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Chapter 10: Problem 52

Iridium (Ir) has a face-centered cubic unit cell with an edge length of \(383.3 \mathrm{pm}\). Calculate the density of solid iridium.

Short Answer

Expert verified
The density of solid iridium is \(22.7\,g/cm^3\).

Step by step solution

01

Determine the number of iridium atoms in a face-centered cubic unit cell

A face-centered cubic unit cell contains atoms at each of its corners and in the center of each face. Therefore, each FCC unit cell consists of: - 8 corner atoms (each 1/8 within the unit cell) - 6 face-centered atoms (each 1/2 within the unit cell) So the total number of atoms in one unit cell is: \(8\, corner\,atoms\cdot\frac{1}{8\,atoms}+6\,face-centered\,atoms\cdot\frac{1}{2\,atoms}=1+3=4\,atoms/unit\ cell\).
02

Calculate the mass of one unit cell

To calculate the mass of one unit cell, we need to find the mass of four iridium atoms. The atomic mass of iridium is 192.22 atomic mass units (amu). Using Avogadro's number (6.022 x 10^23 atoms/mol), we can convert the amu to grams. Mass of 4 iridium atoms = \(\frac{4\,atoms}{1\,mol} \times\frac{192.22\,g}{6.022\times10^{23}\,atoms} = \frac{768.88\,g}{6.022\times10^{23}\,atoms} = 1.277\times 10^{-22}\,g\).
03

Determine the volume of the unit cell

We are given that the edge length of the unit cell is 383.3 pm (picometers). To find the volume of the unit cell, we need to cube the edge length and convert the units from picometers to centimeters. \(1\,pm = 10^{-12}\,m = 10^{-10}\,cm\) Volume of unit cell = \( (383.3\,pm)^3 \times\left(\frac{10^{-10}\,cm}{1\,pm}\right)^3 = (383.3\times10^{-10}\,cm)^3 = 5.633\times10^{-23}\,cm^3\)
04

Calculate the density of solid iridium

Now that we have the mass and volume of one iridium unit cell, we can calculate the density using the following formula: Density = \(\frac{mass}{volume}\) Density of solid iridium = \(\frac{1.277\times 10^{-22}\,g}{5.633\times10^{-23}\,cm^3} = 22.7\,g/cm^3\) The density of solid iridium is 22.7 g/cm³.

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