What is average atomic mass?

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Average Atomic Mass & % Abundance

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Presentation transcript:

What is average atomic mass? Average atomic mass is the weighted average of the atomic masses of the naturally occurring isotopes of an element

Calculating Average Atomic Mass EXAMPLE Boron has two isotopes: B-10 (mass 10.013 amu) 19.8% abundance B-11 (mass 11.009 amu) 80.2% abundance Calculate the average atomic mass. (.198) (10.013) + (.802) ( 11.009) = 1.98 amu + 8.83 amu = 10.81 amu

Calculating Average Atomic Mass Calculate the average atomic mass of Mg. Isotope 1 - 23.985 amu (78.99%) Isotope 2 - 24.986 amu (10.00%) Isotope 3 – 25.982 amu (11.01%) (23.985)(.7899)+(24.986)(.1000)+(25.982)(.1101) 18.95 amu + 2.499 amu + 2.861 amu = 24.31 amu

Average Atomic Mass Helium has two naturally occurring isotopes, He-3 and He-4. The atomic mass of helium is 4.003 amu. Which isotope is more abundant in nature? He-4 is more abundant in nature because the atomic mass is closer to the mass of He-4 than to the mass of He-3.

Calculating % Abundance Set one of the isotope’s relative abundance equal to X. The relative abundances must add up to 1; therefore, the other isotope’s relative abundance would be equal to 1-X. Remember, the relative mass is equal to the relative abundance times the mass number. Remember, the relative masses must add up to the average atomic mass.

Calculating % Abundance Setup an algebraic equation where the relative masses of each isotope is set equal to the average atomic mass for the element. (X • mass number) + [(1-x) • mass number] = avg. atomic mass *When solving for X, keep 4 decimal places. This will allow the % abundance to have 2 decimal places.

Calculating % Abundance X is the relative abundance, so multiply this by 100 to make it % abundance. To obtain the other isotope’s % abundance, the 2 % abundances must add up to 100.

Calculating % Abundance Determine the % abundance for each isotope of antimony. Antimony exists as Sb-121 and Sb-123. Potassium exists as K-39 and K-41. Determine the % abundance for each isotope of potassium.

Isotopic Pennies – number of pre and post 1982 a. Let X be the number of pre-1982 pennies b. Let 10-X be the number of post-1982 pennies c. (X)(3.1g) + (10-X)(2.5g) = mass of 10 pennies pre-82 post-82 EXAMPLE (Mass of a sample of pennies is 31.0g) (X)(3.1g) + [(10-X)(2.5g)] = 31.0 g 3.1X + 25 - 2.5X = 31.0g .6X + 25 = 31.0g .6X = 6.0g X = 6.0g/.6 X = 10 pre-82 pennies 10-X = 0 pre-82 pennies

Isotopic Penny Lab- Average Atomic Mass Calculate percent of pre-82 and post-82 pennies # of pre-82 pennies x 100% # post-82 pennies x 100% 10 10 Calculate the average atomic mass of coinium (% pre-82)(3.1g) + (% post-82)(2.5) = average atomic mass