Average Atomic Mass Chapter 7 4 Average Atomic

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Average Atomic Mass

Average Atomic Mass

Chapter 7. 4 Average Atomic Mass • Objective: (1) To calculate the average atomic

Chapter 7. 4 Average Atomic Mass • Objective: (1) To calculate the average atomic mass for an atom.

Average Atomic Mass • Weighted average of the masses of the existing isotopes of

Average Atomic Mass • Weighted average of the masses of the existing isotopes of an element.

Recall: Atomic Mass Unit • The unit that describes the mass of an atom

Recall: Atomic Mass Unit • The unit that describes the mass of an atom • Symbol: amu

Example #1 • The mass of a Cu-63 atom is 62. 94 amu, and

Example #1 • The mass of a Cu-63 atom is 62. 94 amu, and that of a Cu-65 atom is 64. 93 amu. The percent abundance of Cu-63 is 69. 17% and the percent abundance of Cu-65 is 30. 83%. What is the average atomic mass of Cu?

Example #1 • The mass of a Cu-63 atom is 62. 94 amu, and

Example #1 • The mass of a Cu-63 atom is 62. 94 amu, and that of a Cu-65 atom is 64. 93 amu. The percent abundance of Cu-63 is 69. 17% and the percent abundance of Cu-65 is 30. 83%. What is the average atomic mass? • Step 1: Find the contribution of each isotope: Cu-63: (62. 94 amu) x (0. 6917) = 43. 535598 amu Cu-65: (64. 93 amu) x (0. 3083) = 20. 017919 amu • Step 2: Add the relative abundances from each isotope together. 43. 535598 amu + 20. 017919 amu = 63. 553517 amu Round answer to two numbers after the decimal: 63. 55 amu

Example #2 • Calculate the average atomic mass of chromium. It is made up

Example #2 • Calculate the average atomic mass of chromium. It is made up of isotopes with the following percent compositions and atomic masses: 83. 79% with a mass of 51. 94 amu; 9. 50% with a mass of 52. 94 amu; 4. 35% with a mass of 49. 95 amu; 2. 36 % with a mass of 53. 94 amu.

Example #2 • Calculate the average atomic mass of chromium. It is made up

Example #2 • Calculate the average atomic mass of chromium. It is made up of isotopes with the following percent compositions and atomic masses: 83. 79% with a mass of 51. 94 amu; 9. 50% with a mass of 52. 94 amu; 4. 35% with a mass of 49. 95 amu; 2. 36 % with a mass of 53. 94 amu. • Step 1: Find the contribution of each isotope: (51. 94 amu) x (0. 8379) = 43. 520526 amu (52. 94 amu) x (0. 0950) = 5. 0293 amu (49. 95 amu) x (0. 0435) = 2. 172825 amu (53. 94 amu) x (0. 0236) = 1. 272984 amu • Step 2: Add the relative abundances from each isotope together. 43. 520526 amu + 5. 0293 amu + 2. 172825 amu + 1. 272984 amu = 51. 995635 amu Round answer to two numbers after the decimal: 52. 00 amu

Example #3 • Calculate the average atomic mass of iron if its abundance in

Example #3 • Calculate the average atomic mass of iron if its abundance in nature is 15% iron-55 and 85% iron-56.

Example #3 • Calculate the average atomic mass of iron if its abundance in

Example #3 • Calculate the average atomic mass of iron if its abundance in nature is 15% iron-55 and 85% iron-56. • Step 1: Calculate the contribution of each isotope: Iron-55: (55 amu) x (0. 15) = 8. 25 amu Iron-56: (56 amu) x (0. 85) = 47. 6 amu • Step 2: Add the contribution from each isotope together: 8. 25 amu + 47. 6 amu = 55. 85 amu

Example #4 • Chlorine exists as chlorine-35, which has a mass of 34. 969

Example #4 • Chlorine exists as chlorine-35, which has a mass of 34. 969 amu and makes up 75. 8% of chlorine atoms. The rest of naturally occurring chlorine is chlorine 37, with a mass of 36. 996 amu. What is the average atomic mass of chlorine?

Example #4 • Chlorine exists as chlorine-35, which has a mass of 34. 969

Example #4 • Chlorine exists as chlorine-35, which has a mass of 34. 969 amu and makes up 75. 8% of chlorine atoms. The rest of naturally occurring chlorine is chlorine-37, with a mass of 36. 996 amu. What is the average atomic mass of chlorine? • Step 1: Find the contribution of each isotope: Cl-35: (34. 969 amu) x (0. 758) = 26. 506502 amu Cl-37: (36. 996 amu) x (0. 242) = 8. 953032 amu 1 – 0. 758 • Step 2: Add the relative abundances from each isotope together. 26. 506502 amu + 8. 953032 amu = 35. 459534 amu Round answer to two numbers after the decimal: 35. 46 amu

Example #5 • Element X has two isotopes. One has a mass of 10.

Example #5 • Element X has two isotopes. One has a mass of 10. 0 amu and an abundance of 20. 0%. The other has a mass of 11. 0 amu and an abundance of 80. 0%. Estimate the average atomic mass. What element is it?

Example #5 • Element X has two isotopes. One has a mass of 10.

Example #5 • Element X has two isotopes. One has a mass of 10. 0 amu and an abundance of 20. 0%. The other has a mass of 11. 0 amu and an abundance of 80. 0%. Estimate the average atomic mass. What element is it? • Step 1: Calculate the contribution of each isotope: Isotope 1: (10. 0 amu) x (0. 20) = 2 amu Isotope 2: (11. 0 amu) x (0. 80) = 8. 8 amu • Step 2: Add the contribution from each isotope together: 2 amu + 8. 8 amu = 10. 8 amu The element with an atomic mass of 10. 8 amu is Boron.