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%

Step-by-step explanation:

Mass of isotope boron -10= 10

% abundance of isotope 1 = x% =
`(x)/(100)`

Mass of isotope boron-11 = 11

% abundance of isotope 2 = (100-x)% =
`(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}* {{\text { fractional abundance}})`

`10.8=\sum[(10)* (x)/(100))+(11)* (100-x)/(100)]]`

`x=20`

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