The element boron has three naturally occuring isotopes. One has the atomic mass 10.0129(54.4%), another 11.0093(12.75%). What is the atomic mass of the third isotope if the average atomic mass of boron is 10.811?
So the third isotope has an abundance of
100%-54.4%-12.75% = ? = about 32.85% or 0.3285 as a decimal).
If we let Z = atomic mass of the third isotope we have\
10.0129(0.544) + 11.0093(0.1275) + Z(0.3285) = 10.811
Solve for Z.
Would the correct answer be 12.167?
I came up with 12.0556 which I would round to 12.056 and that's too many significant figures. By the way, I don't see but two isotopes for B; i.e., B-10 and B11. This may be a made up problem. See
http://www.webelements.com/boron/isotopes.html
To find the atomic mass of the third isotope, we can use the formula for calculating the average atomic mass.
The formula for average atomic mass is:
Average Atomic Mass = (Mass of isotope 1 * % abundance of isotope 1) + (Mass of isotope 2 * % abundance of isotope 2) + (Mass of isotope 3 * % abundance of isotope 3)
Given the following information:
Isotope 1: Atomic mass = 10.0129, % abundance = 54.4%
Isotope 2: Atomic mass = 11.0093, % abundance = 12.75%
Average atomic mass of boron = 10.811
Plugging the values into the formula, we have:
10.811 = (10.0129 * 54.4%) + (11.0093 * 12.75%) + (Mass of isotope 3 * % abundance of isotope 3)
Let's solve for the atomic mass of the third isotope:
10.811 = (10.0129 * 0.544) + (11.0093 * 0.1275) + (Mass of isotope 3 * % abundance of isotope 3)
Simplifying further:
10.811 = 5.4443 + 1.4021 + (Mass of isotope 3 * % abundance of isotope 3)
10.811 = 6.8464 + (Mass of isotope 3 * % abundance of isotope 3)
Subtracting 6.8464 from both sides:
3.9646 = (Mass of isotope 3 * % abundance of isotope 3)
To find the mass of the third isotope, we need to divide 3.9646 by the % abundance of the isotope. Since the % abundance is not given in the question, we cannot determine the exact mass of the third isotope.