Question

(a)Two stable isotopes of lithium $_{3}L{{i}^{6}}$ and $_{3}L{{i}^{7}}$ have respective abundances of 7.5% and 92.5%.These two isotopes have masses 6.01512 $\mu$ and 7.01600 $\mu$, respectively. Find the atomic mass of lithium.
(b)Boron has two stable isotopes $_{5}{{B}^{10}}$ and $_{5}{{B}^{11}}$ .Their respective masses are 10.01294 $\mu$ and 11.00931 $\mu$, and the atomic mass of boron is 10.811 $\mu$. Find the abundances of $_{5}{{B}^{10}}$ and $_{5}{{B}^{11}}$

Answer 1

Average atomic mass of an element is obtained by adding, the value obtained by multiplying mass of isotope with their percent abundance. Atoms having the same atomic number but different atomic mass numbers are called isotopes. Sum of percent abundance of all isotopes is 100%.

Complete answer:
In the classes of chemistry, we have studied the concepts which deal with the radioactivity and also some of the related concepts such as the definitions of isotopes, isobars and also some of the related calculations.
Let us see the calculation of mass of lithium as well as abundances of two isotopes of boron atoms based on these concepts.
- Radioactivity or the radioactive decay is the process by which an unstable atomic nucleus loses its energy by radiation and this material which consists of unstable nuclei is called radioactive element.
- Relative abundance is the percentage of an isotope Average atomic mass takes account of mass of each isotope and its percent abundance. Mass of each isotope is multiplied by its percent abundance and added.
So ,Atomic mass of Lithium = $\dfrac{\left( 7.5\times 6.01515 \right)+\left( 92.5\times 7.01600 \right)}{100}=6.94\mu$
Similarly, For Boron, Consider the abundance percentage of boron isotopes, $_{5}{{B}^{10}}$ be $x$ % . As we know the sum of percent abundance of all isotopes is 100%, percent abundance of $_{5}{{B}^{11}}$ will be (100 – x)%.
As given, The atomic mass of boron is 10.811 $\mu$.
so, percent abundances can be obtained as follows:
$10.811\mu =\dfrac{10.01294x+11.0093(100-x)}{100}$
$\Rightarrow 1081.1=10.01294x+1100.931-11.00931x$
$\Rightarrow 19.831=0.99637x$
Thus, x = 19.90%.
So, percentage abundance of $_{5}{{B}^{10}}$ is 19.90% and percentage abundance of $_{5}{{B}^{11}}$ is 80.10%.

Note:
Percent abundance is percent of isotope quantity which is present in nature. Isobars have the same atomic mass but different atomic numbers. Isotones have the same number of neutrons.
Average atomic mass is the sum of atomic mass of all isotopes multiplied by their percent abundances.