Problem 157 - Chemistry and Chemical Reactivity

Question

In an experiment, you need 0.125 mol of sodium metal. Sodium can be cut easily with a knife (Figure \(2.5),\) so if you cut out a block of sodium, what should the volume of the block be in cubic centimeters? If you cut a perfect cube, what is the length of the edge of the cube? (The density of sodium is \(\left.0.97 \mathrm{g} / \mathrm{cm}^{3} .\right)\)

Step-by-Step Solution

Answer

The block should be approximately 2.9637 cm³, with an edge length of 1.43 cm.

Step 1: Convert Moles to Grams

To find out how many grams of sodium you need, use the molar mass of sodium (Na), which is approximately 22.99 g/mol. Multiply the number of moles by the molar mass to find the grams:\[ ext{mass} = 0.125 ext{ mol} \times 22.99 \frac{g}{mol} = 2.87375 ext{ g}.\]

Step 2: Calculate the Volume of Sodium

Using the density formula, \(\text{Density} = \frac{\text{Mass}}{\text{Volume}}\), rearrange to solve for Volume:\[\text{Volume} = \frac{\text{Mass}}{\text{Density}} = \frac{2.87375 \text{ g}}{0.97 \frac{\text{g}}{\text{cm}^3}} \approx 2.9637 \text{ cm}^3.\]

Step 3: Determine the Edge Length of the Cube

For a cube, the volume \(V\) is the cube of the edge length \(s\):\[\text{Volume} = s^3.\]Since we calculated the volume to be approximately 2.9637 cubic centimeters:\[s = \sqrt[3]{2.9637} \approx 1.43 \text{ cm}.\]

Key Concepts

Density

Density is a crucial concept in chemistry, as it connects mass and volume, offering insights into the characteristics of substances. In simple terms, density refers to how much mass is contained within a particular volume of a substance.

Mathematically, density is expressed by the formula:

• \[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \]

This relationship is fundamental when performing chemical calculations or determining the properties of materials.

In this exercise, using the density of sodium, which is \(0.97\, g/cm^3\), helped us calculate the volume of sodium needed for the experiment. This demonstrates how density serves as a bridge between mass and volume, enabling further calculations.

Molar Mass

Molar mass is another foundational concept in stoichiometry. It allows chemists to convert between moles of a substance and its mass in grams. The molar mass is simply the mass of one mole of a specific element or compound and is usually expressed in grams per mole \(g/mol\).

To give an example, sodium (\(Na\)) has a molar mass of approximately \(22.99\, g/mol\). This value helps convert moles to grams in chemical calculations.

• For 0.125 mol of sodium:
• \[ \text{Mass} = 0.125 \text{ mol} \times 22.99 \frac{g}{mol} = 2.87375 \text{ g} \]

The molar mass provides the necessary information to determine the actual mass of sodium required, serving as an essential tool in laboratory settings.

Volume Calculation

Volume calculation plays a vital role when preparing for an experiment or understanding the properties of solid materials. It often involves solving for one of the unknowns using the density formula.

• Here, the formula is rearranged as:
• \[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} \]

This calculation helps us find out how much space a specific mass of a substance will occupy.

For sodium in this exercise, the volume was calculated to be approximately \(2.9637\, cm^3\). Knowing the volume is crucial when cutting out a physical piece from a larger block, ensuring that the correct amount of material is used.

Chemical Calculations

Chemical calculations allow us to predict and quantify various properties of substances in a precise manner. They involve a combination of different concepts such as density, molar mass, and volume calculations.

These calculations are essential for understanding reactions, determining the amount of reagents needed, and predicting reaction outcomes.

For instance, in this exercise:

• We started by using the molar mass to convert moles to grams of sodium.
• Next, we used the density to find the volume of sodium.
• Finally, we calculated the edge length of a cubic piece of sodium.

The ability to perform such calculations precisely underscores the interconnectedness of different concepts in chemistry and ensures that experimental procedures are carried out accurately.