Average Atomic Mass Abundance Average Atomic Mass n
Average Atomic Mass & % Abundance
Average Atomic Mass n The weighted average of the atomic masses of the naturally occurring isotopes of an element n Most elements occur naturally as mixtures of isotopes
Average Atomic Mass n Dependent upon both mass and the relative abundance of each of the elements isotopes
Example Naturally occurring copper exists with the following abundances: n 69. 17% is Cu-63 w/ atomic mass 62. 93 amu n 30. 83% is Cu-65 w/ atomic mass 64. 93 n n (. 6917)x(62. 93) + (. 3083)x(64. 93)= 63. 55 amu
Problem 1 n 3 Isotopes of Ar occur in nature 0. 337% as Ar-36 n 0. 063% Ar-38 n 99. 6% Ar-40 n n Calculate the Average Atomic Mass
Answer Check n (. 00337)x(35. 97) + (. 00063)x(37. 96) + (. 996)x(39. 96)= 39. 95 amu
Problem 2 n 2 Naturally occurring Isotopes of Boron occur with the following abundances: 80. 20% B-11, 11. 01 amu n 19. 80% B-10, 10. 81 amu n n What is the Average Atomic Mass
Answer Check n (. 8020)x(11. 01) + (. 1980)x(10. 81) = 10. 97 amu
Calculating & Abundance Chlorine has two isotopes: chlorine-35 (mass 34. 97 amu) and chlorine-37 (mass 36. 97 amu). n What is the percent abundance of these two isotopes if chlorine's atomic mass is 35. 453? n
Answer Check Part 1 n if 2 isotopes, then the total is 100%. assume one is x% (x), the other is automatically 100 -x%, (1 -x) n x(34. 97) + (1 -x)(36. 97) = 35. 453
Answer Check Part 2 x(34. 97) + (1 -x)(36. 97)=35. 453 n Solve for x n 34. 97 x+36. 97 -36. 97 x=35. 453 n -2 x+36. 97=35. 453 n -2 x=-1. 517 n x=. 7585 n 1 -x=. 2415 n
Answer Check Part 3 n Therefore Cl-35 has a % abundance of 75. 85% and Cl-37 has a % abundance of 24. 15%
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? n The atomic mass of nitrogen is 14. 00674 amu n
Answer Check n The atomic mass of nitrogen is 14. 00674 amu n 14. 00674 = p(14. 003074) + (1 -p)(15. 000108) 14. 00674 = 14. 003074 p + 15. 000108 - 15. 000108 p -0. 997034 p = -0. 993368 n p = 0. 9963 = 99. 63% (N 14) 1 - p = 0. 0037 = 0. 37% (N 15)
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