Question #0b468

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Question #0b468

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Answer 1 · 2015-10-27T23:50:38

$107.9 u$ $"107.9 u"$

Explanation:

The first thing to do here is figure out the percent abundance of the third isotope.

Since you were only told about the existance of three isotopes, it follows that their respective abundances must add up to give $100 %$ $100%$ .

You know that the first one has a percent abundance of $45.3 %$ $45.3%$ , and the second a percent abundance of $12.9 %$ $12.9%$ , which means that you have

$45.3 % + 12.9 % + x % = 100 %$ $45.3% + 12.9% + x% = 100%$

$x = 100 - (45.3 + 12.9) = 41.8 %$ $x = 100 - (45.3 + 12.9) = 41.8%$

Now, the average atomic mass of an element is calculated by doing a weighted average of the atomic masses of its naturally occuring isotopes.

SImply put, each isotope will contribute to the average atomic mass of the element proportionally to its decimal abundance, which is simply its percent abundance divided by $100$ $100$ .

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

In your case, you know that you have

$isotope 1$ $"isotope 1"$ $\to$ $->$ $106.2 u, 45.3 %$ $"106.2 u", "45.3%$

$isotope 2$ $"isotope 2"$ $\to$ $->$ $107.1 u, 12.9 %$ $"107.1 u", "12.9%$

$isotope 3$ $"isotope 3"$ $\to$ $->$ $109.9 u, 41.8 %$ $"109.9 u", "41.8%$

So, the average atomic mass of this element will be

$avg. atomic mass = 106.2 u \times 0.453 + 107.1 u \times 0.129 + 109.9 u + 0.418$ $"avg. atomic mass" = "106.2 u" xx 0.453 + "107.1 u" xx 0.129 + "109.9 u" + 0.418$

$avg. atomic mass = 107.9 u$ $"avg. atomic mass" = color(green)("107.9 u")$

The answer is rounded to three sig figs.

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Question #0b468

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