Why is a very small peak with larger m/z not considered to be the molecular ion?












5












$begingroup$


With regards to the following spectrum:



enter image description here



When asked for the M/Z for the molecular ion, the peak at 58 was taken, not 59.



What causes the peak at 59, and why isn't it taken as the molecular ion?






mass-spectrometry







share|improve this question











$endgroup$








  • 1




    $begingroup$
    The gist here is that nobody can just "know" what the various peaks are. Rather you look at the spectrum and start to look for clues to put together a consistent explanation. So you have a peak at 58 and 43, the difference is 15 where you also have a large peak. 15 is the mass of a methyl group so... and on the analysis goes.
    $endgroup$
    – MaxW
    7 hours ago


















5












$begingroup$


With regards to the following spectrum:



enter image description here



When asked for the M/Z for the molecular ion, the peak at 58 was taken, not 59.



What causes the peak at 59, and why isn't it taken as the molecular ion?






mass-spectrometry







share|improve this question











$endgroup$








  • 1




    $begingroup$
    The gist here is that nobody can just "know" what the various peaks are. Rather you look at the spectrum and start to look for clues to put together a consistent explanation. So you have a peak at 58 and 43, the difference is 15 where you also have a large peak. 15 is the mass of a methyl group so... and on the analysis goes.
    $endgroup$
    – MaxW
    7 hours ago
















5












5








5





$begingroup$


With regards to the following spectrum:



enter image description here



When asked for the M/Z for the molecular ion, the peak at 58 was taken, not 59.



What causes the peak at 59, and why isn't it taken as the molecular ion?






mass-spectrometry







share|improve this question











$endgroup$




With regards to the following spectrum:



enter image description here



When asked for the M/Z for the molecular ion, the peak at 58 was taken, not 59.



What causes the peak at 59, and why isn't it taken as the molecular ion?






mass-spectrometry




mass-spectrometry






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 13 hours ago









orthocresol♦

39.4k7114241




39.4k7114241










asked 13 hours ago









George TianGeorge Tian

677217




677217








  • 1




    $begingroup$
    The gist here is that nobody can just "know" what the various peaks are. Rather you look at the spectrum and start to look for clues to put together a consistent explanation. So you have a peak at 58 and 43, the difference is 15 where you also have a large peak. 15 is the mass of a methyl group so... and on the analysis goes.
    $endgroup$
    – MaxW
    7 hours ago
















  • 1




    $begingroup$
    The gist here is that nobody can just "know" what the various peaks are. Rather you look at the spectrum and start to look for clues to put together a consistent explanation. So you have a peak at 58 and 43, the difference is 15 where you also have a large peak. 15 is the mass of a methyl group so... and on the analysis goes.
    $endgroup$
    – MaxW
    7 hours ago










1




1




$begingroup$
The gist here is that nobody can just "know" what the various peaks are. Rather you look at the spectrum and start to look for clues to put together a consistent explanation. So you have a peak at 58 and 43, the difference is 15 where you also have a large peak. 15 is the mass of a methyl group so... and on the analysis goes.
$endgroup$
– MaxW
7 hours ago






$begingroup$
The gist here is that nobody can just "know" what the various peaks are. Rather you look at the spectrum and start to look for clues to put together a consistent explanation. So you have a peak at 58 and 43, the difference is 15 where you also have a large peak. 15 is the mass of a methyl group so... and on the analysis goes.
$endgroup$
– MaxW
7 hours ago












2 Answers
2






active

oldest

votes


















12












$begingroup$

Without knowing more details, it is hard to guess, but at this $m/z$, it seems likely that the peak is the result of one $^{12}ce{C}$ being substituted by one $^{13}ce{C}$. It is more useful to assume a uniform mass of 12 for carbon when analyzing such a spectrum.



Note that with more carbons in a larger molecule, it becomes more likely that at least one of them is a $^{13}ce{C}$. Thus, the spectra become more complicated that way.



This whole analysis goes out the window if this is not an organic molecule, though similar patterns may arise for other molecules - it depends on the isotopes and their distribution.






share|improve this answer









$endgroup$









  • 4




    $begingroup$
    It looks like the mass spectrum of butane, so I think it's quite a safe assumption that it's organic ;)
    $endgroup$
    – orthocresol♦
    13 hours ago





















8












$begingroup$

The peak at $m/z = 59$ with lower intensity in respect to the one at $m / z =58$ (the molecular ion) is not overseen. Mass spectroscopy is capable to deliver information about the isotopic composition of your sample, too. Since -- accounting over all carbon atoms of your sample -- about 1.1% are of the non-radioactive isotope of $ce{^{13}C}$, you may use this information to determine the carbon atoms in total present.



The story indeed contiues further, as to establish by this "isotope fingerprint" if you have one or two chlorines per molecule present ($ce{^{35}Cl}$ and $ce{^{37}Cl}$), bromines, presence / absence of nitrogen, etc.:



1]
(source)






share|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

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    active

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    12












    $begingroup$

    Without knowing more details, it is hard to guess, but at this $m/z$, it seems likely that the peak is the result of one $^{12}ce{C}$ being substituted by one $^{13}ce{C}$. It is more useful to assume a uniform mass of 12 for carbon when analyzing such a spectrum.



    Note that with more carbons in a larger molecule, it becomes more likely that at least one of them is a $^{13}ce{C}$. Thus, the spectra become more complicated that way.



    This whole analysis goes out the window if this is not an organic molecule, though similar patterns may arise for other molecules - it depends on the isotopes and their distribution.






    share|improve this answer









    $endgroup$









    • 4




      $begingroup$
      It looks like the mass spectrum of butane, so I think it's quite a safe assumption that it's organic ;)
      $endgroup$
      – orthocresol♦
      13 hours ago


















    12












    $begingroup$

    Without knowing more details, it is hard to guess, but at this $m/z$, it seems likely that the peak is the result of one $^{12}ce{C}$ being substituted by one $^{13}ce{C}$. It is more useful to assume a uniform mass of 12 for carbon when analyzing such a spectrum.



    Note that with more carbons in a larger molecule, it becomes more likely that at least one of them is a $^{13}ce{C}$. Thus, the spectra become more complicated that way.



    This whole analysis goes out the window if this is not an organic molecule, though similar patterns may arise for other molecules - it depends on the isotopes and their distribution.






    share|improve this answer









    $endgroup$









    • 4




      $begingroup$
      It looks like the mass spectrum of butane, so I think it's quite a safe assumption that it's organic ;)
      $endgroup$
      – orthocresol♦
      13 hours ago
















    12












    12








    12





    $begingroup$

    Without knowing more details, it is hard to guess, but at this $m/z$, it seems likely that the peak is the result of one $^{12}ce{C}$ being substituted by one $^{13}ce{C}$. It is more useful to assume a uniform mass of 12 for carbon when analyzing such a spectrum.



    Note that with more carbons in a larger molecule, it becomes more likely that at least one of them is a $^{13}ce{C}$. Thus, the spectra become more complicated that way.



    This whole analysis goes out the window if this is not an organic molecule, though similar patterns may arise for other molecules - it depends on the isotopes and their distribution.






    share|improve this answer









    $endgroup$



    Without knowing more details, it is hard to guess, but at this $m/z$, it seems likely that the peak is the result of one $^{12}ce{C}$ being substituted by one $^{13}ce{C}$. It is more useful to assume a uniform mass of 12 for carbon when analyzing such a spectrum.



    Note that with more carbons in a larger molecule, it becomes more likely that at least one of them is a $^{13}ce{C}$. Thus, the spectra become more complicated that way.



    This whole analysis goes out the window if this is not an organic molecule, though similar patterns may arise for other molecules - it depends on the isotopes and their distribution.







    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered 13 hours ago









    TAR86TAR86

    4,88911031




    4,88911031








    • 4




      $begingroup$
      It looks like the mass spectrum of butane, so I think it's quite a safe assumption that it's organic ;)
      $endgroup$
      – orthocresol♦
      13 hours ago
















    • 4




      $begingroup$
      It looks like the mass spectrum of butane, so I think it's quite a safe assumption that it's organic ;)
      $endgroup$
      – orthocresol♦
      13 hours ago










    4




    4




    $begingroup$
    It looks like the mass spectrum of butane, so I think it's quite a safe assumption that it's organic ;)
    $endgroup$
    – orthocresol♦
    13 hours ago






    $begingroup$
    It looks like the mass spectrum of butane, so I think it's quite a safe assumption that it's organic ;)
    $endgroup$
    – orthocresol♦
    13 hours ago













    8












    $begingroup$

    The peak at $m/z = 59$ with lower intensity in respect to the one at $m / z =58$ (the molecular ion) is not overseen. Mass spectroscopy is capable to deliver information about the isotopic composition of your sample, too. Since -- accounting over all carbon atoms of your sample -- about 1.1% are of the non-radioactive isotope of $ce{^{13}C}$, you may use this information to determine the carbon atoms in total present.



    The story indeed contiues further, as to establish by this "isotope fingerprint" if you have one or two chlorines per molecule present ($ce{^{35}Cl}$ and $ce{^{37}Cl}$), bromines, presence / absence of nitrogen, etc.:



    1]
    (source)






    share|improve this answer









    $endgroup$


















      8












      $begingroup$

      The peak at $m/z = 59$ with lower intensity in respect to the one at $m / z =58$ (the molecular ion) is not overseen. Mass spectroscopy is capable to deliver information about the isotopic composition of your sample, too. Since -- accounting over all carbon atoms of your sample -- about 1.1% are of the non-radioactive isotope of $ce{^{13}C}$, you may use this information to determine the carbon atoms in total present.



      The story indeed contiues further, as to establish by this "isotope fingerprint" if you have one or two chlorines per molecule present ($ce{^{35}Cl}$ and $ce{^{37}Cl}$), bromines, presence / absence of nitrogen, etc.:



      1]
      (source)






      share|improve this answer









      $endgroup$
















        8












        8








        8





        $begingroup$

        The peak at $m/z = 59$ with lower intensity in respect to the one at $m / z =58$ (the molecular ion) is not overseen. Mass spectroscopy is capable to deliver information about the isotopic composition of your sample, too. Since -- accounting over all carbon atoms of your sample -- about 1.1% are of the non-radioactive isotope of $ce{^{13}C}$, you may use this information to determine the carbon atoms in total present.



        The story indeed contiues further, as to establish by this "isotope fingerprint" if you have one or two chlorines per molecule present ($ce{^{35}Cl}$ and $ce{^{37}Cl}$), bromines, presence / absence of nitrogen, etc.:



        1]
        (source)






        share|improve this answer









        $endgroup$



        The peak at $m/z = 59$ with lower intensity in respect to the one at $m / z =58$ (the molecular ion) is not overseen. Mass spectroscopy is capable to deliver information about the isotopic composition of your sample, too. Since -- accounting over all carbon atoms of your sample -- about 1.1% are of the non-radioactive isotope of $ce{^{13}C}$, you may use this information to determine the carbon atoms in total present.



        The story indeed contiues further, as to establish by this "isotope fingerprint" if you have one or two chlorines per molecule present ($ce{^{35}Cl}$ and $ce{^{37}Cl}$), bromines, presence / absence of nitrogen, etc.:



        1]
        (source)







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 13 hours ago









        ButtonwoodButtonwood

        9,59211941




        9,59211941






























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