Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# Naturally occurring boron consists of two isotopes, whose atomic masses are $10.01$ and $11.01$. The atomic mass of natural boron is $10.81$. Calculate the percentage of each isotope in natural boron.

Last updated date: 15th Jun 2024
Total views: 402.3k
Views today: 5.02k
Verified
402.3k+ views
Hint: Average atomic mass of an element can be calculated by adding the product of all isotopes with their percentage in which they are present in final form. Average atomic mass basically tells us about the relative natural abundance of that element’s isotopes.

Complete step by step answer: As we know that Boron is a chemical element with atomic number $5$ and electronic configuration $1{s^2}2{s^2}2{p^z}$. It is represented by the symbol $B$. it consists only $0.001$ percent by weight of forth crust. Elemental boron is a metalloid. Industrially, very pure boron is produced with very difficulty because of contamination of carbon or other elements which were removed with difficulty. Boron is used in making borosilicate glass. It is also used in preparation of basic acid. Boron is also used in fertilisers and insecticide. Boron is a useful dopant for such semiconductors as silicon, germanium and silicon carbide.
Now according to the question, suppose the percentage of isotope of $B$ with atomic mass $10.01$ be $'x'$. This implies that the percentage of other isotopes of $B$ with atomic mass $11.01$ be $'100 - x'$ because in calculating the percentage, the sum is always $100\%$.
$\therefore$ Average atomic mass can be calculated by using the formula
Average atomic mass =$\dfrac{{{\text{mass of one isotope }} \times {\text{ it's percentage value + mass of other isotope }} \times {\text{ it's percentage value}}}}{{100}}$ ………… (i)
Now we have a $2$ isotope of boron with atomic mass be $10.01$ and $11.01$ and respective percentages are $'x'$ and $'100 - x'$. And the average atomic mass given is $10.81$. Putting all values in above equation, we get
$\Rightarrow$ $10.81 = \dfrac{{10.01\left( x \right) + 11.01\left( {100 - x} \right)}}{{100}}$
$\Rightarrow$ $1081 = 10.01x + 1101 - 11.01x$
$\Rightarrow$ $1081 = 1101 - x$
Or $x = 20$
Hence, percentage of isotope with atomic mass $10.01 = 20\%$
And percentage of isotope with atomic mass $11.01 = \left( {100 - 20} \right)\% = 80\%$

Note: The average atomic mass is important in order to know the abundance of the isotopes of a particular element. It is measured in atomic mass units or amu. For heavier elements that have larger isotopic distribution, the average atomic mass plays an important role.