1.- Boron has 2 naturally occuring isotopes, which have masses10.0129 and 11.009

Question

1.- Boron has 2 naturally occuring isotopes, which have masses10.0129 and 11.0093 amu, respectively. Given the average atomicmass of boron (10.81amu), determine the % abundance of eachisotope. 2.- The 2 naturally occuring isotopes of antimony,121Sb (57.21%) and 123Sb (42.79%), havemasses of 120.904 and 122.904 amu, respectively. What is theaverage atomic mass of Sb? 1.- Boron has 2 naturally occuring isotopes, which have masses10.0129 and 11.0093 amu, respectively. Given the average atomicmass of boron (10.81amu), determine the % abundance of eachisotope. 2.- The 2 naturally occuring isotopes of antimony,121Sb (57.21%) and 123Sb (42.79%), havemasses of 120.904 and 122.904 amu, respectively. What is theaverage atomic mass of Sb?

Explanation / Answer

1.

Average atomic mass = (Percent abundance * atomic mass) for eachisotope.

.

Let % abundance of Boron- 10.0129 = x %

% abundance of the other isotope = ( 100 - x)

10.81 amu = [( x/ 100) * 10.0129 amu] + [((100-x)/ 100 ) * 11.0093 amu]

Solving the equation for x gives the percent abundance of oneisotope. Subtracting it from 100 will give the percent abundance ofthe other isotope.

2.

Using the formula above,

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