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Chapter 13: Problem 44

Iron has a body-centered cubic unit cell with a cell dimension of \(286.65 \mathrm{pm} .\) The density of iron is \(7.874 \mathrm{g} / \mathrm{cm}^{3} .\) Use this information to calculate Avogadro's number.

Short Answer

Expert verified
Avogadro's number is approximately \(6.022 \times 10^{23}\) atoms/mol.

Step by step solution

01

Understand the Body-Centered Cubic (BCC) Structure

The body-centered cubic (BCC) unit cell of iron contains atoms at each corner and one atom in the center. Every corner atom is shared among 8 adjacent unit cells. Thus, each BCC unit cell effectively contains 2 atoms in total.
02

Convert Cell Dimension to Centimeters

The cell dimension is given as 286.65 pm (picometers). To convert this into centimeters:\[ 286.65 \text{ pm} = 286.65 \times 10^{-12} \text{ m} = 286.65 \times 10^{-10} \text{ cm} = 2.8665 \times 10^{-8} \text{ cm} \]
03

Calculate Volume of the Unit Cell

The unit cell is cubic, so its volume can be calculated as:\[ V_{ ext{cell}} = a^3 = (2.8665 \times 10^{-8} \text{ cm})^3 = 2.3503 \times 10^{-23} \text{ cm}^3 \]
04

Determine Mass of a Unit Cell

Use the density formula, where density \( \rho = \frac{\text{mass}}{\text{volume}} \), to find the mass of one unit cell:\[ \text{mass of unit cell} = \rho \times V_{\text{cell}} = 7.874 \text{ g/cm}^3 \times 2.3503 \times 10^{-23} \text{ cm}^3 = 1.8516 \times 10^{-22} \text{ g} \]Given that the BCC unit cell contains 2 atoms of iron, the mass of 2 atoms is \(1.8516 \times 10^{-22}\) g.
05

Calculate Avogadro's Number

To find Avogadro's number, determine the mass of two iron atoms and relate it to the molar mass of iron (approximately 55.845 g/mol):For 1 mole of iron, there is a corresponding Avogadro's number of atoms, with a total mass of 55.845 g. Therefore:\[ \text{Avogadro's Number} = \text{molar mass} \times \frac{\text{number of atoms per unit cell}}{\text{mass per unit cell}} \]\[ N_A = \frac{55.845 \text{ g/mol} \times 2}{1.8516 \times 10^{-22} \text{ g}} = 6.022 \times 10^{23} \text{ atoms/mol} \]
06

Answer Verification

The calculated Avogadro's number \(6.022 \times 10^{23}\) per mole is a fundamental physical constant and aligns well with the accepted value. This validates our solution given the provided data.

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Key Concepts

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

Body-Centered Cubic Structure
In a body-centered cubic (BCC) structure, each cube has atoms at its corners and one in the center. Imagine a dice where each face is shared with neighboring dice. Similarly, each corner atom in the BCC unit cell is shared among eight adjacent unit cells, contributing only a fraction to a single unit cell. Thus, a BCC unit cell contains effectively 2 atoms:
  • 8 corner atoms contributing 1/8 each.
  • 1 full atom in the center.
This arrangement is efficient and affects iron鈥檚 properties like density, which is explored further in calculations.
Density Calculation
Density is a measure of mass per unit volume. In the case of iron's BCC structure, it connects the mass of atoms in the unit cell to its volume. Given:
  • Density (\( \rho \)) of iron: 7.874 g/cm鲁
  • Volume of unit cell: previously calculated as (\(2.3503 \times 10^{-23} \text{ cm}^3 \))
To find the mass of the unit cell, rearrange the density formula: \[ \text{mass} = \text{density} \times \text{volume} \]Substitute the known values to find how much mass is packed in this tiny cube. The resulting mass will help us relate it to atoms and eventually lead us to Avogadro's number.
Molar Mass
Molar mass is the mass of one mole of a substance, linking mass to the amount of substance through units of grams per mole. For iron, the molar mass is approximately 55.845 g/mol.
  • It tells us how much mass one mole (or 6.022 x 10虏鲁 atoms) of iron weighs.
  • Using the calculated mass of two atoms in a BCC unit cell, we can relate those two atoms to the larger concept of a mole.
This relationship is central to finding Avogadro鈥檚 number, as it bridges the microscale mass of a few atoms to macroscopic units we can measure and understand.
Unit Cell Volume
The volume of a unit cell is key to understanding how atoms are packed. Since the BCC structure is a cube, compute the volume by cubing the edge length.
  • Given: Edge length is (\(286.65 \text{ pm} \)). Convert to cm for consistency in measurements.
  • Use the conversion: (\(1 \text{ pm} = 10^{-12} \text{ m} = 10^{-10} \text{ cm} \)).
  • Calculate volume: (\(V_{\text{cell}} = a^3 = (2.8665 \times 10^{-8} \text{ cm})^3\)).
This volume is crucial for calculating density and ultimately allows us to determine the mass that corresponds to a certain number of atoms, leading us toward calculating Avogadro鈥檚 number.

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Most popular questions from this chapter

The specific heat capacity of silver is \(0.235 \mathrm{J} / \mathrm{g} \cdot \mathrm{K}\) Its melting point is \(962^{\circ} \mathrm{C},\) and its enthalpy of fusion is \(11.3 \mathrm{kJ} / \mathrm{mol} .\) What quantity of energy, in joules, is required to change \(5.00 \mathrm{g}\) of silver from a solid at \(25^{\circ} \mathrm{C}\) to a liquid at \(962^{\circ} \mathrm{C} ?\)

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Define the terms intrinsic semiconductor and extrinsic semiconductor. Give an example of each.
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