1 / 14

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.

Download Presentation

Average Atomic Mass & % Abundance

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  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)

More Related