Isotopic abudance

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To determine the atomic mass of such an element, it is necessary to know not only masses of the individual isotopes but also their atom percents that is isotopic abudances in nature.


Defination: The precentage of the total number of atoms that is composed of a particular isotope is the isotopic or relative abudance of an isotope. For example; the isotopic abudance of the neon isotopes in the naturally occuring neon are neon-20 90.51% ; neon-21 0.27% ; neon-22 9.22% 


General equation for atomic mass calculation: 


Fortunately isotopic abudance as well as isotopic masses can be determined by mass spectrometry.The situation with chorine, which has two stable isotopes is shown below.For chlorine the data obtained from mass spectrometer are


                                    Atomic mass                       Abudance


Cl-35                            34.97 amu                            75.53%


Cl-37                            36.97 amu                            24.47 %


We interpret this to mean that , in elemental chlorine, 75.53% of the atoms have a mass of 34.97 amu and the remaining atoms , 24.47% of the total , have a mass of 36.97 amu . With this information we can readily calculate the atomic mass of chlorine using the general equation:


Atomic mass Y = (atomic mass Y1) x %Y1/100% + (atomic mass Y2) x %Y2/100% + .........


where Y1, Y2 ...... are the isotopes of element Y.


so, atomic mass of Cl = 34.97 amu x 75.53/100 + 36.97 amu x 24.47/100 = 35.46amu


 


One more example;


Sometimes question want to find out the relative abudance when atomic mass of element is given along with each isotope's abudance percentage and atomic mass.Lets see


Bromine (atomic mass = 79.90) consists of two isotopes:Br-79 (78.92 amu) and Br-81 (80.92 amu).What is the abudance of the heavier isotope?


Strategy: The key to solving this problem is to realize that the abudances of the isotopes have to add to 100%.If we let x = abudance of Br-81, then abudance of Br-79 must be (100-x) .


solution: substituting in equation


Atomic mass Y = (atomic mass Y1) x %Y1/100% + (atomic mass Y2) x %Y2/100% + .........


79.90 amu = 78.92amu x (100-x)/100 + 80.92amu (x)/100


                 = 78.92 amu + 2.00 amu (x)/100


solving: x/100 = (79.90 - 78.92)amu/ 2.00 amu  = 0.49 :


                 x = 100(.49) = 49%     


 

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