Problem 75 - Chemistry and Chemical Reactivity

Question

A Dilithium is the fuel for the Starship Enterprise. Because its density is quite low, however, you need a large space to store a large mass. To estimate the volume required, we shall use the element lithium. If you need 256 mol for an interplanetary trip, what must the volume of the piece of lithium be? If the piece of lithium is a cube, what is the dimension of an edge of the cube? (The density for the element lithium is \(0.534 \mathrm{g} / \mathrm{cm}^{3}\) at \(20^{\circ} \mathrm{C} .\) )

Step-by-Step Solution

Answer

The volume is 3325.38 cm³, and the edge length of the cube is approximately 14.93 cm.

Step 1: Calculate the Mass of Lithium Needed

The molar mass of lithium (Li) is approximately 6.94 g/mol. So, we calculate the total mass by multiplying the number of moles by the molar mass: \[\text{Mass} = 256 \text{ mol} \times 6.94 \text{ g/mol} = 1775.84 \text{ g}.\]

Step 2: Determine the Volume of Lithium

Using the density of lithium, we can find the volume required. Density is given as \(0.534 \text{ g/cm}^3\). The volume \(V\) can be calculated using the formula: \[ V = \frac{\text{Mass}}{\text{Density}} = \frac{1775.84 \text{ g}}{0.534 \text{ g/cm}^3} \approx 3325.38 \text{ cm}^3.\]

Step 3: Calculate the Edge Length of the Cube

Since the volume of the lithium piece is that of a cube, the formula for the volume is \(V = a^3\), where \(a\) is the edge of the cube. Solving for \(a\), we get: \[ a = \sqrt[3]{V} = \sqrt[3]{3325.38 \text{ cm}^3} \approx 14.93 \text{ cm}.\]

Key Concepts

Molar Mass Calculation

Molar mass is a fundamental concept in chemistry used to convert moles of an element to mass. It represents the mass of one mole of a given substance and is expressed in grams per mole (g/mol). For lithium (Li), a commonly used element in various fields, its molar mass is approximately 6.94 g/mol.

Calculating the molar mass is vital when you know how many moles of a substance are needed and want to find out its mass. The calculation involves multiplying the number of moles by the molar mass. In our case with lithium, for 256 moles, the calculation involves:

• Given moles of lithium: 256 mol
• Molar mass of lithium: 6.94 g/mol

So, the total mass would be computed as\[\text{Mass} = 256 \text{ mol} \times 6.94 \text{ g/mol} = 1775.84 \text{ g}\]This step ensures you know exactly how many grams of lithium you are dealing with, which is crucial for determining the necessary volume.

Density Formula

Density connects the mass of a substance to its volume and is critical in understanding how much space a certain mass will occupy. The formula to calculate density is given by:

• Density \(D\) = \( \frac{\text{Mass}}{\text{Volume}} \)

For lithium, its density is given as 0.534 g/cmdue to its relatively low density, you'll need to know how much space a specific mass of it will actually fill.

Using this formula, we can rearrange it to find the volume:\[V = \frac{\text{Mass}}{\text{Density}}\]This relationship is especially handy when you have a known mass and density; you simply divide the mass by density to find the volume. For our given scenario:

• Mass of lithium: 1775.84 g
• Density of lithium: 0.534 g/cm³

The volume needed becomes: \[V = \frac{1775.84 \text{ g}}{0.534 \text{ g/cm}^3} \approx 3325.38 \text{ cm}^3\]This calculation tells us the space that 1775.84 grams of lithium will occupy.

Cube Volume and Edge Calculation

When dealing with solids like cubes, knowing how to link the volume of a cube to its edge length is very useful. A cube's volume is determined by the formula:

• \(V = a^3\)

Where \(a\) is the edge length of the cube. To find the edge length when the volume is known, you take the cube root of the volume.

In the case of our lithium problem, the volume of lithium needed is approximately 3325.38 cm³. Therefore, finding the edge \(a\) becomes solving:\[a = \sqrt[3]{V} = \sqrt[3]{3325.38}\]By calculating this, the edge length \(a\) of the cube will be:\[a \approx 14.93 \text{ cm}\]Thus, each edge of the lithium cube you need to store the volume calculated would be approximately 14.93 centimeters long. This not only helps in visualizing the cube but is also crucial for planning storage logistics.