Question

Boron has an atomic mass of 10.8 amu. It is known that naturally occurring boron is composed of two isotopes, boron-10 and boron-11. Determine the percent of each isotope of boron

Answer 1

B-10 = 19.9%
B-11 = 80.1%

Abundance of B10 = x
Abundance of B11= y
You know x+y = 1 because there are only the 2 isotopes.
Y= 1-x

10.01294x + 11.00931 (1-x) = 10.811
10.01294x + 11.00931 - 11.00931x = 10.811 - 0.99016x = -0.198
X = 0.200

Check:
10.01294(0.2) + 11.00931 (0.8) = 10.81

20% B10. 80% B11

Answer 2

Answer: Boron -10 : 20%

Boron -11 : 80%

Explanation:

Mass of isotope  boron -10= 10

% abundance of isotope 1 = x% = $\frac{x}{100}$

Mass of isotope boron-11 = 11

% abundance of isotope 2 = (100-x)% = $\frac{100-x}{100}$

Formula used for average atomic mass of an element :

$\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})$

$10.8=\sum[(10)\times \frac{x}{100})+(11)\times \frac{100-x}{100}]]$

$x=20$

Therefore, percent of boron -10 is 20% and isotope boron-11 is (100-20)= 80%.