Question

Indium (atomic weight=114.82) has two naturally occurring isotopes, the predominant one form has isotopic weight 114.9041 and abundance of 95.75%. Which of the following isotopic weights is the most likely for the other isotope?
A. 112.94
B. 115.90
C. 113.90
D. 114.90

Answer 1

We can calculate the isotopic weight of the other isotope using the atomic mass of isotopes (isotopic weight), the fractional abundance and the atomic weight.

Formula used: We can calculate the isotopic weight of the other isotope using the formula,
Isotopic weight$= \dfrac{{\left( {{\text{Atomic weight}}} \right) - \left( {{\text{Isotopic weight}} \times {\text{Abundance of first isotope}}} \right)}}{{{\text{Abundance of second isotope}}}}$

Complete step by step answer:
Given data contains,
Atomic weight of Indium is 114.82g.
The Isotopic weight of Indium is 114.9041g.
Fractional abundance of the first isotope is 95.75%.
From the fractional abundance of the first isotope, we can calculate the fractional abundance of the second isotope. We have to subtract the fractional abundance of the first isotope from 100 to get the fractional abundance of the second isotope.
Fractional abundance of second isotope$= 100 - {\text{Fractional abundance of first isotope}}$
Let us now substitute the value of fractional abundance of the first isotope.
Fractional abundance of second isotope = $100 - 95.75$
Fractional abundance of second isotope = $4.28$
The calculated value of fractional abundance of second isotope is $4.28\%$.
Let us consider the mass of another isotope to be x.
The average atomic mass of an element is calculated using the formula,
Average atomic mass of an element = $\sum {\left( {{\text{atomic mass of an isotope}} \times {\text{fractional abundance}}} \right)}$
We can substitute the values of isotopic weight, fractional abundance, and atomic mass of an element.
$\left( {114.9041 \times 0.9572} \right) + \left( {x + 0.0428} \right) = 114.82$
Isotopic weight$= \dfrac{{\left( {{\text{Atomic weight}}} \right) - \left( {{\text{Isotopic weight}} \times {\text{Abundance of first isotope}}} \right)}}{{{\text{Abundance of second isotope}}}}$
Substituting the values we get,
$\Rightarrow$ $x = \dfrac{{114.82 - 109.986}}{{0.0428}}$
$\Rightarrow$ $x = 112.94g$
Isotopic weight for the other isotope is 112.94g.

Therefore, the option (A) is correct.

Note:
An isotope of a chemical element is an atom, which has same number of protons and electrons but has a different number of neutrons. Atoms of different elements that contain the same mass number whereas atomic number differs are known as isobars.
Example:
The mass of the fifth isotope of titanium has to be determined.
Given,
Atomic mass of first isotope is $45.953u$
Percentage of first isotope is $8.0\%$
Atomic mass of second isotope is $46.952u$
Percentage of second isotope is $7.3\%$
Atomic mass of third isotope is $47.948u$
Percentage of third isotope is ${\text{73}}{\text{.8\% }}$
Atomic mass of fourth isotope is $48.948u$
Percentage of fourth isotope is $5.5\%$
The average mass of titanium $47.9u$
The masses are summed up together.
Total mass = $45.953u + 46.952u + 47.948u + 48.948u$
Total mass = $189.801u$
Let x be the mass of the fifth isotope.
The mass of fifth isotope is calculated as,
$\dfrac{{189.801 + x}}{5} = 47.9u$
Multiply by ${\text{5}}$ on both sides we get,
$5 \times \dfrac{{189.801u + x}}{5} = 47.9u \times 5$
$\Rightarrow$ $189.801u + x = 239.5u$
$\Rightarrow$ $x = 239.5u - 189.801u$
\Rightarrow $$$x = 49.699u$$ $$x = 50u$$ The mass of the fifth isotope of titanium is50u$.