Why is it necessary to use the average atomic mass of all isotopes, rather than the mass of the most commonly occurring isotope, when referring to the atomic mass of an element?

Md Ariful Islam

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Md Ariful Islam

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Why? Because accounting, and even more so chemistry, is a highly quantitative exercise. Mass is always conserved in a chemical reaction, and accurate masses, which are the weighted averages of the isotopic masses, are necessary to demonstrate this conservation of mass. For the lighter , hydrogen, carbon, and oxygen, this is not so important in that there is normally only the ONE isotope. But for the heavier elements, especially where the isotopic distribution is large, and we get fractional atomic masses, we must account for the isotopic distribution.

PS If my account were debited $£7-50$ instead of the $£7-49$ purchase price, I would raise merry hell, and demand satisfaction from my bank manager.

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An element has three naturally occurring isotopes with the following masses and abundances. How do you calculate the atomic mass of this element? What is the identity of the element?

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2. The element copper has naturally occurring isotopes with mass numbers of 63 and 65. The relative abundance and atomic masses are 69.2% for a mass of 62.93 amu and 30.8% for a mass of 64.93 amu. How do you calculate the average atomic mass of copper?

And thus the $"weighted average"$ is: $(62.93xx69.2%+64.93xx30.8%)*"amu"$ $=$ $63.55$ $"amu"$

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Background Info Most elements have different isotopes, or atoms with the same number of protons, but different numbers of neutrons. Hence, it is possible to have two atoms that are...

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The weighted average of the is equal to $117.3*"amu"$. And thus.........with units of $"amu"$ $[65.43%xx119.8+34.57%xxchi] =117.3$ Where $chi-="mass of the other isotope"$.... And so we solve for $chi$.... $chi=(117.3-0.6543xx119.8)/(0.3457)=112.6*"amu"$

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