Average Atomic Mass & % Abundance

Average Atomic Mass

The weighted average of the atomic masses of the naturally occurring isotopes of an element

Most elements occur naturally as mixtures of isotopes

Average Atomic Mass

Dependent upon both mass and the relative abundance of each of the elements isotopes

Example

Naturally occurring copper exists with the following abundances:

69.17% is Cu-63 w/ atomic mass 62.93 amu

30.83% is Cu-65 w/ atomic mass 64.93

(.6917)x(62.93) + (.3083)x(64.93)= 63.55 amu

Problem 1

3 Isotopes of Ar occur in nature

0.337% as Ar-36, 35.97 amu 0.063% Ar-38, 37.96 amu 99.6% Ar-40, 39.96 amu

Calculate the Average Atomic Mass

Answer Check

(.00337)x(35.97) + (.00063)x(37.96) + (.996)x(39.96)= 39.95amu

Problem 2

2 Naturally occurring Isotopes of Boron occur with the following abundances:

80.20% B-11, 11.01 amu 19.80% B-10, 10.81 amu

What is the Average Atomic Mass

Answer Check

(.8020)x(11.01) + (.1980)x(10.81) = 10.97 amu

Calculating & Abundance

Chlorine has two isotopes: chlorine-35 (mass 34.97 amu) and chlorine-37 (mass 36.97 amu).

What is the percent abundance of these two isotopes if chlorine's atomic mass is 35.453?

Answer Check Part 1

if 2 isotopes, then the total is 100%. assume one is x% (x), the other is automatically 100-x%, (1-x)

x(34.97) + (1-x)(36.97) = 35.453

Answer Check Part 2

x(34.97) + (1-x)(36.97)=35.453 Solve for x 34.97x+36.97-36.97x=35.453 -2x+36.97=35.453 -2x=-1.517 x=.7585 1-x=.2415

Answer Check Part 3

Therefore Cl-35 has a % abundance of 75.85% and Cl-37 has a % abundance of 24.15%

Problem 1

The two naturally occurring isotopes of nitrogen are nitrogen-14, with an atomic mass of 14.003074 amu, and nitrogen-15, with an atomic mass of 15.000108 amu. What are the percent natural abundances of these isotopes?

The atomic mass of nitrogen is 14.00674amu

Answer Check

The atomic mass of nitrogen is 14.00674amu

14.00674 = p(14.003074) + (1 -p)(15.000108)14.00674 = 14.003074p + 15.000108 - 15.000108p-0.997034p = -0.993368

p = 0.9963 = 99.63% (N14)1 - p = 0.0037 = 0.37% (N15)