Problem 13 - Chemistry and Chemical Reactivity

Question

A chemist needs \(2.00 \mathrm{g}\) of a liquid compound with a density of \(0.718 \mathrm{g} / \mathrm{cm}^{3} .\) What volume of the compound is required?

Step-by-Step Solution

Answer

The volume required is approximately 2.785 cm³.

Step 1: Understand the Relationship between Mass, Density, and Volume

The relationship between mass, density, and volume is given by the formula: \( \text{Density} = \frac{\text{Mass}}{\text{Volume}} \). In this problem, we need to rearrange this formula to find the volume: \( \text{Volume} = \frac{\text{Mass}}{\text{Density}} \).

Step 2: Identify Known Values

From the problem, we know the mass of the liquid compound is \(2.00 \, \text{g}\) and the density is \(0.718 \, \text{g/cm}^3\).

Step 3: Substitute Values into the Formula

Substitute the known values into the rearranged formula for volume: \[ \text{Volume} = \frac{2.00 \, \text{g}}{0.718 \, \text{g/cm}^3} \].

Step 4: Perform the Calculation

Calculate the volume by dividing the mass by the density. So, \[ \text{Volume} = \frac{2.00}{0.718} \approx 2.785 \text{ cm}^3 \].

Step 5: Interpret the Result

The volume of the liquid compound required is approximately \(2.785 \, \text{cm}^3\). This means that to obtain 2.00 g of the compound, a volume of 2.785 cm³ is needed based on its density.

Key Concepts

Density Formula

Let's explore the heart of the exercise: the density formula. Density is a physical property that relates the mass of a substance to its volume. It is expressed using the formula:

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

This formula tells us how much mass is contained within a certain volume of a material. To solve problems involving mass, volume, and density, it's crucial to understand that these three quantities are interconnected.
In this particular problem involving a liquid compound, we rearrange the formula to find volume instead of density. By rearranging, we get:

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

This rearrangement is key because it allows us to find how much space the compound occupies when we know its mass and density. This concept is essential not only in theoretical exercises but also in real-world chemistry labs.

Mass and Volume Calculation

Now that we have the density formula rearranged to find the volume, we apply it to our specific case. The given problem provides us with two key values:

• Mass = 2.00 g
• Density = 0.718 g/cm³

By substituting these values into our formula for volume, we perform the calculations:

• Volume = \( \frac{2.00 \, \text{g}}{0.718 \, \text{g/cm}^3} \)
• Volume ≈ 2.785 cm³

The outcome of these calculations provides us with the volume of the compound needed. This is straightforward when you substitute the known values correctly and take direct action by dividing. Remember, the key is to perform the math carefully to avoid errors.

Chemistry Problem Solving

In chemistry, solving problems often involves understanding the relationship between different properties and applying mathematical equations. This particular problem is a classic example of chemistry problem-solving that emphasizes the use of the density formula. These tasks require a systematic approach:

• Firstly, identify what the problem asks you to find. Here, it's the volume.
• Then, determine what information you have and what formulas you might use. This involves knowing the mass and density.
• Next, substitute the known values into your chosen formula, as we did with our rearranged density formula.
• Finally, solve the equation through straightforward calculations.

These problem-solving strategies are not restricted to density but are useful for a variety of chemistry challenges. Understanding and practicing these steps will help you efficiently tackle similar problems both in educational settings and practical applications.