Calculating Aluminum Atoms And Unit Cells In FCC Structure

by Scholario Team 59 views

Hey guys! Let's dive into a cool chemistry problem today that involves calculating the number of aluminum atoms and unit cells in a given mass of aluminum. This problem touches on some fundamental concepts in solid-state chemistry, like crystal structures, unit cells, and Avogadro's number. So, grab your thinking caps, and let's get started!

Understanding the Problem

Our mission, should we choose to accept it, is to figure out how many aluminum (Al) atoms and unit cells are present in 10 grams of aluminum. We know that aluminum has an atomic mass of 27 g/mol and that it crystallizes in a face-centered cubic (FCC) structure. This information is key to cracking the problem. Before we jump into the calculations, let's make sure we're on the same page with some of the core concepts.

Face-Centered Cubic (FCC) Structure: The Basics

So, what exactly is a face-centered cubic structure? In an FCC structure, atoms are arranged in a cube, with one atom at each corner and one atom at the center of each face. Think of it like this: imagine a dice, and then put an extra marble smack-dab in the middle of each of its faces. That's kind of what an FCC structure looks like on the atomic level.

Now, here's the cool part: not all of these atoms fully belong to a single unit cell. The corner atoms are each shared by eight unit cells, so only 1/8 of each corner atom counts towards a single unit cell. The face-centered atoms are shared by two unit cells, so only 1/2 of each face-centered atom belongs to one unit cell. Got it?

Let's break it down mathematically:

  • 8 corner atoms x 1/8 atom per corner = 1 atom
  • 6 face-centered atoms x 1/2 atom per face = 3 atoms

Adding those up, we get a total of 1 + 3 = 4 atoms per unit cell in an FCC structure. This is a crucial piece of information for our calculations.

Molar Mass and Avogadro's Number: The Dynamic Duo

Next up, let's talk about molar mass and Avogadro's number. The molar mass of a substance is the mass of one mole of that substance, usually expressed in grams per mole (g/mol). In our case, the molar mass of aluminum is 27 g/mol. This means that 27 grams of aluminum contain one mole of aluminum atoms.

Avogadro's number (symbolized as Nₐ) is the number of atoms, molecules, or ions in one mole of a substance. It's a gigantic number: approximately 6.022 x 10²³. So, one mole of aluminum contains 6.022 x 10²³ aluminum atoms. Avogadro's number is the bridge that connects the macroscopic world (grams) to the microscopic world (atoms).

Putting It All Together: The Strategy

Okay, we've got our background knowledge sorted. Now, let's map out our strategy for solving the problem. We need to find:

  1. The number of aluminum atoms in 10 grams of aluminum.
  2. The number of unit cells present in 10 grams of aluminum.

Here's the plan:

  1. Calculate the number of moles of aluminum in 10 grams: We'll use the molar mass of aluminum to convert grams to moles.
  2. Calculate the number of aluminum atoms: We'll use Avogadro's number to convert moles to atoms.
  3. Calculate the number of unit cells: We'll use the fact that there are 4 aluminum atoms per FCC unit cell to convert the number of atoms to the number of unit cells.

Sounds like a plan? Let's get calculating!

Step-by-Step Calculations

Alright, buckle up, math time! Don't worry, we'll take it slow and steady.

Step 1: Moles of Aluminum

First, we need to find out how many moles of aluminum are in our 10-gram sample. We'll use the following formula:

  • Moles = Mass / Molar Mass

We know the mass (10 grams) and the molar mass (27 g/mol), so let's plug those values in:

  • Moles of Al = 10 g / 27 g/mol ≈ 0.37 moles

So, we have approximately 0.37 moles of aluminum.

Step 2: Number of Aluminum Atoms

Next, we'll convert moles of aluminum to the number of aluminum atoms using Avogadro's number:

  • Number of Atoms = Moles x Avogadro's Number

Plugging in the values:

  • Number of Al atoms = 0.37 moles x 6.022 x 10²³ atoms/mole ≈ 2.23 x 10²³ atoms

Boom! We've found that there are approximately 2.23 x 10²³ aluminum atoms in 10 grams of aluminum.

Step 3: Number of Unit Cells

Now for the final step: calculating the number of unit cells. Remember that we figured out earlier that there are 4 aluminum atoms per FCC unit cell. We'll use this ratio to convert the number of atoms to the number of unit cells:

  • Number of Unit Cells = Number of Atoms / Atoms per Unit Cell

Plugging in the values:

  • Number of Unit Cells = (2.23 x 10²³ atoms) / (4 atoms/unit cell) ≈ 5.58 x 10²² unit cells

And there you have it! We've calculated that there are approximately 5.58 x 10²² unit cells in 10 grams of aluminum.

The Grand Finale: Putting It All Together

Okay, let's recap what we've found. In 10 grams of aluminum, which crystallizes in a face-centered cubic structure, we have:

  • Approximately 2.23 x 10²³ aluminum atoms
  • Approximately 5.58 x 10²² unit cells

Not too shabby, eh? We took a problem that seemed a bit daunting at first and broke it down into manageable steps. We used our understanding of FCC structures, molar mass, Avogadro's number, and some good ol' math to arrive at the solution.

Why This Matters: Real-World Applications

Now, you might be thinking, "Okay, that's a cool math problem, but why should I care?" Well, understanding crystal structures and how to calculate the number of atoms and unit cells has a ton of real-world applications. It's crucial in materials science for designing new materials with specific properties. For example, the properties of a metal, like its strength, ductility, and conductivity, are heavily influenced by its crystal structure.

Knowing how atoms are arranged in a material allows scientists and engineers to tailor its properties for specific applications. This knowledge is used in everything from designing stronger alloys for aircraft to developing new semiconductors for electronics. Pretty cool, huh?

Key Takeaways and Further Exploration

Before we wrap up, let's highlight the key takeaways from this problem:

  • FCC structures have 4 atoms per unit cell.
  • Molar mass and Avogadro's number are essential for converting between mass, moles, and the number of atoms.
  • Breaking down complex problems into smaller steps makes them much easier to solve.

If you're curious to learn more about solid-state chemistry, I highly recommend exploring different types of crystal structures (like simple cubic and body-centered cubic), as well as concepts like packing efficiency and crystal defects. There's a whole fascinating world of materials science out there just waiting to be discovered!

Wrapping Up

So, there you have it, folks! We've successfully calculated the number of aluminum atoms and unit cells in a 10-gram sample. I hope this walkthrough has been helpful and has given you a better understanding of how chemistry concepts can be applied to solve real-world problems. Keep exploring, keep questioning, and most importantly, keep having fun with science! Peace out! ✌️