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The three naturally occurring isotopes of potassium are $\ce{^39K}$ ($38.963707\:\mathrm{amu}$); $\ce{^40K}$ ($39.963999\:\mathrm{amu}$); and $\ce{^41K}$.

The percent natural abundances of $\ce{^39K}$ and $\ce{^41K}$ are $93.2581\%$ and $6.7302\%$ respectively.

a) What is the natural abundance of $\ce{^40K}$?

b) Determine the isotopic mass of $\ce{^41K}$.

So I started to utilize this equation:

$\mathrm{atomic\:mass = (fractional\:abundance\:isotope\:1 \cdot mass) + (fractional\:abundance\:isotope\:2 \cdot mass)}$

But then I realized that I did not have some values. $\ce{^39K}$ is the only isotope which has two values. $\ce{^40K}$ only has the mass weight, however, if I were to multiply $\ce{^40K}$'s mass with ($1-0.932581$) would that help me arrive at an answer?

Also, would that answer be accurate since it only takes into account that there are only to values that add up to 100%.

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  • $\begingroup$ You can answer a) from the fact that they should add up to 100%. To answer b) you either need to know atomic mass (available from periodic table, if you are allowed to use) or can make e.g an approximation that an added neutron's mass is more or less that same. $\endgroup$ – Greg Sep 29 '14 at 1:15
  • $\begingroup$ All three or just the two? And if so how should I set them up? Thanks in advance. $\endgroup$ – user137452 Sep 29 '14 at 1:22
  • $\begingroup$ Sorry, can you rephrase your question? What is three or two? What do you want to set up? $\endgroup$ – Greg Sep 29 '14 at 1:25
  • $\begingroup$ How do I solve for part a in the question? Do I utilize the formula that I wrote with the question? $\endgroup$ – user137452 Sep 29 '14 at 1:27
  • $\begingroup$ As I said, one way to answer a) is that you assume the three isotope fractions should add up 100%. $\endgroup$ – Greg Sep 29 '14 at 1:31
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A). If you know the percentages of the abundance of $\frac{2}{3}$s of the isotopes then finding the percentage of the $3^{rd}$ is as simple as

$ 100\% - (\%_{\ce{^39K}}+\%_{\ce{^41K}}) = \%_{\ce{^40K}} $

B). Finding the mass of the isotope should be pretty simple. It would be

$(\mathrm{Mass\ of\ proton} \cdot \mathrm{{\#\ of protons}} )+(\mathrm{Mass\ of\ a\ neutron} \cdot \mathrm{{\#\ of neutrons}} ) \approx \mathrm{Mass\ of\ isotope}$

You neglect the weight of electrons because the mass is small relative to the others.

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