Chemistry mass spectrometry .HELP PLS !!!

Bromine is an element in group 7 it has 2 isotopes 79 BR and 81br that naturally occur in roughly equally proportions complete a mass spectrum graph which has relative abundance against mass/ charge ratio and you have to explain the number location and proportion of your peaks

Reply 1

UtterlyUseless69

Original post by Borana34

What have you tried so far?

Hint if you don’t know where to start

Reply 2

Borana34

79x2 and 81x2 and then finding relative abundance both of them at about 50%and I don't know what to do more

Reply 3

Borana34

Original post by Borana34

79x2 and 81x2 and then finding relative abundance both of them at about 50%and I don't know what to do more

Plus I don't have an idea how to do the graph

Reply 4

charco

Study Forum Helper

Original post by Borana34

79x2 and 81x2 and then finding relative abundance both of them at about 50%and I don't know what to do more

Don't forget that there are also ⁷⁹Br⁸¹Br molecules that are statistically twice as likely

Reply 5

Borana34

Oh yeah thanks completely forgot about them

Reply 6

UtterlyUseless69

Original post by Borana34

Plus I don't have an idea how to do the graph

On the x-axis, you plot the m/z values and on the y-axis, you plot the intensities.

Of course, there isn’t really a way as such to put an exact number on what the intensities are, but you can use the probability calculations from GCSE maths to work out the ratios.

They’ve told you that the proportions of each isotope are roughly equal, which implies each one accounts for 50% of all bromine atoms, since there are two isotopes. You could also express the probabilities as decimals (i.e 50% = 0.5), which will probably make the next bit easier.

As such, how would you find the probability that two bromine atoms are 79Br?

(edited 1 year ago)

Reply 7

UtterlyUseless69

Original post by Borana34

79x2 and 81x2 and then finding relative abundance both of them at about 50%and I don't know what to do more

Yes, and as Charco has pointed out, you also have 79Br-81Br, too.

So what would the Mr values for (79Br)2, 79Br-81Br and (81Br)2 be?

Because this is A level (or at least presumed to be), you work under the assumption that the ions have charges of +1 and so the m/z ratios are the Mr values.

(edited 1 year ago)

Reply 8

Borana34

So it's 158,162,and two 160s so it's double

Reply 9

UtterlyUseless69

Original post by Borana34

So it's 158,162,and two 160s so it's double

Yeah, you would have peaks at m/z = 158 and m/z = 162 of equal heights and one peak at m/z = 160 that is twice the height of either.

It’s also probably worth mentioning that there is a small degree of fragmentation, so you will also have two (small) peaks of equal height at m/z = 79 and m/z = 81.