What is the Diameter of an Iridium Atom in Angstroms?

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Discussion Overview

The discussion revolves around calculating the diameter of an iridium atom in angstroms, utilizing the density of iridium and its face-centered cubic (fcc) structure. Participants explore the relationship between atomic mass, density, and atomic arrangement in the fcc lattice.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant notes that iridium has a face-centered cubic unit cell and provides its density and atomic weight as starting points for the calculation.
  • Another participant suggests using Avogadro's Number to determine the number of iridium atoms in one cubic centimeter, indicating a need to consider the arrangement of atoms in the fcc lattice and its relation to the atomic diameter.
  • A participant expresses confusion about the calculation process, proposing to divide the molar mass by Avogadro's Number and then multiply by 4, but is uncertain about converting density into appropriate units.
  • A later reply confirms the initial calculation method but questions what physical quantity is being derived from the proposed calculation involving the molar mass and Avogadro's Number.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the calculation method or the interpretation of the results. There are multiple competing views on how to approach the problem, and the discussion remains unresolved.

Contextual Notes

Participants express uncertainty regarding the conversion of density into appropriate units and the implications of the calculations being performed. There are also unresolved questions about the physical meaning of the quantities being calculated.

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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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).
 
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.
 
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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