Question

Bromine has two naturally occurring isotopes, one is $Br-79$ (mass = 78.9183 u) with abundance of 50.54%. If the average atomic mass of bromine is 79.904, the mass of other isotope will be?

Answer 1

Hint: In order to answer this question, we must recall the concept of average atomic mass and the formulae used in the calculation of Average atomic mass. We will use those formulae and find the correct answer.


Complete solution:
Atomic Mass of an Element:
Every element naturally occurring in nature has several isotopes with different values of natural abundance. Natural abundance is nothing but the percentage share of an isotope in the composition of the element naturally found in nature. The atomic mass of an element is the average atomic mass of all the isotopes of the element. The average atomic mass of an element can be expressed as the sum of the product of the atomic masses of each isotope and its natural abundance.
Average atomic mass is expressed in the atomic mass unit and in short it is expressed as $amu$ or $au$ .
Average atomic mass of an element $m=a_1{m}_1+a_2{m}_2+.....a_n{m}_n$ . In this equation $a_1,a_2,....a_n$ are the fraction representing the natural abudance of the isotopes and $m_1,m_2,....m_n$ are the atomic mass of the isotopes.
Step 1: In this step we will enlist all the given quantities:
Average atomic mass of the element Bromine $m=79.904amu$
There are two naturally occurring isotopes for the element Bromine.
Isotopic mass of $Br-79$ is $m_1=78.9183amu$
Natural abundance of $Br-79$ is $a_1=50.54\mathrm{\%}=0.5054$
Step 2: In this step we will calculate the Average atomic mass of the second isotope od Bromine i.e. $Br-81$ :
Let $a_2,m_2$ be the natural abundance and isotopic mass of the second isotope $Br-81$
Since there are only two isotopes the natural abundance of the second isotope $a_2=1-a_1\Rightarrow a_2=1.0-0.5054\Rightarrow a_2=0.4946$
The atomic mass of the element Bromine can be expressed as $m=a_1{m}_1+a_2{m}_2$
Therefore, the mass of the isotope $Br-81$ :
$\Rightarrow m_2=\frac{m-m_1{a}_1}{a_2}$
$\Rightarrow m_2=\frac{79.904-78.9183\times 0.5054}{0.4946}amu$
$\Rightarrow m_2=80.9112$
Hence, $m_2=80.9112$ is the required atomic mass of the other isotope of Bromine $Br-81$ .


Note:Isotopes are variants of a particular chemical element which differ in neutron number, and consequently in nucleon number. All isotopes of a given element have the same number of protons but different numbers of neutrons in each atom.