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 22.65 g/cm³.

Step by step solution

01

Determine the number of atoms in the unit cell

A face-centered cubic (FCC) unit cell consists of atoms at the 8 corners of the cube and at the centers of each of the 6 faces. Each corner atom contributes 1/8 part of it to the unit cell, and each face-centered atom contributes 1/2 part of it. Therefore, the total number of atoms in an FCC unit cell is: Number of atoms = (8 corner atoms × 1/8) + (6 face-centered atoms × 1/2) Number of atoms = 1 + 3 = 4 atoms
02

Calculate the volume of the unit cell

The unit cell is a cube with an edge length of 383.3 pm. We convert the edge length to meters to make the calculation easier, and then find the volume (V) of the cube using the formula: V = (edge length)^3 Edge length = 383.3 pm = 383.3 × 10⁻¹² m V = (383.3 × 10⁻¹² m)^3 V = 5.6409 × 10⁻²⁶ m³
03

Find the mass of the atoms in the unit cell

First, we'll need the molar mass of Iridium, which is 192.22 g/mol. Next, we'll convert the molar mass to the mass of a single Iridium atom using Avogadro’s number (6.022 × 10²³ atoms/mol): Mass of one Iridium atom = (192.22 g/mol) / (6.022 × 10²³ atoms/mol) = 3.192 × 10⁻²² g Since there are 4 atoms in the unit cell, the total mass of the unit cell is: Mass of the unit cell = 4 atoms × 3.192 × 10⁻²² g = 12.77 × 10⁻²² g
04

Calculate the density of the unit cell

Finally, we calculate the density using the mass and volume of the unit cell: Density = Mass / Volume Density = (12.77 × 10⁻²² g) / (5.6409 × 10⁻²⁶ m³) Density = 22.65 g/cm³ The density of solid Iridium is 22.65 g/cm³.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Face-centered cubic unit cell
A face-centered cubic (FCC) unit cell is a common arrangement of atoms in some metals, including Iridium. This kind of crystalline structure is characterized by atoms located at each of the corners and the centers of all the cube's faces.

Picture a simple cube: it has eight corners and six faces. In an FCC unit cell, every corner has a fraction of an atom that counts as 1/8th when considering the full unit cell because that corner atom is shared by eight surrounding cells. On each face, however, there's a half of an atom, because a face is shared by only two cells. When you add them up, the total is four whole atoms per unit cell. (8 corners × 1/8 + 6 faces × 1/2 = 4 atoms).

Understanding the FCC structure is vital because it affects the properties of the metal, including its density. It provides a framework for visualizing how atoms are densely packed in a given volume which directly leads us to compute the density of the metal.
Avogadro's number
Avogadro's number is a fundamental constant in chemistry, which is the number of constituent particles, usually atoms or molecules, that are contained in the amount of substance given by one mole.

The value of Avogadro's number approximately equals to 6.022 × 1023 particles per mole. This constant allows us to convert between the amount of substance in moles and the number of atoms or molecules it represents. For example, if we have one mole of iridium atoms, we know that we have roughly 6.022 × 1023 iridium atoms.

Appreciating the enormity of Avogadro's number is important in calculations involving the molar mass and the density of substances. It essentially bridges the microscopic world of atoms and the macroscopic world that we can measure and observe.
Molar mass
Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It is a physical property that is defined as the mass of a given substance divided by the amount of that substance, measured in moles.

To determine the molar mass of iridium or any element, you would look at the periodic table, where the molar mass of iridium is listed as 192.22 g/mol. This indicates that one mole of iridium atoms, which is 6.022 × 1023 atoms, weighs 192.22 grams.

Understanding molar mass is critical when calculating the density of a substance, as evident from our Iridium example. We use the molar mass to find out the mass of a single atom, which then helps us to calculate the total mass of atoms within a unit cell. By knowing the mass and volume of a unit cell, the density can be computed with precision.

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Most popular questions from this chapter

\(\mathrm{MnO}\) has either the \(\mathrm{NaCl}\) type structure or the \(\mathrm{CsCl}\) type structure (see Exercise 69). The edge length of the \(\mathrm{Mn} \mathrm{O}\) unit cell is \(4.47 \times 10^{-8} \mathrm{~cm}\) and the density of \(\mathrm{MnO}\) is \(5.28 \mathrm{~g} / \mathrm{cm}^{3}\) a. Does \(\mathrm{MnO}\) crystallize in the \(\mathrm{NaCl}\) or the \(\mathrm{CsCl}\) type structure? b. Assuming that the ionic radius of oxygen is \(140 . \mathrm{pm}\), estimate the ionic radius of manganese.

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