Question

Both $^{18}O$ and $^{16}O$ are found in nature. However, $^{16}O$ is the most common. Why?

Answer 1

Hint: Isotopes are substances that have the same atomic number but the mass number is different, but not all the isotopes of the element are equally present in nature, the presence of the isotope is based on factors like a method of the formation, stability, etc.

Complete answer:
Isotopes are substances that have the same atomic number but the mass number is different, but not all the isotopes of the element are equally present in nature.
$^{18}O$ and $^{16}O$ are the isotopes of oxygen and we know that the atomic number of oxygen is 8. So, both the isotopes are written as \[_{8}^{18}O\] and \[_{8}^{16}O\]. Both the isotopes are not equally found in nature, however, both the isotopes are stable. The \[_{8}^{16}O\] is more common than \[_{8}^{18}O\] because \[_{8}^{16}O\] is a major product formed in the stars.
This can be explained as when the hydrogen runs out in the core of the stars, then it collapses until the temperature of the stars reaches to ${{10}^{8}}$ K.
At this temperature, the alpha-particles undergo a multi-step reaction, in which \[_{8}^{16}O\] is formed by the triple alpha-reaction. The reactions are given below:
\[_{2}^{4}He+_{2}^{4}He\to _{4}^{8}Be\]
\[_{2}^{4}He+_{4}^{8}Be\to _{6}^{12}C\]
\[_{2}^{4}He+_{6}^{12}C\to _{8}^{16}O\]
But the \[_{8}^{18}O\] is only found in massive stars. There are many reactions involved in the formation of \[_{8}^{18}O\], these are given below:
\[_{6}^{12}C\to _{7}^{13}N\to _{6}^{13}C\to _{7}^{14}N\to _{8}^{15}O\to _{7}^{15}N\to _{8}^{16}O\to _{9}^{17}F\to _{8}^{17}O\to _{9}^{18}F\to _{8}^{18}O\]
Therefore, the \[_{8}^{16}O\] is more common than the \[_{8}^{18}O\] in nature.

Note:
There are some isotopes that are equally found in nature, for example, the isotopes of bromine are equally found the nature, its isotopes are $_{35}^{79}Br$ and $_{35}^{81}Br$, and both of them are stable.