The density (in g cm^-3) of the metal which forms a cubic close packed (ccp) lattice with an axial d

Q: Why do we take Z=4 for a ccp lattice?

A cubic close-packed (ccp) lattice has the exact same unit cell geometry as a face-centered cubic (fcc) lattice. In an fcc unit cell, there are 8 corner atoms (each contributing 1/8) and 6 face-centered atoms (each contributing 1/2). Adding these up gives $8 \times (1/8) + 6 \times (1/2) = 4$ atoms per unit cell.

Q: Why did we convert picometers to $10^{-10}$ cm instead of $10^{-12}$ m?

The question specifically asks for the density in $\text{g/cm}^3$. If you convert to meters ($10^{-12}$ m), your final density will be in $\text{g/m}^3$ or $\text{kg/m}^3$, which would require another conversion step at the end. Converting directly to centimeters saves time and reduces errors.

Q: How do you quickly calculate the cube of $400 \times 10^{-10}$ without a calculator?

The trick is to simplify the scientific notation before cubing. Write $400 \times 10^{-10}$ as $4 \times 10^{-8}$. Then, cube the number and the power of ten separately: $4^3 = 64$ and $(10^{-8})^3 = 10^{-24}$. This makes the math much more manageable!

Q: What is the difference between ccp and hcp if they both have the same packing efficiency?

While both ccp and hcp (hexagonal close-packed) have a packing efficiency of 74% and a coordination number of 12, their stacking sequences differ. ccp follows an ABCABC pattern, resulting in a face-centered cubic unit cell. hcp follows an ABAB pattern, resulting in a hexagonal unit cell with $Z=6$.

Q: Why do we divide by Avogadro's number in the density formula?

The molar mass $M$ is the mass of one *mole* of atoms (Avogadro's number of atoms). To find the density of a single unit cell, we need the mass of just the $Z$ atoms inside it. Dividing the molar mass by Avogadro's number gives the mass of a single atom, which we then multiply by $Z$.

Question

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 = $6 	imes 10^{23} 	ext{ mol}^{-1}$

Sigma Percentile · Accepted Answer

The solid state is a fascinating realm where atoms arrange themselves in highly ordered, repeating patterns. In this problem, we are tasked with finding the macroscopic density of a metal by analyzing its microscopic unit cell. This is a classic application of X-ray crystallography principles, bridging the gap between the atomic world and the properties we can measure in a lab.

### Visualizing the Cubic Close-Packed Lattice

Imagine you are shrinking down to the atomic level and looking at a crystal of this metal. It forms a **cubic close-packed (ccp)** lattice. Geometrically, a ccp lattice is identical to a **face-centered cubic (fcc)** lattice.

This means we have atoms at all eight corners of the cube, and atoms right in the center of all six faces.

When we calculate the effective number of atoms per unit cell, the corner atoms contribute $1/8$ each, and the face-centered atoms contribute $1/2$ each.

$$Z = 8 	imes \frac{1}{8} + 6 	imes \frac{1}{2} = 4$$

So, we have exactly $4$ atoms per unit cell.

### The Master Equation for Density

To find the density of this microscopic cube, we use the master equation for solid state density. Density, denoted by $ho$, is simply the mass of the unit cell divided by its volume.

The mass is the number of atoms, $Z$, multiplied by the molar mass, $M$, and then divided by Avogadro's number, $N_A$, to get the mass of a single atom. The volume of a cube is just its edge length, $a$, cubed.

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

This formula is your best friend for these types of problems.

### Navigating the Unit Conversions

Let's substitute our known values into the formula. We know $Z = 4$ and the molar mass $M = 105.6 	ext{ g/mol}$. The edge length is given as $a = 400 	ext{ pm}$.

Here is where many students make a critical error. Density is asked in $	ext{g/cm}^3$. Therefore, we must convert picometers to centimeters. Since $1 	ext{ pm} = 10^{-10} 	ext{ cm}$, we have:

$$a = 400 	imes 10^{-10} 	ext{ cm} = 4 	imes 10^{-8} 	ext{ cm}$$

Now, we can set up our equation:

$$ho = \frac{4 	imes 105.6}{(4 	imes 10^{-8})^3 	imes 6 	imes 10^{23}}$$

### The Final Calculation

Let's tackle the denominator first. Cubing the edge length gives:

$$a^3 = (4 	imes 10^{-8})^3 = 64 	imes 10^{-24} 	ext{ cm}^3$$

Now, multiply this volume by Avogadro's number:

$$a^3 	imes N_A = 64 	imes 10^{-24} 	imes 6 	imes 10^{23} = 384 	imes 10^{-1} = 38.4$$

Next, we calculate the numerator, which is the total mass of the atoms in the unit cell (in amu):

$$Z 	imes M = 4 	imes 105.6 = 422.4$$

Finally, we divide the numerator by the denominator to find the density:

$$ho = \frac{422.4}{38.4} = 11.00 	ext{ g/cm}^3$$

We arrive at a perfectly clean integer answer! Always remember to double-check your unit conversions, as that is the most common trap in solid state calculations.

Enter Numerical Value:

Visualized Solution

$Z$ for ccp lattice

Density Formula

Substitution

Denominator Calculation

Final Density

Atomic Radius Relation

The Sigma Insight: Solid State

Visualizing the Cubic Close-Packed Lattice

The Master Equation for Density

Navigating the Unit Conversions

The Final Calculation

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