Question

Assume Boron has (only) two naturally occurring isotopes, B10 and B11. If Boron has an average atomic mass of 10.811amu, B10 has a mass of 10.0 amu and B11 has a mass of 11.0 amu, what are the abundancies of each isotope

Answer 1

Answer:

B10: 18.9%

B11: 100% - 18.9% = 81.1%

Explanation:

The abundance of the isotope is calculating using;

(M1)(x) + (M2)(1-x) = M(E)

Where;

M1 = Mass of B10 = 10.0 amu

M2 = Mass of B11 = 11 amu

M(E) = Average mass of Boron = 10.811 amu

x = relative abundance

Inserting the values;

10(x)  +  11(1-x) = 10.811

10x + 11 - 11x = 10.811

10x - 11x = 10.811 - 11

-x = -0.189

x = 0.189

Converting to percentage; x = 18.9%

B10: 18.9%

B11: 100% - 18.9% = 81.1%