Answer to Question #129244 in General Chemistry for muhammad mirza

Question

Boron is composed of two isotopes, 10B (10.013 u) and 11B (11.009 u). If the average
atomic mass of Boron is 10.81 u, calculate the percentage abundance of 10B and 11B.

Answer 1

At first we may designate percentage as parts of 1: B10 content as x and B11 content as y. If there are only two isotopes, so their sum will be equal to 1:

x+y=1.

From the average mass we obtain the second equation:

10.013x + 11.009y = 10.81.

We should solve the system of equations:

y=1-x,

10.013x + 11.009(1-x) = 10.81,

10.013x + 11.009 - 11.009x = 10.81,

-0.996x = -0.199,

x=0.2,

y=1-0.2=0.8.

In per cents we have 20% of x (10B) and 80 % of y (11B).

Answer: 10B - 20%, 11B - 80%.