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Ivan
4 years ago
12

Antimony consists of two major isotopes with the following atomic masses: 121Sb (120.904 amu) and 123Sb (122.904 amu). The avera

ge atomic mass of antimony is 121.760 amu. Calculate the percent of each of the isotopes. Report your answer to 3 significant figures.
Physics
1 answer:
Bad White [126]4 years ago
7 0

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\timespercentage 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%

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