Number of Protons = Number of Electrons Number of Protons = Atomic Number Atomic Mass = Protons + Neutrons.

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Presentation transcript:

Number of Protons = Number of Electrons Number of Protons = Atomic Number Atomic Mass = Protons + Neutrons

Average Atomic Mass

  The atomic masses reported in the periodic table represent the weighted average of the masses of the naturally occurring isotopes of that element

Weighted Avg. of Test Grades   Suppose you have: 1 test of 0% and 9 tests of 90% What if I said your average was 45? ( = 90; Divide by 2 = 45)   Would you be happy?

NO!   The long way to do this problem:   ( )  10   Average = 81% I bet you like this a lot better!

The smart way to do this problem!   grades of 90: 9 out of 10 is 90%   grade 0f 0: 1 out of 10 is 10%   So 90% of the test grades are 90   10% of the test grades are 0   (0.90 X 90) + (0.10 X 0) = Average = 81%

Weighted average: Smarter way to calculate the average if you have numerous items to average

Method of Weighted Averages ● ● Convert % to decimal format (Divide % by 100) ● ● Multiply each isotope’s abundance factor by its atomic mass ● ● Find the Sum

Avg. Atomic Mass of Chlorine amu24.230% Isotope amu75.770% Isotope 1 Mass Percent Abundance Chlorine has two isotopes.

Avg. Atomic Mass of Chlorine = (.75770)( amu) + (.24230) ( amu) (.24230) ( amu) = amu amu = amu Convert % to decimals Multiply by appropriate mass Sum

Avg. Atomic Mass of Chlorine   To estimate the answer: 75% of Cl is amu & 25% of Cl is amu   The final answer has to be between amu & amu, but closer to amu  amu

Avg. Atomic Mass of Si 92.21% of Si has a mass of amu 4.70% of Si has a mass of amu 3.09% of Si has a mass of amu

Avg. Atomic Mass of Si X  amu X  amu X  amu X  amu X  amu X  amu amu amu

Avg. Atomic Mass of Pb 1.5% Pb-204  X 204amu: 3.06amu 23.6% Pb-206  X 206amu: 48.62amu 22.6% Pb-207  X 207amu: 46.78amu 52.3% Pb-208  X 208amu:108.78amu Sum amu