Isotope Abundance and Atomic Weight

Ex: 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.

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

Aw = the atomic weight found in the periodic table

% abundance of EACH isotope is converted to decimal ex: 19.9% ÷ 100% = 0.199

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

Aw = [(0.199)(10.013)] + [(0.801)(11.009)]

Aw = [1.992587] + [8.818209]

Aw = 10.810796 so, the atomic weight of B = 10.811

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

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%

PROBLEMS (SOLUTIONS ON BACK)

1. Calculate the weight of silicon using the following data for the percent natural abundance and mass of each isotope:

92.23% ²⁸Si (27.9769 amu); 4.67% ²⁹ Si (28.9765); 3.10% ³⁰Si (29.9738 amu).

2. Thallium has two stable isotopes, ²⁰³Tl and ²⁰⁵Tl. Knowing that the atomic weight of thallium is 204.4,

which isotope is the more abundant of the two?

3. Verify 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%

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

What are the percent abundances of the isotopes?

SOLUTIONS

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

1. Aw= [(0.9223)(27.9769)] + [(0.0467)(28.97650)] + [(0.031) (29.9738)]

Aw = 25.803 + 1.353 + 0.929

Aw for Silicon = 28.085

Hint: double check the periodic table to see if the answer is the same or close

2. ²⁰³Tl % = x

²⁰⁵Tl % = 1 – x

204.4 = [(x)(203)] + [(1- x)(205)]

204.4 = 203x + 205 – 205x

-0.6 = -2x

-0.6 = x

-2

X= 0.3 ²⁰³Tl = 0.3 x 100% = 30 %

²⁰⁵Tl = 1 –x = 1 – 0.3 = 0.7 x 100% = 70%

Therefore ²⁰⁵Tl = 70% is more abundant.

3. Aw= [(0.7899)(23.985042)] + [(0.1)(24.985837)] + [(0.1101)(25.982593)]

Aw = 18.946 + 2.499 + 2.861

Aw for Mg = 24.31, this checks with the Aw that was given in problem #3.

4. 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. Aw for Cu = 63.546

⁶³Cu % = x

⁶⁵Cu % = 1 – x

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

63.546 = 62.9298x + 64.9278 – 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%