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A Level Chemistry Mass Spectrometry

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Reply 20
Here's what the question and markscheme say:

The mass spectrum of butanone shows its main peak at m/z 72. It also has a small signal at m/z 73.
a) Which ionisation technique is likely to have been used?
b) What is the relative formula mass of this compound?
c) Give two reasons for the peak at m/z 73.
d) There would also be tiny peaks at m/z 74, 75, etc. Explain why some ions with these m/z values may be formed
but why their signals may be too small to be seen.

a electron impact
b 72
c molecule containing one 13C, or molecule containing one 2H
d molecules containing more than one 13C or 2H; low abundance so very few of these relative to those without

I'm even more confused than when I first started, is the markscheme wrong?
Original post by usermeme1112
Yes, 72 is the relative formula mass because remember small molecular peaks are usually unstable molecular fragments because if you look at a mass spec graph the final major peak is considered as the relative formula mass of the molecule any peak smaller than the Final major peaks are considered as unstable isotopic peaks.
and it is Electrospray Ionisation because electron impact is only used for molecules with low formula mass and electron spray is used on large formula molecules to prevent a lot of fragmentation which would occur with low formula mass molecules.
Reply 21
Here's the question and the answers
The mass spectrum of butanone shows its main peak at m/z 72. It also has a small signal at m/z 73.
a) Which ionisation technique is likely to have been used?
b) What is the relative formula mass of this compound?
c) Give two reasons for the peak at m/z 73.
d) There would also be tiny peaks at m/z 74, 75, etc. Explain why some ions with these m/z values may be formed
but why their signals may be too small to be seen.

a electron impact
b 72
c molecule containing one 13C, or molecule containing one 2H
d molecules containing more than one 13C or 2H; low abundance so very few of these relative to those without

I'm even more confused than when I first begun, should I assume the markscheme to be incorrect? Thank you for the help
Original post by usermeme1112
Yes, 72 is the relative formula mass because remember small molecular peaks are usually unstable molecular fragments because if you look at a mass spec graph the final major peak is considered as the relative formula mass of the molecule any peak smaller than the Final major peaks are considered as unstable isotopic peaks.
and it is Electrospray Ionisation because electron impact is only used for molecules with low formula mass and electron spray is used on large formula molecules to prevent a lot of fragmentation which would occur with low formula mass molecules.
My teacher says that we don’t need to learn this yet, however I am really confused about how it can be solved. The question is: all particles have the same kinetic energy (=1/2 mv2, where m=mass of the particle and v= velocity) and the velocity of the particles is given by v=d/t (where d=distance travelled and t= time taken). If the time of flight of a 54Cr+ ion is 1.486 x 10-5 seconds, calculate the time of flight of a 50Cr+ ion. Give your answer to the appropriate sig figs. Any help would be appreciated
Original post by FV2
My teacher says that we don’t need to learn this yet, however I am really confused about how it can be solved. The question is: all particles have the same kinetic energy (=1/2 mv2, where m=mass of the particle and v= velocity) and the velocity of the particles is given by v=d/t (where d=distance travelled and t= time taken). If the time of flight of a 54Cr+ ion is 1.486 x 10-5 seconds, calculate the time of flight of a 50Cr+ ion. Give your answer to the appropriate sig figs. Any help would be appreciated

It’s kinda complicated, pm me if you want and I’ll explain it to you 😊

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