Determining the average relative mass of an element from the percentage of isotopes


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> Suppose we have an element $\ce{A}$. Its relative mass is $16$ and it has three isotopes: $\ce{^16A}$ ,$\ce{^17A}$ and $\ce{^18A}$. The available percentage of $\ce{^17A}$ is $0.037\,\%$. What is the available percentage of $\ce{^16A}$ and $\ce{^18A}$?
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Let the percentage of $\ce{^16A}$ be $x\,\%$. Then the percentage of $\ce{^18A}$ is $(100 - (x + 0.037))\,\%$ or $(99.963 - x)\%$. From this we can write


$$\frac{17\times 0.037 + 16x + 18\times (99.963-x)}{100} = 16$$


If I solve this I will get $x = 99.9815$ and the available percentage of $\ce{^18A}$ as $(100 - (99.9815 + 0.037))\,\%$, which is a negative number.
What is wrong in my calculations?




The problem setting ( and your trying) has messed up isotope nucleon numbers, isotope molar masses and element molar mass.


It is rather:
$$\frac{M(\ce{^{17}A}) \times 0.037 + M(\ce{^{16}A}) \times x + M(\ce{^{18}A})\times (99.963-x)}{100} = M(\ce{A})$$


but molar isotope masses are not available in the task.


It is obvious the element is oxygen and the molar mass data can be found, but the task is not about that. Such a task is a shame.


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