Transcript Average Atomic Mass & % Abundance

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)