Question

Three isotopes of Potassium occur naturally:

39_K (mass = 38.9637 amu)
40_K (mass = 39.9640 amu)
41_K (mass = 40.9618 amu)

If the percent abundance of 39_K is 93.26, what is the percent abundance of 41_K?

Answer 1

@ganeshie8

Answer 2

@aaronq

Answer 3

we can find the percent abundance of \(^{41}K\) with the data above and some substitutions.

percent abundance of \(^{40}K\) + percent abundance of \(^{41}K\)= 100%-93.26%=6.74%

i'll re-write in a shorter form: \(\%^{40}K+\%^{41}K=6.74\%\)

The formula for the weighted average is:

\(\sf Atomic~mass_{average}=\sum(isotope~mass_i*\%~abundance_i)\)

Which expands into:

\(\sf Atomic~mass_{average}=\%^{39}K*mass_{^{39}K}+\%^{40}K*mass_{^{40}K}+\%^{41}K*mass_{^{41}K}\)


There are too many unknowns, so we can use the equation above to make a substitution
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\(\%^{40}K+\%^{41}K=6.74\%\rightarrow \%^{40}K=6.74\%-\%^{41}K\)

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Our equation becomes:

\(\sf Atomic~mass_{average}=\%^{39}K*mass_{^{39}K}+(6.74\%-\%^{41}K)*mass_{^{40}K}+\%^{41}K*mass_{^{41}K}\)

You can now solve for the percent abundance of 41K, \((\%^{41}K\)).

Answer 4

6.903% ?????

Answer 5

it cant be more than 6.74%, post what you did?

Answer 6

39.10=(0.9326)*(38.9637)+(0.0674-x)*(39.9640)+(x)*(40.9618)

Answer 7

@aaronq