Question

For every one $^{37}Cl$ isotope there are three $^{35}Cl$ isotopes, in a sample of chlorine. What will be the average atomic mass of chlorine?
(A) 35
(B) 37
(C) 35.5
(D) 35.6

Answer 1

Hint: To answer this question we should have a clear understanding of the basic concepts of stoichiometry. We should also recall how to calculate the average atomic mass for different isotopes.

Complete step by step answer:
We already know that atoms have the same atomic number (number of protons), but different mass numbers (number of protons and neutrons) are called isotopes. Because most elements occur as isotopes and different isotopes have different masses, the atomic mass of an element is the average of the isotopic masses, weighted according to their naturally occurring abundances; this is the mass of each element recorded on the periodic table, also known as the relative atomic mass.
We can calculate the average atomic mass of an element existing in different isotopes by the following formula:
${M_{avg}} = \dfrac{{\sum\limits_{i = 1}^n {{M_i}{A_i}} }}{{\sum\limits_{i = 1}^n {{A_i}} }}$; where M is the atomic mass of isotope and A is the relative abundance.
From the question we understand that,
For $^{37}Cl$: M=37, A=1
For $^{35}Cl$: M=35, A=3
On substituting these values in the formula we get,
${M_{avg}} = \dfrac{{\left( {37 \times 1} \right) + \left( {35 \times 3} \right)}}{{1 + 3}} = 35.5$
Hence, the correct answer is Option (C) 35.5

Note: We should note that chlorine has 25 different isotopes but the most stable ones are mass number 35 and 37. The relative abundance of each isotope can be determined using mass spectrometry. A mass spectrometer ionizes atoms and molecules with a high-energy electron beam and then deflects these ions through a magnetic field based on their mass-to-charge ratios hence the isotopes are separated.