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Average Atomic Mass & % Abundance

Average Atomic Mass & % Abundance. Average Atomic Mass . The weighted average of the atomic masses of the naturally occurring isotopes of an element Most elements occur naturally as mixtures of isotopes. Average Atomic Mass.

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Average Atomic Mass & % Abundance

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  1. Average Atomic Mass & % Abundance

  2. Average Atomic Mass • The weighted average of the atomic masses of the naturally occurring isotopes of an element • Most elements occur naturally as mixtures of isotopes

  3. Average Atomic Mass • Dependent upon both mass and the relative abundance of each of the elements isotopes

  4. Example • Naturally occurring copper exists with the following abundances: • 69.17% is Cu-63 w/ atomic mass 62.93 amu • 30.83% is Cu-65 w/ atomic mass 64.93 • (.6917)x(62.93) + (.3083)x(64.93)= 63.55 amu

  5. Problem 1 • 3 Isotopes of Ar occur in nature • 0.337% as Ar-36, 35.97 amu • 0.063% Ar-38, 37.96 amu • 99.6% Ar-40, 39.96 amu • Calculate the Average Atomic Mass

  6. Answer Check • (.00337)x(35.97) + (.00063)x(37.96) + (.996)x(39.96)= 39.95amu

  7. Problem 2 • 2 Naturally occurring Isotopes of Boron occur with the following abundances: • 80.20% B-11, 11.01 amu • 19.80% B-10, 10.81 amu • What is the Average Atomic Mass

  8. Answer Check • (.8020)x(11.01) + (.1980)x(10.81) = 10.97 amu

  9. Calculating & Abundance • Chlorine has two isotopes: chlorine-35 (mass 34.97 amu) and chlorine-37 (mass 36.97 amu). • What is the percent abundance of these two isotopes if chlorine's atomic mass is 35.453?

  10. Answer Check Part 1 • if 2 isotopes, then the total is 100%. assume one is x% (x), the other is automatically 100-x%, (1-x) • x(34.97) + (1-x)(36.97) = 35.453

  11. Answer Check Part 2 • x(34.97) + (1-x)(36.97)=35.453 • Solve for x • 34.97x+36.97-36.97x=35.453 • -2x+36.97=35.453 • -2x=-1.517 • x=.7585 • 1-x=.2415

  12. Answer Check Part 3 • Therefore Cl-35 has a % abundance of 75.85% and Cl-37 has a % abundance of 24.15%

  13. Problem 1 • The two naturally occurring isotopes of nitrogen are nitrogen-14, with an atomic mass of 14.003074 amu, and nitrogen-15, with an atomic mass of 15.000108 amu. What are the percent natural abundances of these isotopes? • The atomic mass of nitrogen is 14.00674amu

  14. Answer Check • The atomic mass of nitrogen is 14.00674amu • 14.00674 = p(14.003074) + (1 -p)(15.000108)14.00674 = 14.003074p + 15.000108 - 15.000108p-0.997034p = -0.993368 • p = 0.9963 = 99.63% (N14)1 - p = 0.0037 = 0.37% (N15)

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