The unit cell of copper corresponds to a face centered cube of edge length 3.596 A with one copper a

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

In an FCC lattice, there are atoms at the 8 corners and the 6 face centers. Each corner atom is shared by 8 adjacent unit cells, contributing $8 \times \frac{1}{8} = 1$ atom. Each face-centered atom is shared by 2 adjacent unit cells, contributing $6 \times \frac{1}{2} = 3$ atoms. Total $Z = 1 + 3 = 4$.

Q: Why do we multiply the molar mass by $10^{-3}$?

The molar mass is given in grams per mole (g/mol). Since the question asks for density in $\text{kg/m}^3$, which is the standard SI unit, we must convert grams to kilograms by multiplying by $10^{-3}$.

Q: How do I cube $3.596$ quickly without a calculator?

You can approximate $3.596$ as $3.6$. First, square it: $3.6 \times 3.6 \approx 13$. Then multiply by $3.6$ again: $13 \times 3.6 \approx 46.8$. This is very close to the exact value of $46.499$ and is sufficient for finding the nearest integer.

Q: What happens to the powers of 10 in the denominator?

When you cube the edge length, $(10^{-10})^3$ becomes $10^{-30}$. Then, you multiply this by Avogadro's number, which has a power of $10^{23}$. Using exponent rules, $10^{-30} \times 10^{23} = 10^{-7}$.

Q: Can I calculate the density in $\text{g/cm}^3$ first and then convert it?

Yes, absolutely! If you keep $a$ in cm ($3.596 \times 10^{-8} \text{ cm}$) and $M$ in grams, you will get the density in $\text{g/cm}^3$. To convert $\text{g/cm}^3$ to $\text{kg/m}^3$, you simply multiply the result by $1000$.

Question

The unit cell of copper corresponds to a face centered cube of edge length $3.596 	ext{ \AA}$ with one copper atom at each lattice point. The calculated density of copper in $	ext{kg/m}^3$ is ......... . [Molar mass of Cu $= 63.54 	ext{ g}$; Avogadro number $= 6.022 	imes 10^{23}$]

Sigma Percentile · Accepted Answer

### Visualizing the Copper Unit Cell

Imagine you are shrinking down to the atomic level and looking at a tiny, perfect crystal of solid copper. What you would see is a highly organized, repeating pattern of atoms. The smallest repeating unit of this pattern is called a **unit cell**.

For copper, this unit cell has a **Face-Centered Cubic (FCC)** structure. This means if you draw a cube, there is one copper atom at each of the eight corners, and one copper atom right in the middle of each of the six faces.

Because these atoms are shared with neighboring unit cells, we have to calculate the *effective* number of atoms inside just one cube. The corner atoms are shared by 8 cubes, and the face atoms are shared by 2 cubes.

$$Z = \left(8 	imes \frac{1}{8}ight) + \left(6 	imes \frac{1}{2}ight) = 1 + 3 = 4$$

So, there are effectively **4 copper atoms** in every unit cell.

### The Master Equation for Density

To find the density of this crystal, we rely on a fundamental principle: density is simply mass divided by volume.

For a unit cell, the mass is the mass of the $Z$ atoms inside it, and the volume is the volume of the cube, which is $a^3$ (where $a$ is the edge length). The mass of one single atom is its molar mass $M$ divided by Avogadro's number $N_A$.

Combining these gives us our master equation:

$$d = \frac{Z 	imes M}{a^3 	imes N_A}$$

### Navigating the Unit Trap

Here is where many students fall into a trap. The problem gives us the edge length in Angstroms ($	ext{\AA}$) and the molar mass in grams per mole. However, the final answer must be in standard SI units: **$	ext{kg/m}^3$**.

We must carefully convert our given values before substituting them into the equation.

First, let's convert the molar mass from grams to kilograms:

$$M = 63.54 	ext{ g/mol} = 63.54 	imes 10^{-3} 	ext{ kg/mol}$$

Next, let's convert the edge length from Angstroms to meters:

$$a = 3.596 	ext{ \AA} = 3.596 	imes 10^{-10} 	ext{ m}$$

### The Final Calculation

Now, we are ready to substitute our pristine, SI-compliant values into the density formula.

$$d = \frac{4 	imes (63.54 	imes 10^{-3})}{(3.596 	imes 10^{-10})^3 	imes (6.022 	imes 10^{23})}$$

Let's break down the denominator. Cubing the edge length gives us the volume of the unit cell:

$$a^3 \approx 46.499 	imes 10^{-30} 	ext{ m}^3$$

Multiplying this volume by Avogadro's number:

$$a^3 	imes N_A \approx 28.00 	imes 10^{-6}$$

Now, we divide the total mass of the 4 atoms by this value:

$$d = \frac{254.16 	imes 10^{-3}}{28.00 	imes 10^{-6}} \approx 9077 	ext{ kg/m}^3$$

And there we have it! The calculated density of solid copper is **$9077 	ext{ kg/m}^3$**. This elegant calculation shows how macroscopic properties like density emerge directly from the microscopic arrangement of atoms.

Enter Numerical Value:

Visualized Solution

$Z = 4 for FCC$

$Density Formula$

$Unit Conversion$

$Substitution$

$Denominator Calculation$

$Final Density$

$Alternative Applications$

The Sigma Insight: Solid State

Visualizing the Copper Unit Cell

The Master Equation for Density

Navigating the Unit Trap

The Final Calculation

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Enter Numerical Value:

Visualized Solution

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

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

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

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

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

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

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