Question

What is an example of an isotope abundance practice problem?

Answer 1

Hint : The formula to calculate the average atomic mass of the sample consisting of the isotopes is as follows:
\[Average{\text{ }}atomic{\text{ }}mass{\text{ }} = {M_1} \times {(FA)_1} + {M_2} \times {(FA)_2} + ........ + {M_n} \times {(FA)_n}\]
Where, \[{M_1}\], \[{M_2}\],……, \[{M_n}\] are mass of isotopes of 1,2,…..,n
And (FA) is the fractional abundance of isotope 1,2,……,n in the sample.

Complete Step By Step Answer:
Example: What is the average atomic mass of Neon, given that it has 3 isotopes with the follow percent abundances;
\[
  ^{20}Ne{\text{ }} = {\text{ }}19.992{\text{ }}amu{\text{ }}\left( {90.51\% } \right),\; \\
  ^{21}Ne{\text{ }} = {\text{ }}20.993{\text{ }}amu{\text{ }}\left( {0.27\% } \right),\; \\
  ^{22}Ne{\text{ }} = {\text{ }}21.991{\text{ }}amu. \\
 \]
Let \[\% \]abundance of \[^{22}Ne\] be x, then
\[
  90.51\% + 0.27\% + x = 100\% \\
  x = 100 - (90.51 + 0.27)\% \\
  x = 9.22\% \\
 \]
\[
  {(FA)_1} = 90.51\% = 90.51/100 = 0.9051 \\
  {(FA)_2} = 0.27\% = 0.27/100 = 0.0027 \\
  {(FA)_3} = 9.22\% = 9.22/100 = 0.0922 \\
 \]
Now, putting the above values in the formula
\[Average{\text{ }}atomic{\text{ }}mass{\text{ }} = {M_1} \times {(FA)_1} + {M_2} \times {(FA)_2} + ........ + {M_n} \times {(FA)_n}\]
\[
  Average{\text{ }}atomic{\text{ }}mass{\text{ }} = 19.992 \times 0.9051 + 20.993 \times 0.0027 + 21.991 \times 0.0922 \\
  Average{\text{ }}atomic{\text{ }}mass = 20.179 \\
 \]
Hence, the average atomic mass of Neon is 20.179 amu.

Additional Information:
 Isotopes are separated through mass spectrometry; Mass spectroscopy traces show the relative abundance of isotopes vs. mass number (mass : charge ratio).
Some naturally occurring and artificially produced isotopes are radioactive. All atoms heavier than Bismuth \[(_{31}^{209}Bi)\] are radioactive. However, there are many lighter nuclides that are radioactive.

Note :
The sum of the percent natural abundances of all the isotopes of any given element must total 100%. Isotopes of a given element do not all exist in equal ratios.