Naturally occuring boron is composed of only two isoptopes 10/5B and 11/B with isotopic masses of 10.0129 and 11.00931 respectively. To account for the atomic mass of 10.811 what must be the percentage abundance of each isotope?
Let 10B abundance = x
Then 11B abundance = 1-x
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x(10.0129) + (1-x)(11.00931) = 10.811.
Solve for x and 1-x, then convert to percent.
To determine the percentage abundance of each isotope of boron, we can use the atomic mass and the isotopic masses of the isotopes. Let's denote the percentage abundance of 10/5B as x, and the percentage abundance of 11/5B as (100 - x), since the sum of the percentage abundances should be 100%.
We can set up the following equation based on the atomic mass:
(x/100) * 10.0129 + ((100 - x)/100) * 11.00931 = 10.811
Now we can solve this equation to find the value of x, which represents the percentage abundance of 10/5B.
(x/100) * 10.0129 + ((100 - x)/100) * 11.00931 = 10.811
(x * 10.0129) + (100 - x) * 11.00931 = 10.811 * 100
10.0129x + 1100.931 - 11.00931x = 1081.1
-0.99641x = -19.831
x = (-19.831)/(-0.99641) ≈ 19.884
Therefore, the percentage abundance of 10/5B is approximately 19.884% (rounded to three decimal places) and the percentage abundance of 11/5B would be (100 - x) ≈ 80.116%.
Thus, the naturally occurring boron consists of approximately 19.884% of the isotope 10/5B and 80.116% of the isotope 11/5B.