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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.

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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[tex]\times[/tex] percentage abundance of isotope 121Sb+atomic weight of 123Sb[tex]\times [/tex]percentage abundance of isotope 123Sb

Substitute the values

[tex]121.760=120.904x+122.904(1-x)[/tex]

[tex]121.760=120.904x+122.904-122.904x[/tex]

[tex]121.760=-2x+122.904[/tex]

[tex]2x=122.904-121.760[/tex]

[tex]2x=1.144[/tex]

[tex]x=\frac{1.144}{2}=0.572[/tex]

Percentage of 121Sb=[tex]0.572\times 100=[/tex]57.2%

Abundance of isotope 123Sb=1-0.572=0.428

Percentage of isotope 123Sb=[tex]0.428\times 100=[/tex]42.8%

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