Online Introductory Chemistry

Isotopes and average atomic masses

Dr. Walt Volland revised March 31, 2005

Example of a weighted average calculation: Assume you have taken five quizzes with the following scores out of 10 possible points. 10 pts, 5 pts, 10 pts, 10 pts, 10 pts. The average score will be 9 pts. average = (1/5)( 10 pts + 5 pts + 10 pts + 10 pts + 10 pts) = 45 points total / 5 = 9 pts The low score doesn't shift the average down much because the four 10 point scores weight the average toward nine. Here the high scores occurred 4 out of 5 times or 80% of the time. Only one out of five scores was low at 5. The abundant high scores have more influence on the average. You can calculate the weighted average in the following way. Literally what happens in the calculation is average score = [40 points + 5 points][1/5] = [45 points] = 9 points

Example weighted average calculation using isotope abundances: What is the average atomic weight for chlorine if it has two isotopes? The percent abundance for chlorine-35 is 75.53%. The percent abundance for chlorine-37 is 24.4%. The mass for Cl-35 is 35.0 amu and for Cl-37 it is 37.0 amu. Average atomic weight = 26.4 amu + 9.05 amu = 35.5 amu Notice that the average is determined by the more abundant Cl-35. Chlorine-35 is more common so the average is closer to 35 amu than 37 amu. Clearly no atoms of chlorine actually have a mass of 35.5 amu. The tabulated value in the periodic table is a statistical creation.

Exercise: What is the average atomic mass for thallium, Tl, if there are two isotopes with the following masses and abundances? Tl-203 has a mass of 203.059 amu with an abundance of 29.52 % Tl-205 has a mass of 205.059 amu with an abundance of 70.48 % You need to show your work. Solution 1. Convert percents to decimals 29.52 % to 0.2952 and 70.48 % to 0.7048 2. Weighted average = 0.2952 x ( 203.059 amu) + 0.7048 x ( 205.059 amu) 3. Answer: 204.466 amu rounded off to 204.5 amu with 4 significant figures