Sodium only has one naturally occuring isotope, Na23 , with a relative atomic mass of 22.9898 u. A synthetic, radioactive isotope of sodium, Na22 , is used in positron emission tomography. Na22 has a relative atomic mass of 21.9944 u.

A 1.5909 g sample of sodium containing a mixture of Na23 and Na22 has an apparent "atomic mass" of 22.9785 u . Find the mass of Na22 contained in this sample

We can say that the abudandance of [tex]^2^3Na[/tex] is an unknow value "X" and the molar fraction of [tex]^2^2Na[/tex] is "Y". We have to keep in mind that the molar fractions can be added:

Y + X = 1

So, we can put the molar fraction of [tex]^2^2Na[/tex] in terms of [tex]^2^3Na[/tex], so:

Y=1-X

So, we will have the molar fraction of each isotope:

[tex]^2^2Na[/tex]: X-1

[tex]^2^3Na[/tex]: X

And the atomic mass:

[tex]^2^2Na[/tex]: 21.9944

[tex]^2^3Na[/tex]: 22.9898

If we multiply the molar mass by the each atomic mass of each isotope we will have:

[tex] 22.9898*(X)~+~21.9944*(X-1)~=~22.9785[/tex]

Now we can solve for "X" :

[tex]22.9898X~+~21.9944X~-21.9944~=~22.9785[/tex]

[tex]44.9842X-21.9944~=~22.9785[/tex]

[tex]44.9842X~=~22.9785~+~21.9944 [/tex]

[tex]44.9842X~=~44.9729 [/tex]

[tex]X~=~\frac{44.9729}{44.9842}[/tex]

[tex]X~=~0.999749 [/tex]

The molar fraction of [tex]^2^3Na[/tex] is 0.999749. Now we can calculate the molar fraction of [tex]^2^2Na[/tex], so:

[tex]Y~=~1-0.999749~=~0.000251 [/tex]

Now, if we multiply the molar fraction by the mass we can find the mass of [tex]^2^2Na[/tex], so:

[tex]mass~of~^2^2Na~=~1.5909~g*0.000251~=~0.000399316~ g[/tex]

The mass of [tex]^2^2Na[/tex] is 0.000399316 g

I hope it helps!