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 Atomic Mass Calculations from Percent Abundance

Atomic Mass

"An atomic weight (relative atomic mass) of an element from a specified source is the ratio of the average mass per atom of the element to 1/12 of the mass of 12C" in its nuclear and electronic ground state.

A sample of any element consists of one or more isotopes of that element. Each isotope is a different weight. The relative amounts of each isotope for any element represents the isotope distribution for that element. The atomic weight is the average of the isotope weights weighted for the isotope distribution and expressed on the 12C scale as mentioned above.

Atomic Mass Calculations

 Atomic Mass = [(mass of isotope) (%abundance) ] + [(mass of isotope) (%abundance)] + [….] 100%

The equation continues on[….] based on the number of isotopes in the problem.

Example 1 The natural abundance for boron isotopes is: 19.9% 10B (10.013 amu) and 80.1% 11B (11.009amu). Calculate the atomic weight of boron.

Atomic Mass =
 [19.9)(10.013)] + [(80.1)(11.009)] 100%
Atomic Mass =
  +  100%
Atomic Mass = 10.81amu so, the atomic weight of B = 10.81amu

If you look in the periodic table you will be able to check that our answer is correct!

 3Verify that the atomic mass of magnesium is 24.31, given the following :24Mg= 23.985042amu, 78.99%25Mg= 24.985837 amu, 10.00% 26Mg= 25.982593, 11.01% Atomic mass= [(0.7899)(23.985042)] + [(0.1)(24.985837)] + [(0.1101)(25.982593)]Atomic mass = 18.946 + 2.499 + 2.861

AP Chemistry Example 1

Determining the percent abundance of each isotope from atomic mass.

Copper exists as two isotopes: 63Cu (62.9298 amu) and 65Cu (64.9278 amu).

What are the percent abundances of the isotopes?
Since the overall atomic weight for copper is not given in the problem, you must look it up in the periodic table to work this solution. Atomic mass for Cu = 63.546

63Cu % = 1-x                65Cu % =  x

63.546 = [(1-x)(62.9298)] + [(x)(64.9278)]

63.546 = 62.9298 - 62.9298x + 64.9278x

1.3818 = 1.998x

 1.3818 = x 1.998

X = 0.6916           63Cu = 0.6916 x 100% = 69.16%

65Cu = 1 – x = 1 – 0.6916 = 0.3084 x 100% = 30.84%

AP Example 2

The atomic mass of lithium is 6.94, the naturally occurring isotopes are 6Li = 6.015121 amu,  and 7Li = 7.016003 amu.  Determine the percent abundance of each isotope.

Aw = [(%abundance of isotope) (mass of isotope)] + [(%abundance of isotope) (mass of isotope)] + [….]

6.94 =[(% 6Li)(6.015121)] + [(%7Li)(7.016003)]

Since I don’t know what the percentage are, I will have to use variables.

100% of Lithium is determined by these two naturally occurring isotopes.

We will let 6Li = x and 7 Li = 1-x; we use 1 – x instead of 100 – x because the small number is easier to work with.

(in other words we reduced 100% to decimal form 1.00)

 Now let’s plug our variables in: 6.94 =[(% 6Li)(6.015121)] + [(%7Li)(7.016003)] 6.94 = [(x)(6.015121)] +[(1-x)(7.016003)] 6.94 = 6.015121x + 7.016003 – 7.016003x Combine like terms: 6.94 -7.016003 = (6.015121x - 7.016003x) -0.076003 = -1.000882 x Solve for x: -0.076003 = x -1.000882
 X = 0.075936, therefore 6Li = 0.075936 x 100% = 7.59% 1-X = 1 -0.075936 = 0.924064 , therefore 7Li = 0.924064 x 100% = 92.41%