Question

The chlorine present in PVC has two stable isotopes. 35Cl with a mass of 34.97 amu and 37CI with a mass of 36.95 amu. Given chlorine's atomic mass of 35.45, what is the percent abundance of each isotope?

Answer 1

The atomic mass $A$ of chlorine is the weighted average of the atomic masses $A_1 = 34.97$ and $A_2=36.95$ of its isotope. The isotope masses are multiplied by their abundances $f_1$ and $f_2$ (in fraction):

$A = f_1A_1 + f_2A_2$ .

As there are only two stable isotopes, then, inevitably,

$f_1 = 1-f_2$ .

Substituting this into the first equation:

$A = (1-f_2)A_1 + f_2A_2$ .

Therefore, the abundance of the second isotope is:

$f_2 = \frac{A-A_1}{A_2-A_1} = \frac{35.45-34.97}{36.95-34.97} = 0.2424$ , or 24.24%.

Finally, the abundance of the first isotope is:

$f_1 = 1-f_2 = 1-0.2424 = 0.7576$, or 75.76%.

Answer: The percent abundances of the isotopes are 75.76%(34.95 amu) and 24.24%(36.95 amu).