The Mass SpectrometerTopic 2.2

Review of Topic 2.1

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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

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

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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.