Answer
The answer is below.
Work Step by Step
Calculate the number of naturally existing molecules of \[\text{C}{{\text{l}}_{2}}\text{O}\] having different masses with the two isotopes of chlorine and three isotopes of oxygen as follows:
(1) \[{}^{35}\text{Cl}{}^{35}\text{Cl}{}^{16}\text{O}\]
(2) \[{}^{35}\text{Cl}{}^{35}\text{Cl}{}^{17}\text{O}\]
(3) \[{}^{35}\text{Cl}{}^{35}\text{Cl}{}^{18}\text{O}\]
(4) \[{}^{35}\text{Cl}{}^{37}\text{Cl}{}^{16}\text{O}\]
(5) \[{}^{35}\text{Cl}{}^{37}\text{Cl}{}^{17}\text{O}\]
(6) \[{}^{35}\text{Cl}{}^{37}\text{Cl}{}^{18}\text{O}\]
(7) \[{}^{37}\text{Cl}{}^{37}\text{Cl}{}^{16}\text{O}\]
(8) \[{}^{37}\text{Cl}{}^{37}\text{Cl}{}^{17}\text{O}\]
(9) \[{}^{37}\text{Cl}{}^{37}\text{Cl}{}^{18}\text{O}\]
So, there are nine molecules of \[\text{C}{{\text{l}}_{2}}\text{O}\] having different masses.
The most abundant molecules are those which are composed of isotopes with higher abundance. Natural abundance of \[\left( \text{O-17} \right)\] is the highest among the three isotopes of oxygen and the natural abundance of \[\left( \text{Cl-35} \right)\] is higher than that of \[\left( \text{Cl-37} \right)\] for chlorine.
Therefore, \[{}^{35}\text{Cl}{}^{35}\text{Cl}{}^{16}\text{O};\,{}^{35}\text{Cl}{}^{37}\text{Cl}{}^{16}\text{O};\,{}^{37}\text{Cl}{}^{37}\text{Cl}{}^{16}\text{O}\] are the most abundant molecules.
Calculate the mass of \[{}^{35}\text{Cl}{}^{35}\text{Cl}{}^{16}\text{O}\] as follows:
\[\begin{align}
& m\text{(}{}^{35}\text{Cl}{}^{35}\text{Cl}{}^{16}\text{O) = }m\text{(Cl-35)}+m\text{(Cl-35)}+m\text{(}O-16\text{)} \\
& \text{= 2 }\!\!\times\!\!\text{ }\left( \text{34}\text{.9688 amu} \right)\text{+15}\text{.9949 amu} \\
& \text{ =85}\text{.9325 amu}
\end{align}\]
Calculate the mass of \[{}^{35}\text{Cl}{}^{37}\text{Cl}{}^{16}\text{O}\] as follows:
\[\begin{align}
& m\text{(}{}^{35}\text{Cl}{}^{37}\text{Cl}{}^{16}\text{O) = }m\text{(Cl-35)}+m\text{(Cl-37)}+m\text{(}O-16\text{)} \\
& \text{=34}\text{.9688 amu+36}\text{.9659 amu+15}\text{.9949 amu} \\
& \text{ =87}\text{.9296 amu}
\end{align}\]
Calculate the mass of \[{}^{37}\text{Cl}{}^{37}\text{Cl}{}^{16}\text{O}\] as follows:
\[\begin{align}
& m\text{(}{}^{37}\text{Cl}{}^{37}\text{Cl}{}^{16}\text{O) = }m\text{(Cl-37)}+m\text{(Cl-37)}+m\text{(}O-16\text{)} \\
& \text{= 2 }\!\!\times\!\!\text{ }\left( \text{36}\text{.9659 amu} \right)\text{+15}\text{.9949 amu} \\
& \text{ =89}\text{.9267 amu}
\end{align}\]
The number of naturally existing molecules of \[\text{C}{{\text{l}}_{2}}\text{O}\] having different masses is\[\underline{9}\], and the masses of three molecules which are the most abundant are:
\[\underline{m\text{(}{}^{35}\text{Cl}{}^{35}\text{Cl}{}^{16}\text{O)= 85}\text{.9325 amu};\,m\text{(}{}^{35}\text{Cl}{}^{37}\text{Cl}{}^{16}\text{O) =87}\text{.9296 amu};\,m\text{(}{}^{37}\text{Cl}{}^{37}\text{Cl}{}^{16}\text{O) =89}\text{.9267 amu}}\]
