1. Average Atomic Mass & Abundance
Isotopes:
Average Atomic Mass & Abundance

2. Average Atomic Mass The decimal number on periodic table
Weighted average of all isotopes of an element
Depends on percent (relative) abundance and mass of each isotope
Measured in “atomic mass units” (amu)

3. Example Element X has two isotopes #1: #2 What element is it?
Mass=6 amu
Relative Abundance= 7.5% #2
Mass=7 amu
Relative Abundance=92.5%
What element is it?

4. To solve example 1
Average atomic mass= [(mass #1)x(abundance #1)] + [(mass #2)x(abundance #2)] (6 x .075) + (7 x .925)= = amu Closest to ?
Lithium 6.94

5. Finding Percent Abundance
There are two stable isotopes of Chlorine: Chlorine-35 (which weighs amu) and Chlorine-37 (which weighs amu). If the relative atomic mass of Chlorine is amu, what is the percent abundance of each isotope?

6. X Y ? Having 2 variables would be much harder.
Chlorine-35 (which weighs amu) & Chlorine-37 (which weighs amu). X
Cl-35= ? %
Cl-37= ? %
Y ?
Having 2 variables would be much harder.
We want to express Cl-37 abundance using X
.6 & & .2

7. X 1-X Cl-35 = 34.97 amu Cl-37 = 36.97 amu Cl-35= ? % Cl-37= ? %
Chlorine-35 (which weighs amu) & Chlorine-37 (which weighs amu). Chlorine relative atomic mass amu. X
Cl-35 = amu
Cl-37 = amu
Cl-35= ? %
Cl-37= ? %
1-X
35.45
34.97 X
36.97
1-X
(34.97X) + (36.97)-(36.97X) = 35.45
-2.00X = 35.45
-2.00X = -1.52
2.00X = 1.52
X= 1.52/2.00
X = 0.76

8. Cl-35 = X = .76 = 76% Cl-37 = (1-X) = (1-.76) = .24 = 24%
Chlorine-35 (which weighs amu) & Chlorine-37 (which weighs amu). Chlorine relative atomic mass amu.
Cl-35 = X = .76 = 76%
Cl-37 = (1-X) = (1-.76) = .24 = 24%

9. Isotope Practice Packet
Turn in if finished today—if not it is homework due at the beginning of class tomorrow
Don’t forget to sign up for and tell your parents about remind 101