Question

Calculate the atomic mass (average) of chlorine using the following data:

	%Natural Abundance	Molar mass
$^{35}Cl$	$75.77$	$34.9689$
$^{37}Cl$	$24.23$	$36.9659$

Answer 1

Hint: In nature an element is formed in the form of different isotopes. There are two main isotopes of chlorine. Average atomic mass of isotopes is the sum of the product of abundance $\left( {percentage/100} \right)$ and molar mass for all isotopes.

Complete step by step answer:
Most of the elements in periodic tables have more than one isotope due to which we sometimes use the average molar mass of all isotopes for some calculation and properties. Due to these isotopes, atomic mass of some element is in fraction of atomic mass unit.
In given data we also have some more isotopes of chlorine but these are present in very less amount due to which their effect can be neglected. Also some isotopes are radioactive in nature and unstable.
In given data there are two isotopes of chlorine, that are $^{35}Cl$ and $^{37}Cl$ with percentage abundance of $75.77$ and $24.23$ respectively.
Molar masses of $^{35}Cl$ and $^{37}Cl$ are $34.9689$ and $36.9659$ respectively.
Average atomic mass of chlorine is given by the sum of the product of abundance $\left( {percentage/100} \right)$ and molar mass for all isotopes.
${M_{Avg}} = \sum\limits_{}^{} {\left( {\dfrac{{{\text{% Natural Abundance}}}}{{100}} \times M{\text{molar Mass}}} \right)} $
Here summation is applied on two isotopes of chlorine.
Then, \[{M_{Avg}} = \left( {\dfrac{{74.77}}{{100}} \times 34.9689} \right) + \left( {\dfrac{{24.23}}{{100}} \times 36.9659} \right) \simeq 35.5\]
Hence the average molar mass of chlorine is $35.5gram$.

Note:Average molar mass is different for different compositions. For example, we take a sample in which both isotopes of chlorine are present in half. Then the average molar mass of chlorine in this sample is different from natural average molar mass.