Question

What will be the ratio of $C{l^{35}}$ and $C{l^{37}}$ respectively in chlorine if the average atomic mass of chlorine is 35.5?
A. $\;1:1$
B. $2:1$
C. $3:1$
D. $3:2$

Answer 1

The average atomic mass of an element is the sum of the masses of its isotopes multiplies by their respective abundances. In case of only two isotopes of an atom, it can be calculated by $1 = x + (1 - x)$ , where x is the abundance of any one isotope.

Complete step-by-step answer:
Average atomic mass of an element is the sum of the masses of its isotopes, each multiplied by its respective natural abundance. The natural abundance of a particular isotope is the amount of it present naturally on the planet. It is found in the periodic table under the elemental symbol.
Different isotopes of the same elements have different atomic mass but same atomic number due to difference in their number of neutrons. For atoms that have only two isotopes, their abundance can be calculated by $M = {m_1}x + {m_2}(1 - x)$.
Let, x be the fraction of $C{l^{37}}$ in $C{l^{35.5}}$
Using the formula $M = {m_1}x + {m_2}(1 - x)$ and putting the values in it, we get
So, $37x + 35(1 - x) = 35.5$
$37x + 35 - 35x = 35.5$
$2x = 0.5$
Or $x = 0.25$
Thus, $25\%$ of $C{l^{37}}$ present in $C{l^{35.5}}$
From this we can calculate the percentage of $C{l^{35}}$ , which is
$100 - x = \% \,C{l^{35}}$
So, $C{l^{35}}$ is $75\%$.
The ratio of both the isotopes will be $C{l^{35}}:C{l^{37}}::3:1$.

Hence, the correct option is (C).

Note: The formula used above $1 = x + (1 - x)$ is valid only for those atoms which have just two isotopes available. For more number of isotopes, the formula used is-
average atomic mass = ${N_1}{m_1} + {N_2}{m_2} + {N_3}{m_3} +$ …… where N is the fraction representing the natural abundance of the isotope and m is the weight of the isotope.