Question

The isotopic abundance of $C - 12$ and $C - 14$ is $98\%$ and $2\%$ respectively. What would be the number of $C - 14$ isotope in $12g$ carbon sample?
A.) $1.032 \times {10^{22}}$
B.) $3.01 \times {10^{23}}$
C.) $5.88 \times {10^{23}}$
D.) $6.02 \times {10^{23}}$

Answer 1

In this question, to find number of isotopes of $C - 14$ isotope in $12g$ carbon sample find out the average atomic mass of the carbon sample and then find the $2\%$ of the average atomic mass of the sample.

Complete step by step answer:
In this question, the isotopes of an element are those elements which have the same atomic number but have different atomic masses. The isotopes have the same number of protons but different number of neutrons in each atom. The carbon has $15$ known isotopes among them most common isotopes are carbon$- 12$, carbon$- 13$ and carbon$- 14$.
The average atomic mass of the sample consisting of isotopes can be given as:
${\text{Average Atomic mass = (MM}}{{\text{)}}_{12}} \times {\text{ (Abundance}}{{\text{)}}_{12}}{\text{ + (MM}}{{\text{)}}_{14}} \times {\text{ (Abundance}}{{\text{)}}_{14}}$ $- (1)$
Where, ${{\text{(MM)}}_{12}} =$ molecular mass of carbon$- 12$ (that is $12g$)
${{\text{(MM)}}_{14}} =$ molecular mass of carbon$- 14$ (that is $14g$)
${{\text{(Abundance)}}_{12}} =$ Abundance of carbon$- 12$ in the sample (given in question $98\%$)
${{\text{(Abundance)}}_{14}} =$ Abundance of carbon$- 14$ in the sample (given in question $2\%$)
Now, by putting all values in equation $- (1)$ we get,
${\text{Average atomic mass = 12}} \times \dfrac{{98}}{{100}}{\text{ + 14}} \times \dfrac{2}{{100}} \\\ = \dfrac{{1176 \times 28}}{{100}} \\\ = \dfrac{{1204}}{{100}} \\\ = 12.04g \\\$
As we know that carbon sample contains $2\%$ of carbon$- 14$.
That is $100g$ of carbon sample contains $2g$ of carbon$- 14$
So, $1g$ of carbon sample contains $\dfrac{2}{{100}}g$ of carbon$- 14$
Therefore, $12.04 g$ of carbon sample contains $\dfrac{2}{{100}} \times 12.04 g$ of carbon$- 14$.
That is, $12.04g$ of carbon sample contains $0.2408g$ of carbon$- 14$.
Now, as we know that for any atom,
$1mol = {N_A} = 6.02 \times {10^{23}}{\text{ atoms}} =$ molecular mass of atom (that is $14$ for carbon-$14$)
$14g = 6.02 \times {10^{23}}{\text{ atoms}} \\\ {\text{1g}} = \dfrac{{6.02 \times {{10}^{23}}}}{{14}}{\text{ atoms}} \\\$
For $0.2408g = \dfrac{{6.02 \times {{10}^{23}}}}{{14}} \times 0.2408{\text{ atoms}}$
$= 1.032 \times {10^{22}}{\text{ atoms}}$
Therefore, the number of atoms of carbon$- 14$ isotopes is $= 1.032 \times {10^{22}}{\text{ atoms}}$

Hence, option A.) is the correct answer.

Note:
Always remember that if there is a sample given which includes the isotopes of an element then to find average atomic mass we will not just add the molecular mass of the but we will add the product of the molecular mass and abundance for each isotope as the formula is given above.