Problem 117 - Chemistry and Chemical Reactivity

Question

The common chemical compound alum has the formula \(\mathrm{KAl}\left(\mathrm{SO}_{4}\right)_{2} \cdot 12 \mathrm{H}_{2} \mathrm{O} .\) An interesting characteristic of alum is that it is possible to grow very large crystals of this compound. Suppose you have a crystal of alum in the form of a cube that is \(3.00 \mathrm{cm}\) on each side. You want to know how many aluminum atoms are contained in this cube. Outline the steps to determine this value, and indicate the information that you need to carry out each step.

Step-by-Step Solution

Answer

There are approximately \(6.00 \times 10^{22}\) aluminum atoms in the alum cube.

Step 1: Calculate the Volume of the Cube

The crystal is in the form of a cube with each side measuring \(3.00 \ \mathrm{cm}\). To find the volume \(V\), use the formula for the volume of a cube: \[ V = s^3 \]where \(s\) is the side length of the cube. Here:\[ V = (3.00 \ \mathrm{cm})^3 = 27.0 \ \mathrm{cm}^3 \]

Step 2: Determine the Number of Moles of Alum

Find the density of the alum crystal in \(\mathrm{g/cm}^3\) (a common density for alum is approximately \(1.75 \ \mathrm{g/cm}^3\)). Use this to find the mass \(m\) of the alum crystal. \[ m = \text{density} \times V = 1.75\ \mathrm{g/cm}^3 \times 27.0 \ \mathrm{cm}^3 = 47.25 \ \mathrm{g} \]Next, calculate the number of moles \(n\) of alum using its molar mass. The molar mass of alum \(\mathrm{KAl(SO}_4\mathrm{)_2 \cdot 12H}_2\mathrm{O}\) is approximately \(474.39 \ \mathrm{g/mol}\): \[ n = \frac{m}{\text{molar mass}} = \frac{47.25 \ \mathrm{g}}{474.39 \ \mathrm{g/mol}} \approx 0.0996 \ \text{mol} \]

Step 3: Calculate the Number of Aluminum Atoms

Each formula unit of alum contains one Aluminum atom. Use Avogadro's number (\(6.022 \times 10^{23}\) atoms/mol) to calculate the number of Aluminum atoms:\[ \text{Number of Aluminum atoms} = n \times 6.022 \times 10^{23} \]Substitute the value from step 2:\[ 0.0996 \ \text{mol} \times 6.022 \times 10^{23} \ \text{atoms/mol} \approx 6.00 \times 10^{22} \ \text{Al atoms} \]

Key Concepts

Molar Mass Calculation

To calculate the molar mass of any compound, you add up the atomic masses of all the atoms in its formula. Alum, represented as \( \mathrm{KAl(SO}_4\mathrm{)_2 \cdot 12H}_2\mathrm{O} \), requires careful calculation. Here's a breakdown on how to do it:

1. **Identify the Elements and Their Amounts:**

• Potassium (K): 1 atom
• Aluminum (Al): 1 atom
• Sulfur (S): 2 atoms
• Oxygen (O): 8 from \(\mathrm{(SO}_4\mathrm{)_2}\) plus 12 from \(\mathrm{12 H}_2\mathrm{O}\), totaling 20 atoms
• Hydrogen (H): 24 atoms from \(\mathrm{12 H}_2\mathrm{O}\)

2. **Use Atomic Masses from the Periodic Table:** - Potassium (K): approx 39.10 amu
- Aluminum (Al): approx 26.98 amu
- Sulfur (S): approx 32.07 amu
- Oxygen (O): approx 16.00 amu
- Hydrogen (H): approx 1.01 amu

3. **Calculate:** - Total Molar Mass = (1 × 39.10) + (1 × 26.98) + (2 × 32.07) + (20 × 16.00) + (24 × 1.01)
- This adds up to approximately 474.39 g/mol, the molar mass of alum.

In molar mass calculations, careful tracking of each element's amount is key. It provides the bridge between macroscopic measurements and atomic scale understanding.

Avogadro's Number

Avogadro's number is a fundamental constant that describes the number of constituent particles, usually atoms or molecules, that are contained in one mole of a substance. It is approximately \(6.022 \times 10^{23}\). In the context of alum crystals or any chemical substance, this constant allows us to relate moles to actual numbers of atoms or molecules.

To use Avogadro's number:

• Determine the number of moles you have of a compound using its mass and molar mass.
• Multiply the moles by Avogadro's number to convert this quantity into the number of atoms or molecules.

For alum, one mole of alum contains an Avogadro's number of each type of molecule. Each formula unit of alum corresponds to one aluminum atom, so the number of aluminum atoms is directly calculated by multiplying the moles of alum by \(6.022 \times 10^{23}\).

Take for example, from our exercise, 0.0996 moles of alum is involved, leading to a calculation of \(0.0996 \times 6.022 \times 10^{23} \approx 6.00 \times 10^{22}\) aluminum atoms. This offers students a tangible way to understand the connection between moles and actual atomic counts.

Density of Solids

The density of a solid is a property that describes how much mass is contained in a given volume. It is often expressed in units of grams per cubic centimeter (g/cm³). The formula for density \( \rho \) is given by the equation:

\[ \rho = \frac{m}{V}, \]
where \(m\) is mass and \(V\) is volume.

Knowing the density of a material allows one to calculate either the mass if the volume is known, or the volume if the mass is known.To utilize density in problems like calculating elements in alum crystals:

• Start with the volume, often easily calculated for geometric shapes like cubes.
• Multiply the volume by the density to find the mass.

In our exercise example, the alum cube:- Volume = \(3.00^3 = 27.0\,\)cm³.
- Given Density: 1.75 g/cm³.
Mass is thus calculated as \(27.0 \times 1.75\, \)g, resulting in 47.25 g.

Understanding density in this way highlights how materials differ in the same volume and aids in linking mass and molar calculations for further chemical analysis.