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












6












$begingroup$


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
    Mar 10 at 18:16


















6












$begingroup$


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
    Mar 10 at 18:16
















6












6








6





$begingroup$


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 Mar 10 at 12:27









orthocresol♦

39.6k7114242




39.6k7114242










asked Mar 10 at 12:08









George TianGeorge Tian

717218




717218








  • 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
    Mar 10 at 18:16
















  • 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
    Mar 10 at 18:16










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
Mar 10 at 18:16






$begingroup$
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
Mar 10 at 18:16












2 Answers
2






active

oldest

votes


















13












$begingroup$

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♦
    Mar 10 at 12:27





















10












$begingroup$

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

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

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    13












    $begingroup$

    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♦
      Mar 10 at 12:27


















    13












    $begingroup$

    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♦
      Mar 10 at 12:27
















    13












    13








    13





    $begingroup$

    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 Mar 10 at 12:25









    TAR86TAR86

    4,91911031




    4,91911031








    • 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♦
      Mar 10 at 12:27
















    • 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♦
      Mar 10 at 12:27










    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♦
    Mar 10 at 12:27






    $begingroup$
    It looks like the mass spectrum of butane, so I think it's quite a safe assumption that it's organic ;)
    $endgroup$
    – orthocresol♦
    Mar 10 at 12:27













    10












    $begingroup$

    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









    $endgroup$


















      10












      $begingroup$

      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









      $endgroup$
















        10












        10








        10





        $begingroup$

        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 Mar 10 at 12:34









        ButtonwoodButtonwood

        9,61211942




        9,61211942






























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