REMINDERS &
Element Quiz #2 tomorrow
 Atomic Structure Test Friday

DAILY SCHEDULE
Review Isotope/Different Element Worksheet
 Remind 101
 Isotope Abundance/Average Atomic Mass Notes
 Isotope Practice Packet

ISOTOPES:
Average Atomic Mass & Abundance
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)
EXAMPLE

Element X has two isotopes

#1:
Mass=6 amu
 Relative Abundance= 7.5%


#2
Mass=7 amu
 Relative Abundance=92.5%


What element is it?
TO SOLVE EXAMPLE 1
Average atomic mass=
[(mass #1)x(abundance #1)] + [(mass #2)x(abundance #2)]
(6 x .075) + (7 x .925)=
0.45 + 6.475=
6.925 amu
Closest6.94
to ?
Lithium
FINDING PERCENT ABUNDANCE
There are two stable isotopes of Chlorine:
Chlorine-35 (which weighs 34.97 amu)
and
Chlorine-37 (which weighs 36.97 amu).
If the relative atomic mass of Chlorine is
35.45 amu, what is the percent abundance
of each isotope?
Chlorine-35 (which weighs 34.97 amu) &
Chlorine-37 (which weighs 36.97 amu).

Cl-35= ?X
%

Cl-37= ?Y%?
• Having 2 variables would be much harder.
• We want to express Cl-37 abundance using X
20%
40%
60%
Cl-37
.6 & .4
Cl-35
Cl-35
80%
.8 & .2
Cl-37
Chlorine-35 (which weighs 34.97 amu) & Chlorine-37 (which
weighs 36.97 amu). Chlorine relative atomic mass 35.45 amu.

Cl-35= ?X
%
• Cl-35 = 34.97 amu

Cl-37= ?1-X
%
• Cl-37 = 36.97 amu
35.45
34.97
36.97
X
(34.97X) + (36.97)-(36.97X) = 35.45
-2.00X + 36.97 = 35.45
-2.00X = -1.52
2.00X = 1.52
X= 1.52/2.00
X = 0.76
1-X
Chlorine-35 (which weighs 34.97 amu) & Chlorine-37 (which
.
weighs 36.97 amu). Chlorine relative atomic mass 35.45 amu

Cl-35 = X = .76 = 76%

Cl-37 = (1-X) = (1-.76) = .24 = 24%
Chlorine Isotopes Abundance
24%
Cl-35
Cl-37
76%
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