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

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Presentation on theme: "Average Atomic Mass & % Abundance. Average Atomic Mass The weighted average of the atomic masses of the naturally occurring isotopes of an element Most."— 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)


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