Question 1 (1 point)
Chlorine has two stable isotopes of mass 34.97 u and 36.97 u. Suppose the percent abundance of the lighter isotope were 64.75%. What would be the atomic mass of Cl? (The actual abundance is 75.77%.) I need the procces to get to the answer
2007-09-18 05:31:58 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
Average atomic mass = %abundance1*mass1 + %abundance2*mass2
If the abundance of the lighter isotope (34.97) is 64.97%, then the abundance of the heavier isotope (36.97) is going to be: 100 - 64.75 = 35.25%
Then,
Average atomic mass = 0.6475*34.97 + 0.3525*36.97
= 22.6430 + 13.0319
= 35.68 a.m.u
Hope this helps...
2007-09-18 05:49:50 · answer #1 · answered by Anonymous · 0⤊ 1⤋
(0.6475 * 34.97) + (0.3525 * 36.97) = 22.643 + 13.032 = 35.675 amu
Rounding up to the correct number of significant digits, that becomes 35.68 amu.
2007-09-18 12:52:28 · answer #2 · answered by Dave_Stark 7 · 0⤊ 0⤋
Frank says hi...
But please, next time do use the course notes located in webCt under:
GENERAL CHEMISTRY I-->Course Notes-->Ch 2 Average Mass Isotopes
Doubt you can keep on bugging online users to do your homework
So please try to do your homework ahead of time so that one can ask questions in class if a problem arises.
2007-09-19 19:48:47 · answer #3 · answered by SANiK 1 · 0⤊ 0⤋
100.00% - 64.75% = 35.25% (heavier isotope)
64.75% x 34.97 = 22.64
35.25% x 36.97 = 13.03
____________________
35.67
2007-09-18 12:49:37 · answer #4 · answered by steve_geo1 7 · 1⤊ 0⤋