Question #09099

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Question #09099

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Answer 1 · 2015-09-14T00:34:54

Element $X$ $"X"$ is silicon.

Explanation:

The idea here is that you need to use the atomic masses of the isotopes and their respective abundances to find the average atomic mass of element $X$ $"X"$ .

Even without doing any calculations, you can guess that the average atomic mass of elemnt $X$ $"X"$ will be very close to 28.00 amu due to the very high abundance of the $^{28} X$ $""^28"X"$ isotope.

Its high abundance tells you that it contributes most to the average atomic mass, so you can expect the atomic mass of element $X$ $"X"$ to be close to that value.

You can use fractional abundances to calculate the contribution each isotope has to the average atomic mass

$For^{28} X : \frac{92.23}{100} \cdot 27.977 amu = 25.80319 amu$ $"For " ""^28"X":" " 92.23/100 * "27.977 amu" = "25.80319 amu"$

$For^{29} X : \frac{4.67}{100} \cdot 28.976 amu = 1.35318 amu$ $"For " ""^29"X":" "4.67/100 * "28.976 amu" = "1.35318 amu"$

$For^{30} X : \frac{3.10}{100} \cdot 29.974 amu = 0.92919 amu$ $"For " ""^30"X":" "3.10/100 * "29.974 amu" = "0.92919 amu"$

The average atomic mass of the element will be the sum of each isotope's contribution

$avg. atomic mass = \sum isotope contribution$ $"avg. atomic mass" = sum"isotope contribution"$

This means that you have

$25.80319 + 1.35318 + 0.92919 = 28.08556 amu$ $25.80319 + 1.35318 + 0.92919 = color(green)("28.08556 amu")$

A quick look in the periodic table will reveal that element $X$ $"X"$ is silicon.

![http://www.ciaaw.org/http://silicon.htm](https://useruploads.socratic.org/ukM1bs4SQ02iG12hXg4B_Si.png)

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Question #09099

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