Question

The atomic mass of silver found in nature is $\mathrm{Ar}\left(\mathrm{Ag}\right)=107.983$. This silver is made of $107\mathrm{Ag}$ and $109\mathrm{Ag}$ Isotopes. Calculate the proportion of the mass of $107\mathrm{Ag}$ isotope which is in natural silver atoms.

Answer 1

If the fractions of $107\mathrm{Ag}$ and $109\mathrm{Ag}$ are x and y, respectively,

Then,

$$\begin{gathered} (107\times \mathrm{x})+(109\times \mathrm{y})=107.983.......\left(1\right) \\ \mathrm{x}+\mathrm{y}=1.......\left(2\right) \\ \Rightarrow \mathrm{x}=(1-\mathrm{y}) \end{gathered}$$

Substituting the value of x in equation (1),

$$\begin{gathered} 107(1-\mathrm{y})+109\mathrm{y}=107.983 \\ 107-107\mathrm{y}+109\mathrm{y}=107.983 \\ 2\mathrm{y}=107.983-107 \\ 2\mathrm{y}=0.983 \\ \mathrm{y}=0.4915 \\ \left(1\right)\Rightarrow \mathrm{x}=1-0.4915 \\ =0.5085 \end{gathered}$$

Writing in percentage,

$$\begin{gathered} {\mathrm{Ag}}_{107}=50.85\% \\ {\mathrm{Ag}}_{109}=49.15\% \end{gathered}$$

Hence, the proportion of the mass of $107\mathrm{Ag}$ isotope which is in natural silver atoms is 50.85 %.