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    Methods of analysis and detection, a2 chem.

    Relative abundance of the M peak is 8.8
    Calculate the relative abundance of the M+1 peak.

    Equation is


    n = height of (M+1) peak x 100 / height of m peak x 1.1

    There's a mass spectrum etc.

    The peak stuff makes sense. I just don't get how you figure out what n is? According to the mark scheme it's = 2.

    Why?!
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    The peak of the (m+1) peak is going to be [n x 1.1/100] as large as the m peak. Where n is the number of carbons.

    This is beacuse 13C is 1.1% of carbon.

    So if you have C40H70 then the (m+1) peak is going to be 44% as high as the m peak.

    Or C2H4 then (m+1) peak is going to be 2.2% as high as the m peak.
 
 
 
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