# Mass Spectrum

I am stuck on this question. How do you know what the numbers are on the y axis and how do you go about working it out. This is from pmt and i have attached the link below. The question I'm referring to is 1b. Thanks

Original post by Sansy786
I am stuck on this question. How do you know what the numbers are on the y axis and how do you go about working it out. This is from pmt and i have attached the link below. The question I'm referring to is 1b. Thanks

You don’t need to know them - you can just count the number of divisions (e.g at m/z = 124, the peak is 2 divisions high).
Original post by TypicalNerd
You don’t need to know them - you can just count the number of divisions (e.g at m/z = 124, the peak is 2 divisions high).

Thanks! I don't know how that didn't occur to me. Lol
Original post by Sansy786
Thanks! I don't know how that didn't occur to me. Lol

And as for the second part of your question, do you know how to calculate relative atomic mass when you have intensities rather than percentage abundances?
Original post by TypicalNerd
And as for the second part of your question, do you know how to calculate relative atomic mass when you have intensities rather than percentage abundances?

by intensities do you mean abundance? If you mean intensities then i don't know. I know percentage abundance and abundance calculations, cos i think you just input the values into the equation RAM= (mass x abundance) + (mass x abundance) then divide by the total abundance rather than 100%. Also I am stuck on another question I dont know how you find molecular mass in mass spectroscopy. Stuck on question 5 and 6.1
https://filestore.aqa.org.uk/resources/chemistry/AQA-7404-7405-SG-TOFMS-QA.PDF
(edited 8 months ago)
Original post by Sansy786
by intensities do you mean abundance? If you mean intensities then i don't know. I know percentage abundance and abundance calculations, cos i think you just input the values into the equation RAM= (mass x abundance) + (mass x abundance) then divide by the total abundance rather than 100%. Also I am stuck on another question I dont know how you find molecular mass in mass spectroscopy. Stuck on question 5 and 6.1
https://filestore.aqa.org.uk/resources/chemistry/AQA-7404-7405-SG-TOFMS-QA.PDF

Intensity in this context refers to the height of
the peak, which in turn is proportional to the percentage abundance of the isotope.

With Q’s 5 and 6.1, you need to be aware that the m/z ratio = the relative molecular mass of the ion(s) responsible for the peaks (since at A level, you assume the ions have a charge of +1).

For Q5, what are the mass numbers of the two isotopes of chlorine? How might this help in light of the above?

For Q6.1, when you ionise larger molecules, they often fragment into smaller ions. Which peak is most likely caused by the unfragmented molecule? How might that help?
(edited 8 months ago)
Original post by Sansy786
X

Hi @Sansy786,

Thank you for posting in the Chemistry Forum!

I've updated the title of your thread to 'Mass Spectrum' for better clarity.

5hyl33n
Original post by TypicalNerd
Intensity in this context refers to the height of
the peak, which in turn is proportional to the percentage abundance of the isotope.

With Q’s 5 and 6.1, you need to be aware that the m/z ratio = the relative molecular mass of the ion(s) responsible for the peaks (since at A level, you assume the ions have a charge of +1).

For Q5, what are the mass numbers of the two isotopes of chlorine? How might this help in light of the above?

For Q6.1, when you ionise larger molecules, they often fragment into smaller ions. Which peak is most likely caused by the unfragmented molecule? How might that help?

Ok so for Q5, I would divde 70, 72, 74 all by 2, as you need to divide the mass by the charge which is 2 to get 35, 36, 37
And for 6.1 so do you look at the highest m/z value, which is the unfragmented one, which is 56, so it wont be 57 as thats probably fragmentated.
Thank you for this, I had to read the questions over over again until i understood
(edited 8 months ago)
Original post by Sansy786
Ok so for Q5, I would divde 70, 72, 74 all by 2, as you need to divide the mass by the charge which is 2 to get 35, 36, 37
And for 6.1 so do you look at the highest m/z value, which is the unfragmented one, which is 56, so it wont be 57 as thats probably fragmentated.
Thank you for this, I had to read the questions over over again until i understood

For Q5, the ions only have a charge of +1 (remember this is A level, so all the ions are assumed to have charges of +1). Therefore the m/z values must correspond to the relative masses of the ions. So we must have three Cl2^+ ions, each with different masses. Can you represent the formulae of each? (For an idea of how you might express the formula, an oxygen molecule made up of one atom of 16O and one atom of 18O would be 16O=18O)

For Q6.1, you look at the highest m/z values. I just re-read the question myself and was surprised. It should be 56. The peak at 57 is the so called M+1 peak, which is caused by the presence of 13C (so molecules that have not fragmented but contain a slightly heavier isotope of carbon than the usual 12C). I didn’t think the M+1 peak was on the syllabus for AQA, but the way you distinguish it from the M+ peak is by comparing the heights. As it is such a tiny peak at m/z = 57, you can be sure it’s the M+1 peak.
(edited 8 months ago)