What is atomic concentration and how can it be found for different elements?

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    Atomic Concentration
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SUMMARY

Atomic concentration refers to the ratio of the number of atoms of a specific element to the total number of atoms in a sample. For pure elements, this concentration is 100%. In the case of alloys, such as Fe90Zr10, atomic concentrations are expressed in units of atoms per cubic meter (m^-3), calculated by dividing mass density (kg m^-3) by atomic weight (g mol^-1) and multiplying by Avogadro's number (6.022 x 10^23 atoms mol^-1). This method provides precise values for the atomic concentration of elements within a compound.

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  • Knowledge of Avogadro's number and its significance
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Chemists, materials scientists, and students studying atomic theory and alloy compositions will benefit from this discussion.

y4ku24
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Hello everyone.

Just a quick question: Could somebody explain to me what atomic concentration is, and whether there is a list of all the atomic concentration of the elements somewhere in the internet.

The 1st time I heard the term, I typed it on wikipedia and google, surprisingly the search did not come out with a straight answer of what it is. Is there another word for it?

Thank you in advance.
 
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The atomic concentration of X is the ratio of the number of atoms of X to the total number of atoms. The atomic concentration of any pure element is 100%.
 
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Dear Mapes,

Thank you for your feedback. However, I am reading a paper stating that a Fe90Zr10 alloy has an atomic concentration of Fe: 8.5x10^28 m^-3 and Zr: 4.29x10^28 m^-3

If it is a ratio, why is there a unit, and how did they come up with the estimated values above?
 
y4ku24 said:
Dear Mapes,

Thank you for your feedback. However, I am reading a paper stating that a Fe90Zr10 alloy has an atomic concentration of Fe: 8.5x10^28 m^-3 and Zr: 4.29x10^28 m^-3

If it is a ratio, why is there a unit, and how did they come up with the estimated values above?

OK, this is also a way of reporting concentration. They're dividing the mass density (kg m-3) by the atomic weight (g mol-1) and multiplying by Avogadro's number (6.022 x 1023 atoms mol-1). Does this help?
 
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Yup, very much! thanks again.
 

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