Question

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

Answer 1

Answer:

0.000399316  g

Explanation:

We can start with the molar fraction for each isotope:

We can say that the abudandance of $^2^3Na$ is an unknow value "X" and the molar fraction of $^2^2Na$ 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 $^2^2Na$ in terms of $^2^3Na$, so:

Y=1-X

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

$^2^2Na$: X-1

$^2^3Na$: X

And the atomic mass:

$^2^2Na$: 21.9944

$^2^3Na$: 22.9898

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

$22.9898*(X)~+~21.9944*(X-1)~=~22.9785$

Now we can solve for "X" :

$22.9898X~+~21.9944X~-21.9944~=~22.9785$

$44.9842X-21.9944~=~22.9785$

$44.9842X~=~22.9785~+~21.9944$

$44.9842X~=~44.9729$

$X~=~\frac{44.9729}{44.9842}$

$X~=~0.999749$

The molar fraction of $^2^3Na$ is 0.999749. Now we can calculate the molar fraction of $^2^2Na$, so:

$Y~=~1-0.999749~=~0.000251$

Now, if we multiply the molar fraction by the mass we can find the mass of $^2^2Na$, so:

$mass~of~^2^2Na~=~1.5909~g*0.000251~=~0.000399316~ g$

The mass of $^2^2Na$ is 0.000399316  g

I hope it helps!