This page looks at the information you can get from the mass spectrum of an element. It shows how you can find out the masses and relative abundances of the various isotopes of the element and use that information to calculate the relative atomic mass of the element. It also looks at the problems thrown up by elements with diatomic molecules - like chlorine, Cl
Monatomic elements include all those except for things like chlorine, Cl
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Note: If you need to know how this diagram is obtained, you should read the page describing how a mass spectrometer works. | |||||||||||||||||||||||

The two peaks in the mass spectrum shows that there are 2 isotopes of boron - with relative isotopic masses of 10 and 11 on the | |||||||||||||||||||||||

Notes: are atoms of the same element (and so with the same number of protons), but with different masses due to having different numbers of neutrons.
IsotopesWe are assuming (and shall do all through this page) that all the ions recorded have a charge of 1+. That means that the mass/charge ratio (m/z) gives you the mass of the isotope directly. The ^{12}C isotope weighs exactly 12 units. | |||||||||||||||||||||||

The relative sizes of the peaks gives you a direct measure of the relative abundances of the isotopes. The tallest peak is often given an arbitrary height of 100 - but you may find all sorts of other scales used. It doesn't matter in the least. You can find the relative abundances by measuring the lines on the stick diagram. In this case, the two isotopes (with their relative abundances) are:
The relative atomic mass (RAM) of an element is given the symbol
A "weighted average" allows for the fact that there won't be equal amounts of the various isotopes. The example coming up should make that clear. Suppose you had 123 typical atoms of boron. 23 of these would be The total mass of these would be (23 x 10) + (100 x 11) = 1330 The average mass of these 123 atoms would be 1330 / 123 = 10.8 (to 3 significant figures). 10.8 is the relative atomic mass of boron. Notice the effect of the "weighted" average. A simple average of 10 and 11 is, of course, 10.5. Our answer of 10.8 allows for the fact that there are a lot more of the heavier isotope of boron - and so the "weighted" average ought to be closer to that.
The 5 peaks in the mass spectrum shows that there are 5 isotopes of zirconium - with relative isotopic masses of 90, 91, 92, 94 and 96 on the
This time, the relative abundances are given as percentages. Again you can find these relative abundances by measuring the lines on the stick diagram. In this case, the 5 isotopes (with their relative percentage abundances) are:
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Note: You almost certainly wouldn't be able to measure these peaks to this degree of accuracy, but your examiners may well give you the data in number form anyway. We'll do the sum with the more accurate figures. | |||||||||||||||||||||||

Suppose you had 100 typical atoms of zirconium. 51.5 of these would be | |||||||||||||||||||||||

Note: If you object to the idea of having 51.5 atoms or 11.2 atoms and so on, just assume you've got 1000 atoms instead of 100. That way you will have 515 atoms, 112 atoms, etc. Most people don't get in a sweat over this, and just use the numbers as they are! | |||||||||||||||||||||||

The total mass of these 100 typical atoms would be (51.5 x 90) + (11.2 x 91) + (17.1 x 92) + (17.4 x 94) + (2.8 x 96) = 9131.8 The average mass of these 100 atoms would be 9131.8 / 100 = 91.3 (to 3 significant figures). 91.3 is the relative atomic mass of zirconium. | |||||||||||||||||||||||

Note: If you want further examples of calculating relative atomic masses from mass spectra, you might like to refer to my book, Calculations in A level Chemistry. | |||||||||||||||||||||||

Chlorine is taken as typical of elements with more than one atom per molecule. We'll look at its mass spectrum to show the sort of problems involved. Chlorine has two isotopes, You would be wrong! The problem is that chlorine consists of molecules, not individual atoms. When chlorine is passed into the ionisation chamber, an electron is knocked off the molecule to give a _{2}^{+}. These ions won't be particularly stable, and some will fall apart to give a chlorine atom and a Cl^{+} ion. The term for this is fragmentation.If the Cl The Cl Think about the possible combinations of chlorine-35 and chlorine-37 atoms in a Cl Both atoms could be 35 + 35 = 70 35 + 37 = 72 37 + 37 = 74
That means that you would get a set of lines in the m/z = 70 region looking like this: These lines would be The relative heights of the 70, 72 and 74 lines are in the ratio 9:6:1. If you know the right bit of maths, it's very easy to show this. If not, don't worry. Just remember that the ratio is 9:6:1. What you can't do is make any predictions about the relative heights of the lines at 35/37 compared with those at 70/72/74. That depends on what proportion of the molecular ions break up into fragments. That's why you've got the chlorine mass spectrum in two separate bits so far. You must realise that the vertical scale in the diagrams of the two parts of the spectrum isn't the same. The overall mass spectrum looks like this: | |||||||||||||||||||||||

Note: This is based on information from the NIST Chemistry WebBook. NIST is the US National Institute of Standards and Technology. | |||||||||||||||||||||||

**Where would you like to go now?****To the mass spectrometry menu . . .**
© Jim Clark 2000 (last modified February 2014) |