The isotope 121sb makes up 57.21% of naturally occurring antimony. The only other naturally occurring isotope is 123sb. Deduce the relative atomic mass of antimony according to this data. How do I do that?

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Answer 1 · 2017-10-23T15:31:57

The ponderate media between both isotopes gives 121.86.

In each 100 $m o l$ $mol$ of Sb you have:

$57.21$ $57.21$ mol of $_^{121} S b$ $_ ^(121)Sb$ and $100 - 57.21 = 42.79 m o l$ $100-57.21=42.79 mol$ of $_^{123} S b$ $_ ^(123)Sb$

Assuming the mass of the isotopes is integer (they are close to it) we may calculate the corresponding mass of that amount:

$57.21 \times 121 + 42.79 \times 123 = 12185.58 g$ $57.21xx121+42.79xx123=12185.58 g$

As this correspond to 100 mol the atomica mass is:

$\frac{12185.58}{100} = 121.86 g / (m o l)$ $12185.58/100=121.86 g//(mol)$ .

The real value is $121.76 g / (m o l)$ $121.76 g//(mol)$ for comparison.

As you are talking about relative atomic mass, take the units out, so the answer is only 121.86