Problem 103 - Chemistry The Molecular Science

Question

Ethyl alcohol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\), has a density of \(0.789 \mathrm{~g} / \mathrm{mL}\) at \(25^{\circ} \mathrm{C}\). Water weighs \(1.00 \mathrm{~kg} / \mathrm{L}\) at \(25^{\circ} \mathrm{C}\). Calculate the volume of ethanol that contains the same number of molecules as \(1.00 \mathrm{~L}\) water.

Step-by-Step Solution

Answer

The volume of ethanol required is approximately 3.242 L.

Step 1: Calculate Moles of Water

First, find the moles of water in 1 liter (1,000 mL) using its density. Water's density is \(1.00 \text{ kg/L}\), meaning \(1.00 \text{ kg}\) or \(1,000 \text{ g}\) of water per liter; the molar mass of water (\(\mathrm{H}_{2}\mathrm{O}\)) is \(18.015 \text{ g/mol}\). Therefore, the moles of water is calculated by dividing the mass by the molar mass:\[\text{Moles of water} = \frac{1000 \text{ g}}{18.015 \text{ g/mol}} \approx 55.5 \text{ mol}\]

Step 2: Calculate Number of Water Molecules

Use Avogadro's number (\(6.022 \times 10^{23} \text{ molecules/mol}\)) to convert moles to number of molecules. Multiply the moles of water by Avogadro's number:\[\text{Number of water molecules} = 55.5 \text{ mol} \times 6.022 \times 10^{23} \text{ molecules/mol} \approx 3.34 \times 10^{25} \text{ molecules}\]

Step 3: Calculate Moles of Ethanol

As water and ethanol should have the same number of molecules, ethanol should also have \(3.34 \times 10^{25}\) molecules. Using Avogadro's number, calculate the moles of ethanol required:\[\text{Moles of ethanol} = \frac{3.34 \times 10^{25} \text{ molecules}}{6.022 \times 10^{23} \text{ molecules/mol}} \approx 55.5 \text{ mol}\]

Step 4: Calculate Mass of Ethanol

Use the number of moles and the molar mass of ethanol (\( \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} \), molar mass \(46.068 \text{ g/mol}\)) to find the mass:\[\text{Mass of ethanol} = 55.5 \text{ mol} \times 46.068 \text{ g/mol} \approx 2557.8 \text{ g}\]

Step 5: Calculate Volume of Ethanol

Divide the mass of ethanol by its density (\(0.789 \text{ g/mL}\)) to find the volume:\[\text{Volume of ethanol} = \frac{2557.8 \text{ g}}{0.789 \text{ g/mL}} \approx 3242 \text{ mL}\] Convert to liters: 3242 mL = 3.242 L.

Key Concepts

Understanding the Mole Concept

The mole concept is a fundamental aspect of chemistry that allows scientists to count particles at the atomic and molecular level by using a standard unit called a "mole." A mole is defined as exactly \( 6.022 \times 10^{23} \) entities, which can be atoms, molecules, or even ions, and this is known as Avogadro's number.

When calculating the amount of substance, chemists use moles to easily convert between the mass of a substance and the number of molecules or atoms within it. This conversion is crucial because it's often impractical to count individual atoms or molecules due to their minuscule size.

For example, in the exercise above, the number of moles of water was calculated by dividing the mass of water by its molar mass (i.e., \( \frac{1000 \text{ g}}{18.015 \text{ g/mol}} \)). Moles are directly related to the concept of Avogadro's number, allowing the conversion from moles to molecules and vice versa, which streamlines the calculations in chemical reactions and experiments.

The Basics of Density Calculations

Density is a key physical property that defines how much mass is contained within a certain volume of a substance, typically expressed in grams per milliliter (g/mL) or kilograms per liter (kg/L). In the case of liquids like water and ethanol, understanding density helps to determine how substances will interact and occupy space.

In the exercise, water was defined to have a density of \(1.00 \text{ kg/L}\), meaning that in each liter there’s exactly 1 kilogram of water. This property allows for quick calculations of the mass when the volume is known and vice versa.

Ethanol, on the other hand, has a density of \(0.789 \text{ g/mL}\), which means it is less dense than water. This lower density affects its calculations, as seen when the mass (2557.8 g) is divided by this density to determine that the volume of ethanol is about 3242 mL or 3.242 liters. Mastering density calculations is crucial for many practical applications, from chemical engineering to laboratory experiments.

Molecular Conversions and Practical Applications

Molecular conversions refer to the process of converting between different measures in chemical reactions, such as moles to molecules, and are essential for a thorough understanding of how much of a substance is present.

In the given scenario, after determining the moles of both water and ethanol to be 55.5, the task was to calculate the number of molecules using Avogadro's number. By doing this, you can confirm that equivalent amounts of different substances contain the same number of molecules or moles. This equality is crucial for the principle of stoichiometry, which states that chemical reactions are balanced in terms of moles.

In practical terms, molecular conversions are used ubiquitously in scientific calculations. These conversions ensure accurate predictions and measurements, critical for fields such as pharmaceuticals, environmental science, and materials science, where precise molecular interrelations define product effectiveness and efficacy.