Answer to Question #255065 in General Chemistry for mark

Question

Boron (B; Z = 5) has two naturally occurring isotopes. Find the percent abundances of ¹⁰B and ¹¹B given the atomic mass of B = 10.81 amu, the isotopic mass of ¹⁰B = 10.0129 amu, and the isotopic mass of ¹¹B = 11.0093 amu. (Hint: The sum of the fractional abundances is 1. If x = abundance of ¹⁰B, then 1 - x = abundance of ¹¹B.) 

Answer 1

Let the fractional abundance of B-10 be x

And fractional abundance of B-11 be (1- x)

Average atomic mass of boron = 10.81 amu

Atomic mas of B-10 = 10.0129 amu

Atomic mas of B-11 = 11.0093 amu

To determine an average atomic mass of an element we use::

x = 0.200 = 0.200 × 100= 20%

(1-x) = 0.800 =0.800 × 100= 80%

The percent abundance of isotopes B-10 and B-11 is 20% and 80% respectively.