The atomic mass of chlorine is listed as 35.454 amu on the periodic table. It has two naturally occurring isotopes, CI-35 and CI-37. What are the approximate percentage of each?
If you want REAL numbers you can get them from www.webelements.com.
If you want to calculate them, you need to know the masses of the 35 and the 37.
x = fraction of the 35.
1-x = fraction of the 37
x(mass 35) + (1-x)(mass 37) = 35.454.
To determine the approximate percentage of each naturally occurring isotope of chlorine, you can use the atomic mass and the concept of weighted average.
1. Start by noting the atomic masses of each isotope:
- CI-35 has a mass of 35 amu
- CI-37 has a mass of 37 amu
2. Next, note the atomic mass listed on the periodic table, in this case, 35.454 amu.
3. Since the atomic mass listed on the periodic table is a weighted average of the isotopes' masses, we can set up the following equation:
[(percentage of CI-35) × (mass of CI-35)] + [(percentage of CI-37) × (mass of CI-37)] = (atomic mass listed)
4. To find the percentages, let's assume that x represents the percentage of CI-35 and (100 - x) represents the percentage of CI-37.
5. Substitute the values into the equation:
(x/100) × 35 amu + [(100 - x)/100] × 37 amu = 35.454 amu
6. Solve the equation for x:
(35x + 3700 - 37x)/100 = 35.454
-2x + 3700 = 3545.4
-2x = -154.6
x ≈ 77.3
7. The percentage of CI-35 is approximately 77.3%, and therefore the percentage of CI-37 is approximately (100 - 77.3) = 22.7%.
Therefore, the approximate percentage of CI-35 is 77.3% and CI-37 is 22.7%.