A Level Chemistry Mass Spectrometry

OP

Original post by knksj112

I don't understand the answer to this question
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?

For a) I wrote electron impact but I don't know if my reasoning is correct- is it because fragmentation has occurred to produce two peaks rather than one which would be seen from electrospray ionisation

For b) I thought that the relative formula mass would be 73 as the greatest m/z peak value equals the relative formula mass but the mark scheme says it's 72?

Could someone please explain where I've gone wrong in part a and b? Thank you

Reply 2

Prestige XV

The peak at 73 is due to the butanone with one of the carbon atoms being a carbon 13 isotope. This is a low chance so it’s a small peak. But the RFM is just the same as butanone as the formula hasn’t changed.

Reply 3

AlevelMBC

10

what spec are you doing? because if youre aqa, fragmentation isnt on spec anymore

Reply 4

OP

Thank you.
The presence of C13 or H2 isn't mentioned on the spec? How would I figure out whether the peak is the molecular ion peak or due to C13?

Original post by Prestige XV

The peak at 73 is due to the butanone with one of the carbon atoms being a carbon 13 isotope. This is a low chance so it’s a small peak. But the RFM is just the same as butanone as the formula hasn’t changed.

Reply 5

Thomas_Pickin10

I am an A Level student so and i haven't heard of 'ionisation techniques'. However, i think where you have gone wrong on part B is that the relative formula mass of Butanone is 72 (C4H8O), that's why it's not 73. I actually don't know why there would be a peak at 73. Just out of interest, is this GCSE?

Original post by knksj112

I don't understand the answer to this question
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?

For a) I wrote electron impact but I don't know if my reasoning is correct- is it because fragmentation has occurred to produce two peaks rather than one which would be seen from electrospray ionisation

For b) I thought that the relative formula mass would be 73 as the greatest m/z peak value equals the relative formula mass but the mark scheme says it's 72?

Could someone please explain where I've gone wrong in part a and b? Thank you

Reply 6

OP

Oh I didn't know this. My teacher said that you do need to know it for AQA..

Original post by AlevelMBC

what spec are you doing? because if youre aqa, fragmentation isnt on spec anymore

Reply 7

OP

No it's A Level. There's electron impact and electrospray ionisation

Original post by Thomas_Pickin10

I am an A Level student so and i haven't heard of 'ionisation techniques'. However, i think where you have gone wrong on part B is that the relative formula mass of Butanone is 72 (C4H8O), that's why it's not 73. I actually don't know why there would be a peak at 73. Just out of interest, is this GCSE?

Reply 8

AlevelMBC

10

its definitely a level, the 3rd post was correct in stating the peak at 73 is due to a carbon 13 being present

Original post by Thomas_Pickin10

I am an A Level student so and i haven't heard of 'ionisation techniques'. However, i think where you have gone wrong on part B is that the relative formula mass of Butanone is 72 (C4H8O), that's why it's not 73. I actually don't know why there would be a peak at 73. Just out of interest, is this GCSE?

Reply 9

Prestige XV

Original post by knksj112

Thank you.
The presence of C13 or H2 isn't mentioned on the spec? How would I figure out whether the peak is the molecular ion peak or due to C13?

The molecular ion peak is much larger than the M+1 peak. As I said the probability of the compound contain a carbon 13 is much smaller, so the relative intensity of the peak is much smaller than the molecular ion peak.

Reply 10

Thomas_Pickin10

Original post by knksj112

No it's A Level. There's electron impact and electrospray ionisation

Oh okay. I do OCR A Chemistry and we don't do that. Are you on AQA?

Reply 11

AlevelMBC

10

Yep, no longer on our spec. All you need to know now is electrospray ionisation, and its method in gaining a H+ ion. and electron gun, where high energy electrons bombard an atom to knock an electron off

Original post by knksj112

Oh I didn't know this. My teacher said that you do need to know it for AQA..

Reply 12

OP

Sorry but what do you mean by the M+1 peak?
Isn't the molecular ion peak usually smaller anyway? How does the peak with C13 differ to that of the molecular ion peak?

Original post by Prestige XV

The molecular ion peak is much larger than the M+1 peak. As I said the probability of the compound contain a carbon 13 is much smaller, so the relative intensity of the peak is much smaller than the molecular ion peak.

Reply 13

OP

Yep

Original post by Thomas_Pickin10

Oh okay. I do OCR A Chemistry and we don't do that. Are you on AQA?

Reply 14

anaby01

in electrospray ionization, the compound gains a proton so it becomes M + 1 and that’s why the peak is at 73 rather than 72.

(edited 4 years ago)

Reply 15

OP

Well that makes revision a bit easier. Thank you!

Original post by AlevelMBC

Yep, no longer on our spec. All you need to know now is electrospray ionisation, and its method in gaining a H+ ion. and electron gun, where high energy electrons bombard an atom to knock an electron off

Reply 16

OP

The markscheme says electron impact

Original post by anaby01

in electrospray ionization, the compound gains a proton so it becomes M + 1 and that’s why the peak is at 73 rather than 72.

Reply 17

anaby01

but if it’s electron impact electrons are knocked off so there’s no change in molecular mass?

Original post by knksj112

The markscheme says electron impact

Reply 18

Habib-Reh2

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.

Original post by knksj112

I don't understand the answer to this question
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?

For a) I wrote electron impact but I don't know if my reasoning is correct- is it because fragmentation has occurred to produce two peaks rather than one which would be seen from electrospray ionisation

For b) I thought that the relative formula mass would be 73 as the greatest m/z peak value equals the relative formula mass but the mark scheme says it's 72?

Could someone please explain where I've gone wrong in part a and b? Thank you

Reply 19

Habib-Reh2