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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 $\ce{^18A}$?

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?

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    $\begingroup$ What you have done is correct but there is something mistake in average mass. $\endgroup$
    – Infinite
    Commented May 14, 2022 at 13:53
  • $\begingroup$ @Infinite , The formula I used to calculate average relative mass, does that work for 3 or more isotopes? I mean is something like this correct? $$\frac{((p * a) + (q * b) + (r * c) + (s * d) )}{100} = relative-atomic-mass$$. where p,q,r, and s are the atomic masses of isotopes and a,b,c, and d are there's percentages of abundance? $\endgroup$
    – Sayem Rahman
    Commented May 14, 2022 at 14:12
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    $\begingroup$ If you use a relative atomic mass of 15.9980, does it work? $\endgroup$
    – ACR
    Commented May 14, 2022 at 15:48
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    $\begingroup$ Your calculations are fine, but the scenario is impossible. $\endgroup$
    – Scott Cookson
    Commented May 17, 2022 at 2:58
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    $\begingroup$ Yeah, what @ScottCookson said. If the average mass of an element is 16, and the isotopes have weights of 16, 17, and 18, then there can't be any 17-A or 18-A at all—their abundances have to be 0%. I.e., the only way you can get an average mass of 16 is if the sample is 100% 16-A. Any amount of 17-A or 18-A will increase the average mass above 16. So if you actually do have some 17-A then, mathematically, the only way you can get an average mass of 16 is to have a negative amount of 18-A! That's why you're getting that result. Bottom line: The problem wasn't set up correctly. $\endgroup$
    – theorist
    Commented May 17, 2022 at 4:48

1 Answer 1

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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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