What is the average atomic mass of silver if 13 out of 25 atoms are silver-107 and 12 out of 25 atoms are silver-109?

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What is the average atomic mass of silver if 13 out of 25 atoms are silver-107 and 12 out of 25 atoms are silver-109?

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Answer 1 · 2015-12-03T20:27:16

$107.96 u$ $"107.96 u"$

Explanation:

As you know, the average atomic mass of an element is calculated by taking the weighted average of the atomic masses of its naturally occurring isotopes.

In simple terms, each isotope will contribute to the average mass of the element proportionally to its percent abundance.

$avg. atomic mass = \sum i ({isotope}_{i} \times {abundance}_{i})$ $color(blue)("avg. atomic mass" = sum_i ("isotope"_i xx "abundance"_i))$

For the actual calculations, it's easier to use decimal abundances, which are simply percent abundances divided by $100$ $100$ .

So, what would the decimal abundances of the two isotopes of silver be?

Well, you know that for every $25$ $25$ atoms of silver, you get

$13$ $13$ atoms of silver-107

$12$ $12$ atoms of silver-109

This means that the decimal abundance of silver-107 will be equal to $\frac{13}{25}$ $13/25$ and the decimal abundance of silver-109 will be equal to $\frac{12}{25}$ $12/25$ .

Now, if the atomic masses of the two isotopes were not provided by the problem, you can assume them to be

$^{107} Ag \to 107 u$ $""^107"Ag" -> "107 u"$

$^{109} Ag \to 109 u$ $""^109"Ag" -> "109 u"$

So, the average atomic mass of silver will be

$avg. atomic mass = 107 u \times \frac{13}{25} + 109 u \times \frac{12}{25}$ $"avg. atomic mass" = "107 u" xx 13/25 + "109 u" xx 12/25$

$avg. atomic mass = 107.96 u$ $"avg. atomic mass " = color(green)(" 107.96 u")$

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What is the average atomic mass of silver if 13 out of 25 atoms are silver-107 and 12 out of 25 atoms are silver-109?

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