Download - The Mass Spectrometer
XMass Number (A)
protons + neutrons
A
Atomic Number (Z)number of protons
Zn+/n-
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.The symbol (X) for a
given element
• 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
• magnesium results from the mass spectrometer:– 80% 24Mg– 10% 25Mg– 10% 26Mg
Calculate the relative atomic mass of magnesium with the provided data. (2.2.3)
• 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
– (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
Be clear with your answer and state
the percent of each isotope.