Question

4. The element boron has two naturally occurring isotopes. One has atomic mass 10.0129 u, while the other has atomic mass 11.0093 u. The average atomic mass of boron is 10.811 u. Determine the percentage occurrence of each isotope.

Answer 1

Given:

m1 = 10.0129 amu

m2 = 11.0093 amu

m = 10.811 amu

use:

atomic mass = sum of (mass of isotope * abundance) / 100

m = (m1*A1 + m2*A2)/100

10.811 = ( 10.0129*A1 + 11.0093*(100-A1))/100

10.811 = ( 10.0129*A1 + 1100.93- 11.0093*A1))/100

A1 = 19.9 %

Use:

A2 = 100 - A1

= 100 - 19.9

= 80.1 %

Answer: 19.9 % and 80.1 %