At 100°C, copper (Cu) has FCC unit cell structure with cell edge length of x Å. What is the approxim

Q: Why is $Z = 4$ for an FCC unit cell?

In an FCC unit cell, there are 8 corner atoms (each shared by 8 adjacent cells, contributing $8 \times \frac{1}{8} = 1$ atom) and 6 face-centered atoms (each shared by 2 adjacent cells, contributing $6 \times \frac{1}{2} = 3$ atoms). Total $Z = 1 + 3 = 4$.

Q: Why do we convert Angstroms to centimeters instead of meters?

The standard unit for density in chemistry is $\text{g cm}^{-3}$. Since the molar mass is in grams, converting the edge length to centimeters directly gives the volume in $\text{cm}^3$, making the final density unit $\text{g cm}^{-3}$.

Q: What is the value of $1 \text{ \AA}$ in centimeters?

$1 \text{ \AA}$ (Angstrom) is equal to $10^{-10}$ meters. Since $1 \text{ m} = 100 \text{ cm}$, $1 \text{ \AA} = 10^{-10} \times 10^2 \text{ cm} = 10^{-8} \text{ cm}$.

Q: Can I use $N_A = 6 \times 10^{23}$ for faster calculation?

While $6 \times 10^{23}$ is a good approximation, using $6.022 \times 10^{23}$ or $6.023 \times 10^{23}$ (as given in some contexts) ensures you don't fall into trap options that are very close to each other. Here, the options are spaced out, so approximation works, but it's safer to use the exact value.

Q: What if the temperature was different, say $200^\circ\text{C}$?

The temperature is given to specify the state of the crystal. While thermal expansion could slightly change the edge length $x$, the formula and the method remain exactly the same. The value of $x$ provided at that specific temperature is all you need.

Question

At $100^\circ	ext{C}$, copper (Cu) has FCC unit cell structure with cell edge length of $x 	ext{ \AA}$. What is the approximate density of Cu (in $	ext{g cm}^{-3}$) at this temperature?
[Atomic mass of Cu $= 63.55 	ext{ u}$]

Options:

(A) $\frac{211}{x^3}$

(B) $\frac{205}{x^3}$

(C) $\frac{105}{x^3}$

(D) $\frac{422}{x^3}$

Sigma Percentile · Accepted Answer

The Density of a Copper Crystal: Unpacking the FCC Unit Cell

### Analyzing the Setup

Imagine a tiny, perfect cube of copper. This is its unit cell, and since it crystallizes in a Face-Centered Cubic (FCC) structure, it contains atoms at all eight corners and right in the middle of all six faces.

Because corner atoms are shared by eight adjacent cells and face atoms are shared by two, the total number of effective atoms in one FCC unit cell is **$Z = 4$**. This is the foundational geometric fact we need to unlock the crystal's density.

### The Master Equation

To find the density of this crystal, we rely on a very famous and elegant formula. Density is simply the mass of the unit cell divided by its volume.

For a crystal lattice, we express this as:
$$d = \frac{Z \cdot M}{N_A \cdot a^3}$$
Here, $Z$ is the number of atoms, $M$ is the molar mass, $N_A$ is Avogadro's number, and $a^3$ is the volume of the cubic unit cell.

### The Unit Conversion Trap

There is a catch here, and it is a classic trap where many students make a silly mistake. The edge length $a$ is given as $x 	ext{ \AA}$ (Angstroms), but we need our final density in $	ext{g cm}^{-3}$.

We must convert Angstroms to centimeters before doing anything else. Since $1 	ext{ \AA} = 10^{-8} 	ext{ cm}$, our edge length becomes:
$$a = x 	imes 10^{-8} 	ext{ cm}$$

### Final Calculation

Now, we substitute all our known values into the master equation. We plug in $4$ for $Z$, $63.55$ for $M$, $6.023 	imes 10^{23}$ for $N_A$, and $(x 	imes 10^{-8})^3$ for the volume.

$$d = \frac{4 	imes 63.55}{6.023 	imes 10^{23} 	imes (x 	imes 10^{-8})^3}$$

Let's simplify the denominator first. Cubing $10^{-8}$ gives $10^{-24}$. Multiplying this by Avogadro's $10^{23}$ leaves us with a neat $10^{-1}$.

$$d = \frac{254.2}{6.023 	imes 10^{-1} 	imes x^3}$$

In the numerator, $4 	imes 63.55$ gives $254.2$. Dividing this by $0.6023$ yields approximately **$422$**.

$$d \approx \frac{422}{x^3} 	ext{ g cm}^{-3}$$

And there we have it! The density is elegantly expressed in terms of the unknown edge length $x$.

Animated Solution for Chemistry - States of Matter: At $10 0^{\circ} C$ , copper (Cu) has FCC unit cell structure with cell edge length of $x \overset{˚}{A}$ . What is the approximate density of Cu (in $g cm^{- 3}$ ) at this temperature? [Atomic mass of Cu $= 63.55 u$ ]

Select Answer:

Visualized Solution

$Z = 4$

$d = \frac{Z \cdot M}{N _{A} \cdot a ^{3}}$

$a = x \times 1 0^{- 8} cm$

$d = \frac{4 \times 63.55}{6.023 \times 1 0 ^{23} \times ( x \times 1 0 ^{- 8} ) ^{3}}$

$1 0^{23} \times 1 0^{- 24} = 1 0^{- 1}$

$d \approx \frac{422}{x ^{3}}$

$Z_{B C C} = 2$

The Sigma Insight: Solid State

Analyzing the Setup

The Master Equation

The Unit Conversion Trap

Final Calculation

Similar Questions

The unit cell of copper corresponds to a face centered cube of edge length $3.596 \overset{˚}{A}$ with one copper atom at each lattice point. The calculated density of copper in $kg/m^{3}$ is ......... . [Molar mass of Cu $= 63.54 g$ ; Avogadro number $= 6.022 \times 1 0^{23}$ ]

Copper crystallises in fcc with a unit cell length of $361 pm$ . What is the radius of copper atom?

Copper crystallises in fcc lattice with a unit cell edge of 361 pm. The radius of copper atom is

The density (in g cm $^{- 3}$ ) of the metal which forms a cubic close packed (ccp) lattice with an axial distance (edge length) equal to 400 pm is ______. Use: Atomic mass of metal = 105.6 amu and Avogadro's constant = $6 \times 1 0^{23} mol^{- 1}$

Total volume of atoms present in a face-centred cubic unit cell of a metal is ( $r$ is atomic radius)

An element with molar mass $2.7 \times 1 0^{- 2} kg mol^{- 1}$ forms a cubic unit cell with edge length $405 pm$ . If its density is $2.7 \times 1 0^{3} kg m^{- 3}$ , the radius of the element is approximately ............ $\times 1 0^{- 12} m$ (to the nearest integer).

A metal crystallises in a face centred cubic structure. If the edge length of its unit cell is ' $a$ ', the closest approach between two atoms in metallic crystal will be

A crystalline solid of a pure substance has a face-centred cubic structure with a cell edge of $400 pm$ . If the density of the substance in the crystal is $8 g cm^{- 3}$ , then the number of atoms present in $256 g$ of the crystal is $N \times 1 0^{24}$ . The value of N is :

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Z=4

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d=NA​⋅a3Z⋅M​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a=x×10−8 cm

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d=6.023×1023×(x×10−8)34×63.55​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 1023×10−24=10−1

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d≈x3422​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ZBCC​=2