Question

Carbon occurs in nature as a mixture of $^{{\text{12}}}{\text{C}}$ and $^{{\text{13}}}{\text{C}}$ . The average atomic mass of carbon is ${\text{12}}{\text{.011}}$ . What is the % abundance of $^{{\text{12}}}{\text{C}}$ in nature?
(A) {\text{88}}{\text{.9% }}
(B) {\text{98}}{\text{.9% }}
(C) {\text{89}}{\text{.9% }}
(D) {\text{79}}{\text{.9% }}

Answer 1

In the above question, carbon’s two isotopes are given along with the average atomic mass. We have to find the percentage of abundance of these isotopes. We can put the value of two isotopes' atomic mass in the average atomic mass equation to get the percentage of two isotopes.

Formula used:
${\text{A = }}\sum\limits_{{\text{i = 1}}}^{\text{n}} {{{\text{a}}_{\text{i}}}{{\text{p}}_{\text{i}}}}$
Where A = atomic mass of the element
${{\text{a}}_{\text{i}}}$ =atomic mass of ${{\text{i}}^{{\text{th}}}}$ isotope.
${{\text{p}}_{\text{i}}}$ = percentage of occurrence of ${{\text{i}}^{{\text{th}}}}$ isotope.
n= number of isotopes.

Complete step by step solution:
Isotope of a particular element can be defined as the element having the same atomic number but different atomic mass. In other words, we can say that isotopes have the same number of protons but they differ in the number of neutrons.
Carbon has two isotopes C-12 and C-13.
Atomic mass of C-12 is 12 and atomic mass of C-13 is 13. Let the percentage of C-12 be x and the percentage of C-13 be ${\text{1 - x}}$ .
The average atomic mass can be written as:
${\text{A = }}\sum\limits_{{\text{i = 1}}}^{\text{n}} {{{\text{a}}_{\text{i}}}{{\text{p}}_{\text{i}}}}$
Since, carbon has only 2 isotopes and hence we can replace n with 2.
${\text{A = }}\sum\limits_{{\text{i = 1}}}^{\text{2}} {{{\text{a}}_{\text{i}}}{{\text{p}}_{\text{i}}}} {\text{ = }}{{\text{a}}_{\text{1}}}{{\text{p}}_{\text{1}}}{\text{ + }}{{\text{a}}_{\text{2}}}{{\text{p}}_{\text{2}}}$
Substituting the values, we get:
$12.011 = 12 \times x + 13 \times (1 - x)$
$\Rightarrow 12.011 = 12x + 13 - 13x$
Rearranging the equation, we get:
$x = 13 - 12.011 = 0.989$
Therefore, the percentage of abundances of C-12 is {\text{98}}{\text{.9% }} .
So, the correct option is option B.

Note:
In these types of questions, where we have to calculate the percentage abundance of the isotopes of an element we have to take their atomic mass of their isotopes. In case, we have only 2 isotopes, we can take the percentage of one isotope as x and the other as ${\text{1 - x}}$ .