Question

The element boron exists in nature as two isotopes: 10B has a mass of 10.0129 u, and 11B has a mass of 11.0093 u. The average atomic mass of boron is 10.81 u. Calculate the relative abundance (as percentages) of the two boron isotopes.

Answer 1

Answer:

Percentage abundance of B 10 is = 20 %

Percentage abundance of B 11 is = 80 %

Explanation:

The formula for the calculation of the average atomic mass is:

$Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})$

Given that:

Since the element has only 2 isotopes, so the let the percentage of first be x and the second is 100 -x.

For first isotope, B 10:

% = x %

Mass = 10.0129 u

For second isotope, B 11:

% = 100  - x  

Mass = 11.0093 u

Given, Average Mass = 10.81 u

Thus,  

$10.81=\frac{x}{100}\times {10.0129}+\frac{100-x}{100}\times {11.0093}$

$10.0129x+11.0093\left(100-x\right)=1081$

Solving for x, we get that:

x = 20 %

Thus percentage abundance of B 10 is = 20 %

Percentage abundance of B 11 is = 100 - 20 %  = 80 %

Answer 2

Answer:

The relative abundance of 10B = 0.20 (20%)

The relative abundance of 11B = 0.80 (80%)

Explanation:

Step 1: Data given

Boron has 2 natural isotopes

⇒ 10B has a mass of 10.0129 u

⇒ 11B has a mass of 11.0093 u

Average atomic mas of Boron = 10.81

Step 2:

10B has an abundance of X %

11B has an abundance of Y %

X+ Y = 1

X = 1 - Y

10.81 = 10.0129* (1 - Y) + 11.0093*Y

10.81 = 10.0129 - 10.0129Y + 11.0093Y

0.7971 = 0.9964Y

Y = 0.80

X = 1.0 - 0.80 = 0.20

10.0129*0.20 + 11.0093*0.80 = 2.00258 +8.8044 = 10.81

The relative abundance of 10B = 0.20 (20%)

The relative abundance of 11B = 0.80 (80%)