Question

A solution is made containing \(14.6 \mathrm{~g}\) of \(\mathrm{CH}_{3} \mathrm{OH}\) in \(184 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}\). Calculate (a) the mole fraction of \(\mathrm{CH}_{3} \mathrm{OH},(\mathbf{b})\) the mass percent of \(\mathrm{CH}_{3} \mathrm{OH},(\mathrm{c})\) the molality of \(\mathrm{CH}_{3} \mathrm{OH}\).

Short answer

(a) The mole fraction of CH3OH is 0.04268. (b) The mass percent of CH3OH is 7.348%. (c) The molality of CH3OH is 2.477 mol/kg.

Step-by-step solution

Find the moles of CH3OH and H2O

Given the mass of each component, we can calculate the number of moles by dividing the mass by the molar mass. Recall that the molar mass of CH3OH is 12.01 (C) + 1.01 (H) * 4 + 15.99 (O) = 32.04 g/mol and the molar mass of H2O is 18.015 g/mol. So, we can find the moles of CH3OH and H2O as follows: Moles of CH3OH = \( \frac{14.6 \ \text{g}}{32.04 \ \text{g/mol}} = 0.4559 \ \text{mol} \) Moles of H2O = \( \frac{184 \ \text{g}}{18.015 \ \text{g/mol}} = 10.219 \ \text{mol} \)

Calculate the mole fraction of CH3OH

The mole fraction of CH3OH is the ratio of the moles of CH3OH to the total moles in the solution. We can calculate this as follows: Mole fraction of CH3OH = \( \frac{\text{moles of CH3OH}}{\text{moles of CH3OH} + \text{moles of H2O}} \) Mole fraction of CH3OH = \( \frac{0.4559 \ \text{mol}}{0.4559 \ \text{mol} + 10.219 \ \text{mol}} = 0.04268 \)

Calculate the mass percent of CH3OH

The mass percent of CH3OH is the ratio of the mass of CH3OH to the total mass of the solution, multiplied by 100. We can calculate this as follows: Mass percent of CH3OH = \( \frac{\text{mass of CH3OH}}{\text{mass of CH3OH} + \text{mass of H2O}} \times 100 \) Mass percent of CH3OH = \( \frac{14.6 \ \text{g}}{14.6 \ \text{g} + 184 \ \text{g}} \times 100 = 7.348 \% \)

Calculate the molality of CH3OH

Molality is defined as the number of moles of solute per kilogram of solvent. In this case, CH3OH is the solute and H2O is the solvent. We can calculate the molality as follows: Molality of CH3OH = \( \frac{\text{moles of CH3OH}}{\text{mass of H2O (in kg)}} \) Molality of CH3OH = \( \frac{0.4559 \ \text{mol}}{184 \ \text{g} \times \frac{1 \ \text{kg}}{1000 \ \text{g}}} = 2.477 \ \text{mol/kg} \) To summarize, we have calculated (a) the mole fraction of CH3OH as 0.04268, (b) the mass percent of CH3OH as 7.348%, and (c) the molality of CH3OH as 2.477 mol/kg.

Key concepts

Mole Fraction

The mole fraction is a way to express the concentration of a component in a mixture or solution. It's represented as the ratio of the number of moles of the substance of interest to the total number of moles of all substances in the mixture. For example, if we have a solution with two substances A and B, with \(n_A\) moles of A and \(n_B\) moles of B, the mole fraction of A \(X_A\) is calculated using the formula:
\[ X_A = \frac{n_A}{n_A + n_B} \]
Mole fraction is a dimensionless quantity and it is particularly useful in cases where the ratio of the components is more important than their absolute quantity. Since mole fraction accounts only for the number of particles and not their weight or volume, it is an ideal way to describe concentration for calculations involving gas mixtures or reactions based on stoichiometry.

Mass Percent

Mass percent, also known as weight percent, is a common concentration measure that expresses the mass of a solute as a percentage of the total mass of the solution.
\[ Mass \(\%\) = \frac{mass \(\text{of solute}\)}{mass \(\text{of solute}\) + mass \(\text{of solvent}\)} \times 100 \]
This is an intuitive way of understanding concentration because it describes how many grams of the solute are present in every 100 grams of the solution. Mass percent is often used in chemistry labs and industry where the precise ratios of components are crucial. Whether you are dissolving a solid in a liquid or mixing two liquids, knowing the mass percent can help ensure consistency and reproducibility in experiments and products.

Molality

Molality is a concentration term that describes the amount of solute in a solution relative to the mass of the solvent. It is defined as the number of moles of solute per kilogram of solvent and is expressed as:
\[ Molality \(m\) = \frac{n \(\text{of solute in moles}\)}{mass \(\text{of solvent in kilograms}\)} \]
Unlike molarity, which takes into account the volume of the solution, molality is temperature-independent because it relies on mass. Therefore, molality is particularly suitable for experiments conducted at varying temperatures where volume might change due to thermal expansion. Molality is often used in calculations involving colligative properties, such as boiling point elevation and freezing point depression, where the effect depends on the number of solute particles rather than their volume.

Molar Mass

Molar mass is a fundamental concept in chemistry, defined as the mass of one mole of a given substance. It is typically expressed in units of grams per mole (g/mol). The molar mass of an element is the mass number listed on the periodic table, while for a compound, it is the sum of the molar masses of its constituent atoms.
\[ Molar \(\text{Mass}\) = \sum(\text{molar masses of all atoms in the compound}) \]
Molar mass links the macroscopic world we observe to the microscopic realm of atoms and molecules. For instance, when you weigh out \(32.04 \(\text{g}\) \) of methanol (\(\text{CH}_3\text{OH}\)), you’re actually counting out a mole (approximately \(6.022 \times 10^{23}\) molecules) of methanol by mass. Knowing the molar mass is essential in chemical reactions to quantify how much of a substance is needed or produced. It is the cornerstone for converting between mass and moles, a key step in stoichiometry.