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

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?

• What you have done is correct but there is something mistake in average mass. May 14, 2022 at 13:53
• @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? May 14, 2022 at 14:12
• If you use a relative atomic mass of 15.9980, does it work? May 14, 2022 at 15:48
• Your calculations are fine, but the scenario is impossible. May 17, 2022 at 2:58
• 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. May 17, 2022 at 4:48

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