What is the Diameter of an Iridium Atom in Angstroms?

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In summary, Iridium has a face-centered cubic unit cell with a density of 22.61 grams per centimeter cubed. To find the diameter of the iridium atom, one can use Avogadro's Number to calculate the number of atoms in one cubic cm and then use the relationship between lattice parameter and atomic diameter/radius. This can be done by dividing the molar mass by 6.022 x 10^23 and multiplying it by 4, as the fcc lattice is composed of four atoms. However, this calculation only describes the quantity of molar mass divided by Avogadro's Number, and further steps are needed to convert the density into g/angstroms.
  • #1
parwana
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Iridium has a face-centered cubic unit cell. The density of iridium is 22.61 grams per centimeter cubed. What is the diameter, in angstroms, of the iridium atom ?

Atomic weight of Iridium = 192.1
 
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  • #2
What work have you done.

With the information given and Avogadro's Number, one can find the number of atoms of iridium in one cubic cm.

Think how the atoms are arraged in the fcc lattice and how the lattice parameter relates to the atomic diameter (or radius).
 
  • #3
wouldnt u just divide the molar mass by 6.022 X 10^23 and times it by 4 since the fcc is made like that. After that I have no clue. Dont I have to convert density into g/angstroms?? Please help with this equation, I am going crazy over this.
 
  • #4
parwana said:
wouldnt u just divide the molar mass by 6.022 X 10^23 and times it by 4 since the fcc is made like that.
So far, this is correct. But what quantity is it that you have calculated by doing this ?

ie : What quantity does [tex]4* \frac{molar~ mass}{6.022 \cdot 10^{23}} [/tex] describe ?
 

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