Question

A particular element exists in two stable isotopic forms. One isotope has a mass of 78.9183 amu (50.69% abundance). The other isotope has a mass of 80.9163 amu (49.31% abundance). Calculate the average mass of the element and determine its identity.

Answer 1

Answer:

Average atomic mass = 79.9035 amu.

The given element is bromine.

Explanation:

Given data:

Mass of 1st isotope = 78.9183 amu

Percent abundance of 1st isotope = 50.69%

Mass of 2nd isotope = 80.9163 amu

Percent abundance of 1st isotope = 49.31%

Average atomic mass = ?

Solution:

Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass)  / 100

Average atomic mass  = (50.69×78.9183 )+(49.31× 80.9163) /100

Average atomic mass =  4000.3686+ 3989.9828 / 100

Average atomic mass  = 7990.3514 / 100

Average atomic mass = 79.9035 amu.

The given element is bromine.