An element has two naturally occurring isompes widx atomic masses of 10.01 amu and 11.01 amu. The relative abundances of these elements are 19.8% and 80.2%, respectively. What is the average atomic mass of the element?

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Answer 1 · 2015-11-29T14:51:18

$10.812 u$ $10.812u$

Now I hope you have been exposed to the concept of weighted average;
If not here is a video

Now lets get our data;

Lets call our element x; Let its existence and abundance be divided in 100 parts

so

$10.01$ $10.01$ amu is present in 19.8parts

Total atomic mass in this isotope is;

$10.01 \cdot 19.8 = 198.198$ $10.01 *19.8 = 198.198$ mass units

$11.01$ $11.01$ amu is present in 80.2parts

Total atomic mass in this isotope is;

$11.01 \cdot 80.2 = 883.002$ $11.01 *80.2= 883.002$ mass units

Now total number of mass units $= 883.002 + 198.198 = 1081.2$ $= 883.002 +198.198= 1081.2$

But now the total number of parts $= 100$ $= 100$

So all that's left is the weighted average;

$= \frac{1081.2}{100} = 10.812 u$ $= 1081.2/100= 10.812 u$