet and
electromagnetic field
• deflections depends on:
– lighter particles deflect more
– higher positive charged particles deflect more
–detection
• particles with different masses will be
detected at different points at the end
deflection
Carbon- 12 as a standard
• carbon- 12
– ALL masses on the periodic table are based on
their relationship to carbon-12
• the C-12 atom has been given the atomic weight of
exactly 12.000000000 and is used as the basis upon
which the atomic weight of other isotopes is
determined
Calculate the relative atomic mass of
magnesium with the provided data. (2.2.3)
• magnesium results
from the mass
spectrometer:
– 80%
– 10%
– 10%
24Mg
25Mg
26Mg
• just a simple weighted mean
– .80(24) + .10(25) + .10(26) = 24.3 amu
Calculate the abundance (the % of each isotope
found in nature) of each isotope (2.2.3)
• Rubidium (Rb) has relative atomic mass of
85.47 and two isotopes
– rubidium- 85 and rubidium- 87
• make rubidium 85 = x
• make rubidium 87 = y
– (x · 85) + (y · 87) = 85.47
• x+y=1
• therefore substitute (1 – x) for y
Be clear with your
answer and state
the percent of each
isotope.
– (x · 85) + ((1-x) · 87) = 85.47
• solve for x
• x = .765 or 76.5% for rubidium- 85
• therefore y = .235 or 23.5% for rubidium- 87