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    I am doing a past paper and this question came up:

    A mass spectrometer was used to analyse a sample of bromine, Br2, with only the Br-79 and Br-81 isotopes present. Explain why a very small peak occurs at m/z = 80.

    The mark scheme says:

    An explanation that makes reference to the following points:

     Br2 has a mass of 160 (and 158 and 162) (1)
     Br22+ has a m/z = 80 (1)

    Allow molecule made of one atom 79Br and one atom 81Br


    Please can someone explain this as I am confused.
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    (Original post by Sam_C_)
    I am doing a past paper and this question came up:

    A mass spectrometer was used to analyse a sample of bromine, Br2, with only the Br-79 and Br-81 isotopes present. Explain why a very small peak occurs at m/z = 80.

    The mark scheme says:

    An explanation that makes reference to the following points:

     Br2 has a mass of 160 (and 158 and 162) (1)
     Br22+ has a m/z = 80 (1)

    Allow molecule made of one atom 79Br and one atom 81Br


    Please can someone explain this as I am confused.
    All particles can be ionised in the MS and produce a trace at M+ = m/z

    Occasionally a particle can be doubly ionised, M2+, in which case it has an m/z value equal to a particle with half the mass which is singly ionised.

    So in a sample of bromine, the possible m/z values for the molecular ions are Br2+ = 158, 160 and 162

    If one of those is doubly ionised it will have an m/z vaue of 79, 80 and 81 respectively.
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    (Original post by charco)
    All particles can be ionised in the MS and produce a trace at M+ = m/z

    Occasionally a particle can be doubly ionised, M2+, in which case it has an m/z value equal to a particle with half the mass which is singly ionised.

    So in a sample of bromine, the possible m/z values for the molecular ions are Br2+ = 158, 160 and 162

    If one of those is doubly ionised it will have an m/z vaue of 79, 80 and 81 respectively.

    Ok thanks I think I get it now, it is just slightly complicated.
 
 
 
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