Question

Antimony consists of two major isotopes with the following atomic masses: 121Sb (120.904 amu) and 123Sb (122.904 amu). The average atomic mass of antimony is 121.760 amu. Calculate the percent of each of the isotopes. Report your answer to 3 significant figures.

Answer 1

Answer:

121Sb=57.2%

123Sb=42.8%

Explanation:

We are given that

Atomic mass of 121Sb=120.904 amu

Atomic mass of 123Sb=122.904 amu

Average atomic mass of antimony=121.760 amu

We have to find the percent of each of the isotope.

Let x be the percent of 121Sb and 1-x be the percent of 123Sb.

Using formula of average atomic weight

Average atomic weight=atomic weight of 121Sb$\times$ percentage abundance of isotope 121Sb+atomic weight of 123Sb$\times$percentage abundance of isotope 123Sb

Substitute the values

$121.760=120.904x+122.904(1-x)$

$121.760=120.904x+122.904-122.904x$

$121.760=-2x+122.904$

$2x=122.904-121.760$

$2x=1.144$

$x=\frac{1.144}{2}=0.572$

Percentage of 121Sb=$0.572\times 100=$57.2%

Abundance of isotope 123Sb=1-0.572=0.428

Percentage of isotope 123Sb=$0.428\times 100=$42.8%