## Chemistry: Molecular Approach (4th Edition)

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: $\underline{m\text{(}{}^{35}\text{Cl}{}^{35}\text{Cl}{}^{16}\text{O)= 85}\text{.9325 amu};\,m\text{(}{}^{35}\text{Cl}{}^{37}\text{Cl}{}^{16}\text{O) =87}\text{.9296 amu};\,m\text{(}{}^{37}\text{Cl}{}^{37}\text{Cl}{}^{16}\text{O) =89}\text{.9267 amu}}$