Problem 21
Question
An fec cubic cell contains eight \(\mathrm{X}\) atoms at the cell and six \(\mathrm{Y}\) atoms at the faces. What is the cmpirical formula of the solid? (a) \(\mathrm{X}_{3} \mathrm{Y}_{4}\) (b) \(X_{3} Y\) (c) \(\mathrm{XY}_{3}\) (d) \(\mathrm{X}_{4} \mathrm{Y}_{3}\)
Step-by-Step Solution
Verified Answer
(c) \( XY_3 \)
1Step 1: Determine the Contribution of Atom X
An fcc cubic cell contains eight corner atoms of type \( X \). In a face-centered cubic lattice, each corner atom is shared among eight different unit cells. Hence, the contribution of each corner atom to a single unit cell is \( \frac{1}{8} \). Therefore, the total contribution from \( X \) atoms is \( 8 \times \frac{1}{8} = 1 \).
2Step 2: Determine the Contribution of Atom Y
The fcc unit cell has six face-centered \( Y \) atoms. Each face-centered atom is shared between two unit cells. Thus, the contribution of each face-centered atom to one unit cell is \( \frac{1}{2} \). So, the total number of \( Y \) atoms in a single unit cell is \( 6 \times \frac{1}{2} = 3 \).
3Step 3: Determine the Empirical Formula
Now, we know there is 1 atom of \( X \) and 3 atoms of \( Y \) in each unit cell. The empirical formula is derived from the ratio of \( X \) to \( Y \) atoms. Therefore, the empirical formula of the compound is \( XY_3 \).
Key Concepts
Face-Centered CubicUnit CellAtom ContributionEmpirical Ratio
Face-Centered Cubic
In chemistry and crystallography, the term "Face-Centered Cubic" (often abbreviated as FCC) describes one of the most efficient ways that atoms or molecules can pack together in a solid. An FCC unit cell is a cubic structure where each corner and the center of each face of the cube is occupied by an atom.
Each corner atom in this structure is shared by eight adjacent unit cells, while each atom at the center of a face is shared between two neighboring cells. Because of this shared nature, the contribution of atoms in a face-centered cubic unit cell differs based on their position—corner atoms contribute less than face-centered atoms.
This arrangement allows for maximum space utilization, contributing to the material's overall strength and density. The FCC structure is common in many metals and alloys, making it a fundamental concept in material science.
Each corner atom in this structure is shared by eight adjacent unit cells, while each atom at the center of a face is shared between two neighboring cells. Because of this shared nature, the contribution of atoms in a face-centered cubic unit cell differs based on their position—corner atoms contribute less than face-centered atoms.
This arrangement allows for maximum space utilization, contributing to the material's overall strength and density. The FCC structure is common in many metals and alloys, making it a fundamental concept in material science.
Unit Cell
A unit cell is the smallest repeating unit in a crystal lattice that, when repeated in three dimensions, forms the entire lattice. Think of it as a building block of the crystal structure, crystalline solids, or metals. Its dimensions and the arrangement of atoms within it define the overall properties of the crystal.
In the case of the FCC structure, the unit cell is cubic in shape. It is essential for determining how atoms are packed in a solid and plays a critical role in calculating properties like density, atomic packing factor, and coordination number.
Understanding the unit cell's structure and dimensions helps in predicting material behavior, making it a crucial aspect of material science and chemistry.
In the case of the FCC structure, the unit cell is cubic in shape. It is essential for determining how atoms are packed in a solid and plays a critical role in calculating properties like density, atomic packing factor, and coordination number.
Understanding the unit cell's structure and dimensions helps in predicting material behavior, making it a crucial aspect of material science and chemistry.
Atom Contribution
When analyzing a unit cell like the FCC, each atom's contribution to the unit cell is based on its position—whether it's a corner atom or a face-centered atom affects its calculation.
- Corner Atoms: There are eight atoms located at the corners of the FCC unit cell. Each of these atoms is shared by eight unit cells, so each contributes only \( \frac{1}{8} \) of an atom to the cell.
- Face-Centered Atoms: Six atoms occupy the centers of the faces. Since each of these atoms is shared between two adjacent cells, they contribute \( \frac{1}{2} \) of an atom to the unit cell.
Empirical Ratio
An empirical ratio defines the simplest whole-number ratio of the different types of atoms in a compound. In the exercise, we are given an FCC unit cell with atoms of type \( X \) and \( Y \), and asked to find the empirical formula.
From the step-by-step solution, we learned:
Empirical ratios are essential for understanding the chemical composition, enabling scientists to determine how atoms combine to form complex substances. They simplify complex crystalline or molecular information into understandable, manipulable forms, crucial for chemical analysis and synthesis.
From the step-by-step solution, we learned:
- The total contribution from \( X \) atoms (corners): 8 \( \times \frac{1}{8} = 1 \).
- The total contribution from \( Y \) atoms (faces): 6 \( \times \frac{1}{2} = 3 \).
Empirical ratios are essential for understanding the chemical composition, enabling scientists to determine how atoms combine to form complex substances. They simplify complex crystalline or molecular information into understandable, manipulable forms, crucial for chemical analysis and synthesis.
Other exercises in this chapter
Problem 20
How many formula units of \(\mathrm{CsCl}\) are present in the unit cell ? (a) 1 (b) 2 (c) 4 (d) 6
View solution Problem 20
The rms speed of \(\mathrm{N}_{2}\) molecules in a gas is \(u\). If the temperature is doubled and the nitrogen molecules dissociate into nitrogen atoms, the rm
View solution Problem 21
Oxygen (one \(\mathrm{dm}^{3}\) at 1 atm) takes 2 min to effuse through an orifice. How much time (in minutes) will be taken by \(\mathrm{N}_{2}, \mathrm{He}\)
View solution Problem 22
A solid has a structure in which \(\mathrm{W}\) atoms are located at the corners of the cubic lattice, \(\mathrm{O}\) atoms at the center of the edges and \(\ma
View solution