Question

The two naturally occurring isotopes of chlorine (35Cl and 37Cl) have masses of 34.9689 and 36.9658 u, respectively. Determine the percentage of 37Cl in naturally occurring chlorine. Calculate the value to the maximum precision possible with the given data (correct number of significant figures is required).

Answer 1

Atomic mass of 35Cl isotope = 34.9689

Atomic mass of 37Cl isotope = 36.9658

Let the percentage of 37Cl in naturally occurring chlorine be x.

then 100 - x = the percentage of 35Cl isotope.

We know that,

Atomic mass of natural occuring chlorine = 35.4527 amu

(100 - x)*34.9689 + x*36.9658 = 100*35.4527

3496.89 - x*34.9689 + x*36.9658 = 3545.27

1.9969*x = 48.38

x = 24.2275 %

So, percentage of 37Cl in naturally occurring chlorine = 24.2275 %