


The relative atomic mass quoted on the periodic table is a weighted average of all the isotopes R.A.M = (isotopic mass x % abundance) 100 24 25 26 20 40 60 80 100 % abundance m/z 78.70% 10.13% 11.17% Fig: spectra for Magnesium from mass spectrometer Use these equations to work out the R.A.M For above example of Mg R.A.M = [(78.7 x 24) + (10.13 x 25) + (11.17 x 26)] /100 = 24.3 3 If asked to give the species for a peak in a mass spectrum then give charge and mass number e.g. 24Mg+ 24Mg+ 25Mg+ 26Mg+ R.A.M = (isotopic mass x relative abundance) total relative abundance If relative abundance is used instead of percentage abundance use this equation N Goalby chemrevise.org Example: Calculate the relative atomic mass of Tellurium from the following abundance data: 124-Te relative abundance 2; 126-Te relative abundance 4; 128-Te relative abundance 7; 130-Te relative abundance 6 R.A.M = [(124×2) + (126×4) + (128×7) + (130×6)] 19 = 127.8 Example: Copper has two isotopes 63-Cu and 65-Cu. The relative atomic mass of copper is 63.5. Calculate the percentage abundances of these two isotopes. 63.55 = yx63 + (1-y)x65 63.55 = 63y +65 -65y 63.55 = 65 -2y 2y = 1.45 y = 0.725 %abundance 63-Cu =72.5% %abundance 65-Cu = 27.5%
Cl has two isotopes Cl35 (75%) and Cl37(25%) Br has two isotopes Br79 (50%) and Br81(50%) These lead to the following spectra caused by the diatomic molecules 70 72 74 m/z relative abundance Cl35Cl35 +Cl35Cl37 +Cl37Cl37 + 158 160 162 m/z relative abundance Br79Br79 + Br79Br81 + Br81Br79 + Br81Br81 + Mass spectrometers have been included in planetary space probes so that elements on other planets can be identified. Elements on other planets can have a different composition of isotopes Measuring the Mr of a molecule If a molecule is put through a mass spectrometer with an Electron impact ionisation stage it will often break up and give a series of peaks caused by the fragments. The peak with the largest m/z, however, will be due to the complete molecule and will be equal to the relative molecular mass , Mr ,of the molecule. This peak is called the parent ion or molecular ion Molecular ion C4H10 + Spectra for C4H10 58 Mass spectrum for butane 43 29 Mass spectra for Cl2 and Br2 If a molecule is put through a mass spectrometer with Electro Spray Ionisation then fragmentation will not occur. There will be one peak that will equal the mass of the MH+ ion. It will therefore be necessary to subtract 1 to get the Mr of the molecule. So if a peak at 521.1 is for MH+ , the relative molecular mass of the molecule is 520.1. The 160 peak has double the abundance of the other two peaks because there is double the probability of 160 Br79 -Br81 + as can be Br79-Br81 and Br81-79
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3.1.1.2 Mass number and isotopes
The mass spectrometer gives accurate information about relative isotopic mass and also about the relative abundance of isotopes. Mass spectrometry can be used to identify elements. Mass spectrometry can be used to determine relative molecular mass. Students should be able to: • interpret simple mass spectra of elements • calculate relative atomic mass from isotopic abundance, limited to mononuclear ions.