Boron's Percent abundance

boron has two naturally occurring isotopes, boron 10 with a mass of 10.0129 amu and boron-11 with a mass of 11.00391 amu. the average atomic mass of boron is 10.811. what is the percent abundance of the boron 10 isotope

Hello,

You treat this problem like a math problem with x.

You know that to find the average atomic mass, you multiply the percent of the isotope by its amu mass and add it to the other isotope.

So, first set the percent abundance of boron-10 as x. The percent abundance of boron-11 will be 100 - x .

Here is your equation:

(x)(10.0129) + (1-x)(11.00391) = 10.811
10.0129x + 11.00391 - 11.00391x = 10.811
-0.99101x = -0.19291
x = 0.1947 = 19.47% <<

x(10.0129) + (1-x)(11.00391) = 10.811

10.0129x -11.00391x + 11.00391 = 10.811

-0.89101x = -0.19281

x = 0.19281 / 0.89101 = 21.639% B-10
1-x = 78.361% B-11.