Problem 135 - Chemistry Matter and Change

Question

Acne Benzoyl peroxide \(\left(\mathrm{C}_{14} \mathrm{H}_{10} \mathrm{O}_{4}\right)\) is a substance used as an acne medicine. What is the mass in grams of \(3.50 \times 10^{-2} \mathrm{mol} \mathrm{C}_{14} \mathrm{H}_{10} \mathrm{O}_{4} ?\)

Step-by-Step Solution

Answer

The mass of \(3.50 \times 10^{-2}\) mol of Benzoyl peroxide is 8.48 grams.

Step 1: Determine Molecular Weight

First, find the molecular weight of the compound Benzoyl peroxide \( (\mathrm{C}_{14} \mathrm{H}_{10} \mathrm{O}_{4}) \) by adding together the atomic weights of each element in the compound: - Carbon (C): 14 atoms, each weighing approximately 12.01 g/mol, contributing \( 14 \times 12.01 \) - Hydrogen (H): 10 atoms, each weighing approximately 1.01 g/mol, contributing \( 10 \times 1.01 \) - Oxygen (O): 4 atoms, each weighing approximately 16.00 g/mol, contributing \( 4 \times 16.00 \). Thus, the molecular weight \( = 14 \times 12.01 + 10 \times 1.01 + 4 \times 16.00 = 242.23 \text{ g/mol}.\)

Step 2: Use the Molar Mass to Find Mass

Now, calculate the mass for \(3.50 \times 10^{-2} \text{ mol}\) of Benzoyl peroxide. Use the formula: \[ \text{mass (g)} = \text{moles} \times \text{molar mass (g/mol)} \].With \(3.50 \times 10^{-2} \text{ mol}\) and a molar mass of \(242.23 \text{ g/mol}\), the mass is \(3.50 \times 10^{-2} \times 242.23 = 8.478 \text{ grams} \).

Step 3: Round Your Answer

Round the answer to an appropriate number of significant figures based on the given data. The moles value \(3.50 \times 10^{-2}\) has three significant figures, so round the mass to three significant figures: \(8.478 \to 8.48 \text{ grams}.\)

Key Concepts

Molecular Weight

Understanding molecular weight is crucial for performing chemical calculations, especially when you are required to find the mass of a given number of moles of a substance. Molecular weight, also known as molecular mass, is the sum of the atomic weights of all atoms in a molecule.

Let's break down the steps for calculating the molecular weight of Benzoyl peroxide \( (\mathrm{C}_{14} \mathrm{H}_{10} \mathrm{O}_{4}) \).

• For Carbon (C): There are 14 carbon atoms, each with an atomic weight of approximately 12.01 g/mol. So, the total weight contribution from carbon is \( 14 \times 12.01 = 168.14 \text{ g/mol} \).
• For Hydrogen (H): There are 10 hydrogen atoms, each weighing about 1.01 g/mol. Thus, their total contribution is \( 10 \times 1.01 = 10.1 \text{ g/mol} \).
• Lastly, Oxygen (O): There are 4 oxygen atoms, each having an atomic weight of about 16.00 g/mol, contributing a total of \( 4 \times 16.00 = 64.00 \text{ g/mol} \).

Adding these values gives the total molecular weight of Benzoyl peroxide, \( 168.14 + 10.1 + 64 = 242.23 \text{ g/mol} \). Understanding how to sum these values is crucial for later calculations.

Significant Figures

When working with measurements in chemistry, significant figures are vital for expressing precision. They indicate the reliability of a measurement and ensure that calculations reflect the precision of the data used.

In our example, the number of moles provided is \(3.50 \times 10^{-2}\), which consists of three significant figures. Consequently, any calculated result should also reflect this precision.

Calculating the mass using the molar mass involves this important concept. Suppose you calculate the mass as \(8.478 \text{ grams} \), based on the determined molar mass \(242.23 \text{ g/mol} \). This raw result must then be adjusted to maintain the three significant figures originally provided:

• The final value should be rounded to \(8.48 \text{ grams} \), ensuring it aligns with the precision of the initial measurement.

This practice of rounding based on significant figures not only keeps calculations consistent but also better communicates the certainty of the final result.

Chemical Calculations

Chemical calculations often involve determining masses, moles, and molecular weights. Understanding the relationship between these quantities is fundamental.

In chemistry, to determine the mass from moles, the following relation is utilized:\[ \text{mass (g)} = \text{moles} \times \text{molar mass (g/mol)}\]This formula allows you to convert between moles, which are a measure of quantity, and mass. In our scenario with Benzoyl peroxide, we demonstrated this by first calculating the molecular weight, then using it to find the gram mass from a given number of moles.

Here’s how it is practically applied:

• We know the moles of Benzoyl peroxide are \(3.50 \times 10^{-2}\).
• The molar mass has been calculated to be \(242.23 \text{ g/mol} \).
• Multiplying these values, \(3.50 \times 10^{-2} \times 242.23\), gives the mass \(8.478\) grams, which is then rounded to \(8.48\) grams.

By grasping the concepts of mole-mass conversion and the necessary operations, students can solve a variety of chemical calculation problems with confidence.