Percent Composition & Average Atomic Mass (AAM)
Assume a new element E has 3 isotopes; E-a, E-b, E-c
a,b,c are the mass numbers of each isotope
(or aE, bE, cE)
A sample of E contains the following isotopes: (E is the ‘element’, not an isotope)
aE
= x grams
= y grams
cE = z grams
bE
The total amount of the sample = x+y+z
sumxyz = x+y+z
Percent Composition
Percent of aE =
x
(x+y+z)
x 100; the value is in % (% aE);
y
Percent of bE = (x+y+z) x 100; the value is in %; (% bE);
z
Percent of cE = (x+y+z) x 100; the value is in %; (% cE);
x
(x+y+z)
y
(x+y+z)
z
(x+y+z)
gives the decimal form
gives the decimal form
gives the decimal form
NOTE: Any time you multiply a decimal by 100, you get %
To convert %  decimal, divide % by 100
(% aE) + (% bE) + (% cE) = 100 (or extremely close)
x
(x+y+z)
+
y
z
(x+y+z)
+ (x+y+z) = 1 (or extremely close)
Average Atomic Mass of E (AAM)
Given:
(AMI is atomic mass of the isotope)
a
AMI of E = q
AMI of bE = r
AMI of cE = s
x
AAM = q * (x+y+z)
+
y
r * (x+y+z) + s *
z
(x+y+z)
Final Check: The AAM will be between the lowest & highest AMI
If it doesn’t, you did something wrong.
Note: I used 3 isotopes… there could be only 2, there could be more.
Depends on the element & the number of naturally occurring isotopes in the given problem
© 2022 David Lerner, ITutorMathandScience.com