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The Mass Spectrometer Topic 2.2

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Review of Topic 2.1

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X Mass Number (A) protons + neutrons A Atomic Number (Z) number of protons Z n+/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

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Mass Spectrometer (Topic 2.2)

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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

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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

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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

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– 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

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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

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magnesium results from the mass spectrometer: – 80% 24 Mg – 10% 25 Mg – 10% 26 Mg 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

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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.

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11 Na 12 Mg 1s 2 2s 2 2p 4 Wrong 22 Ti [Ar] 4s 2 3d 10 4p 2 ⃝ 17 Cl [Ar] 4s 1 3d 5 1s 2 2s 2 2p 6 3s 1 S orbital 1s 2 2s 2 2p 6 3s 2 [Ar] 4s 2 3d 9 p.

11 Na 12 Mg 1s 2 2s 2 2p 4 Wrong 22 Ti [Ar] 4s 2 3d 10 4p 2 ⃝ 17 Cl [Ar] 4s 1 3d 5 1s 2 2s 2 2p 6 3s 1 S orbital 1s 2 2s 2 2p 6 3s 2 [Ar] 4s 2 3d 9 p.

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