the mass spectrometer topic 2.2. review of topic 2.1
TRANSCRIPT
![Page 4: The Mass Spectrometer Topic 2.2. Review of Topic 2.1](https://reader035.vdocuments.net/reader035/viewer/2022062308/56649ccb5503460f94994f5f/html5/thumbnails/4.jpg)
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
![Page 6: The Mass Spectrometer Topic 2.2. Review of Topic 2.1](https://reader035.vdocuments.net/reader035/viewer/2022062308/56649ccb5503460f94994f5f/html5/thumbnails/6.jpg)
• 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
![Page 7: The Mass Spectrometer Topic 2.2. Review of Topic 2.1](https://reader035.vdocuments.net/reader035/viewer/2022062308/56649ccb5503460f94994f5f/html5/thumbnails/7.jpg)
• 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
![Page 8: The Mass Spectrometer Topic 2.2. Review of Topic 2.1](https://reader035.vdocuments.net/reader035/viewer/2022062308/56649ccb5503460f94994f5f/html5/thumbnails/8.jpg)
![Page 9: The Mass Spectrometer Topic 2.2. Review of Topic 2.1](https://reader035.vdocuments.net/reader035/viewer/2022062308/56649ccb5503460f94994f5f/html5/thumbnails/9.jpg)
• 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
![Page 10: The Mass Spectrometer Topic 2.2. Review of Topic 2.1](https://reader035.vdocuments.net/reader035/viewer/2022062308/56649ccb5503460f94994f5f/html5/thumbnails/10.jpg)
–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
![Page 11: The Mass Spectrometer Topic 2.2. Review of Topic 2.1](https://reader035.vdocuments.net/reader035/viewer/2022062308/56649ccb5503460f94994f5f/html5/thumbnails/11.jpg)
deflection
![Page 13: The Mass Spectrometer Topic 2.2. Review of Topic 2.1](https://reader035.vdocuments.net/reader035/viewer/2022062308/56649ccb5503460f94994f5f/html5/thumbnails/13.jpg)
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
![Page 14: The Mass Spectrometer Topic 2.2. Review of Topic 2.1](https://reader035.vdocuments.net/reader035/viewer/2022062308/56649ccb5503460f94994f5f/html5/thumbnails/14.jpg)
• 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
![Page 15: The Mass Spectrometer Topic 2.2. Review of Topic 2.1](https://reader035.vdocuments.net/reader035/viewer/2022062308/56649ccb5503460f94994f5f/html5/thumbnails/15.jpg)
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.