Isotopes of chlorine 35 and 378/13/2023 You might suppose that the mass spectrum would look like this: We'll look at its mass spectrum to show the sort of problems involved.Ĭhlorine has two isotopes, 35Cl and 37Cl, in the approximate ratio of 3 atoms of 35Cl to 1 atom of 37Cl. 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.Ĭhlorine is taken as typical of elements with more than one atom per molecule. The average mass of these 100 atoms would be 9131.8 / 100 = 91.3 (to 3 significant figures).ĩ1.3 is the relative atomic mass of zirconium. The total mass of these 100 typical atoms would be Most people don't get in a sweat over this, and just use the numbers as they are! That way you will have 515 atoms, 112 atoms, etc. 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. 51.5 of these would be 90Zr, 11.2 would be 91Zr and so on. Suppose you had 100 typical atoms of zirconium. We'll do the sum with the more accurate figures. 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. In this case, the 5 isotopes (with their relative percentage abundances) are: zirconium-90 Again you can find these relative abundances by measuring the lines on the stick diagram. This time, the relative abundances are given as percentages. 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 12C scale. 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. A simple average of 10 and 11 is, of course, 10.5. Notice the effect of the "weighted" average. The average mass of these 123 atoms would be 1330 / 123 = 10.8 (to 3 significant figures).ġ0.8 is the relative atomic mass of boron. The total mass of these would be (23 x 10) + (100 x 11) = 1330 23 of these would be 10B and 100 would be 11B. Suppose you had 123 typical atoms of boron. The example coming up should make that clear. The relative atomic mass (RAM) of an element is given the symbol A r and is defined as: The relative atomic mass of an element is the weighted average of the masses of the isotopes on a scale on which a carbon-12 atom has a mass of exactly 12 units.Ī "weighted average" allows for the fact that there won't be equal amounts of the various isotopes. In this case, the two isotopes (with their relative abundances) are: boron-10 You can find the relative abundances by measuring the lines on the stick diagram. The tallest peak is often given an arbitrary height of 100 - but you may find all sorts of other scales used. The relative sizes of the peaks gives you a direct measure of the relative abundances of the isotopes. The carbon-12 scale is a scale on which the mass of the 12C isotope weighs exactly 12 units. That means that the mass/charge ratio (m/z) gives you the mass of the isotope directly. We are assuming (and shall do all through this page) that all the ions recorded have a charge of 1+. Notes: Isotopes 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. 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 12C scale. Note: If you need to know how this diagram is obtained, you should read the page describing how a mass spectrometer works. Monatomic elements include all those except for things like chlorine, Cl 2, with molecules containing more than one atom. It also looks at the problems thrown up by elements with diatomic molecules - like chlorine, Cl 2. 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. This page looks at the information you can get from the mass spectrum of an element.
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