Question

An element is found in nature in two isotopic forms with mass numbers (A-1) and (A+3). If the average atomic mass of the element is found to be A, then the relative abundance of the heavier isotope in the nature will be:
(A) $ 66.6\% $
(B) 75%
(C) 25%
(D) $ 33.3\% $

Answer 1

Hint: To answer this question, you must recall the formula for the average atomic mass of an element from the mass and abundance of its isotopes. Isotopes are various forms of elements which differ in the atomic mass.
Formula used: $ {\text{M}} = {{\text{f}}_{\text{1}}}{{\text{M}}_{\text{1}}} + {{\text{f}}_{\text{2}}}{{\text{M}}_{\text{2}}} + ......... + {{\text{f}}_{\text{n}}}{{\text{M}}_{\text{n}}} $
Where, $ {{\text{f}}_{\text{1}}}{\text{, }}{{\text{f}}_{\text{2}}}{\text{,}}......{{\text{f}}_{\text{n}}} $ are the fractions of the natural abundance of the various isotopes.
And $ {{\text{M}}_{\text{1}}}{\text{,}}{{\text{M}}_{\text{2}}}{\text{,}}..........{{\text{M}}_{\text{n}}} $ are the atomic masses of the isotopes.

Complete step by step solution
First we must discuss what average atomic mass of an element is. The average atomic mass of any element is represented by the sum of the masses of all its isotopes and each mass is multiplied with its abundance on the earth. We generally use the average atomic masses of elements for all calculations and considerations.
In this question, we are given the mass numbers of the two isotopic forms an element. The mass numbers of these isotopes are (A - 1) and (A + 3).
Now, let the abundance of the lighter isotope, i.e. (A – 1) be $ x $
So the abundance of the heavier isotope, i.e. (A + 3) will be $ \left( {1 - x} \right) $
The average atomic mass of the element is given to be A. So using the formula for the calculation of average atomic mass, $ {\text{M}} = {{\text{f}}_{\text{1}}}{{\text{M}}_{\text{1}}} + {{\text{f}}_{\text{2}}}{{\text{M}}_{\text{2}}} + ......... + {{\text{f}}_{\text{n}}}{{\text{M}}_{\text{n}}} $
Substituting the values, we get, $ \left( {A - 1} \right)x + \left( {A + 3} \right)\left( {1 - x} \right) = A $
 $ \Rightarrow - 4x + 3 = 0 $
 $ \Rightarrow x = 3/4 $
So, the abundance of the heavier isotope is $ \left( {1 - x} \right) = \left( {1 - 3/4} \right) = 1/4 $
Thus, the correct answer is C.

Note
We know that isotopes of an element have different mass numbers. This arises a problem when we need to write the weight of a sample of the element as to which mass number must be used. Thus, we take an average of the isotopic masses along with taking the abundances into consideration.