Problem 112 - Chemistry: An Atoms-Focused Approach

Question

Ethylene glycol is the common name for the liquid used to keep the coolant in automobile cooling systems from freezing. It is \(38.7 \%\) carbon, \(9.7 \%\) hydrogen, and \(51.6 \%\) oxygen by mass. Its molar mass is \(62.07 \mathrm{g} / \mathrm{mol}\) and its density is \(1.106 \mathrm{g} / \mathrm{mL}\) at \(20^{\circ} \mathrm{C}\) a. What is the empirical formula of ethylene glycol? b. What is the molecular formula of ethylene glycol? c. In a solution prepared by mixing equal volumes of water and ethylene glycol, which ingredient is the solute and which is the solvent?

Step-by-Step Solution

Answer

Question: Determine the empirical and molecular formulas of ethylene glycol and identify the solute and solvent when equal volumes of water and ethylene glycol are mixed. Answer: The empirical formula of ethylene glycol is CH3O, and the molecular formula is C2H6O2. When equal volumes of water and ethylene glycol are mixed, the solute is ethylene glycol and the solvent is water.

Step 1: Convert given percentages to grams

Assuming we have 100 g of ethylene glycol, we get 38.7 g of carbon, 9.7 g of hydrogen, and 51.6 g of oxygen, according to the given mass percentages.

Step 2: Convert the mass of elements to moles

Divide the grams of each element by their respective molar masses to find the number of moles. Carbon: \(\frac{38.7 \mathrm{g}}{12.01 \mathrm{g/mol}}= 3.22 \mathrm{mol}\) Hydrogen: \(\frac{9.7 \mathrm{g}}{1.008 \mathrm{g/mol}}= 9.62 \mathrm{mol}\) Oxygen: \(\frac{51.6 \mathrm{g}}{16.00 \mathrm{g/mol}}= 3.23 \mathrm{mol}\)

Step 3: Determine the ratio of moles

Set the lowest number of moles (3.22) equal to 1, and then find the ratio by dividing each moles value by 3.22. Carbon: \(\frac{3.22}{3.22}=1\) Hydrogen: \(\frac{9.62}{3.22}=2.99 \approx 3\) Oxygen: \(\frac{3.23}{3.22}=1\) Thus, the empirical formula is CH\(_3\)O. b. Molecular Formula:

Step 1: Calculate the molar mass of the empirical formula

Determine the molar mass of CH\(_3\)O by adding the molar masses of each element. Molar Mass of CH\(_3\)O: \((1 \times 12.01) + (3 \times 1.008) + (1 \times 16.00) = 31.03 \mathrm{g/mol}\)

Step 2: Calculate the ratio of the molecular formula's molar mass to the empirical formula's molar mass

Divide the molecular formula's molar mass given in the problem (62.07 g/mol) by the empirical formula's molar mass (31.03 g/mol) to find the ratio. Ratio: \(\frac{62.07}{31.03} = 2\)

Step 3: Multiply the empirical formula by the ratio

Multiply each element in the empirical formula (CH\(_3\)O) by the ratio (2) to determine the molecular formula. Molecular Formula: C\(_2\)H\(_6\)O\(_2\) c. Solute and Solvent:

Step 1: Compare the densities of water and ethylene glycol

Given the density of ethylene glycol is 1.106 g/mL and the density of water is 1.000 g/mL.

Step 2: Identify the solute and solvent

The solute is the substance present in a smaller amount in a solution. In the case of 1:1 by volume, since ethylene glycol has a higher density, it has more mass per unit volume than water. Thus, ethylene glycol has a smaller volume and is the solute, while water, having a larger volume, is the solvent.

Key Concepts

Ethylene Glycol

Ethylene glycol is a versatile chemical compound primarily used in the automotive industry to prevent the freezing of coolant in vehicle cooling systems. Its properties, such as a molar mass of 62.07 grams per mole and a density of 1.106 grams per milliliter at room temperature (20°C), make it an effective antifreeze agent. Ethylene glycol is also known for its sweet taste, although it is highly toxic if ingested. This compound's effective antifreeze capability is due to its ability to lower the freezing point of liquid, making it crucial for use in colder climates.

In addition to its primary application as antifreeze, ethylene glycol is used in various industrial processes, including the production of polyester fibers and resins. Its chemical stability and solubility in water make it suitable for these diverse applications.

Chemical Composition

Understanding the chemical composition of a substance like ethylene glycol is essential for determining its empirical and molecular formulas. Ethylene glycol is composed of 38.7% carbon, 9.7% hydrogen, and 51.6% oxygen by mass. These percentages translate into actual mass when considering a sample size, such as 100 grams, allowing us to calculate the number of moles of each element.

• The number of moles for each element can be calculated by dividing its mass by its atomic mass.
• For carbon: 38.7 grams / 12.01 grams/mol = 3.22 moles.
• For hydrogen: 9.7 grams / 1.008 grams/mol = 9.62 moles.
• For oxygen: 51.6 grams / 16.00 grams/mol = 3.23 moles.

These calculations yield the empirical formula CH extsubscript{3}O by setting the lowest number of moles as a reference. The molecular formula is determined by comparing the molar mass of the empirical formula and the actual molar mass of the compound. In this case, the ratio between the molar masses (62.07 g/mol divided by 31.03 g/mol) indicates that the molecular formula is C extsubscript{2}H extsubscript{6}O extsubscript{2}.

Solute and Solvent Identification

In chemistry, identifying the solute and solvent in a solution is crucial for understanding how substances interact. The solvent is typically the component present in a greater amount, whereas the solute is present in a smaller amount. This context is crucial when mixing substances like water and ethylene glycol. Given the density content, ethylene glycol's density is measured at 1.106 grams per milliliter compared to water's 1.000 grams per milliliter. When mixing equal volumes, because ethylene glycol is denser, it comprises a greater mass in the same volume compared to water.

Despite this, when considering the typical roles each substance plays in a solution, water often serves as the solvent, and ethylene glycol, due to its smaller volumetric presence, behaves as the solute. This identification assists in predicting properties of the solution, such as boiling and freezing points, which are tailored by the presence of ethylene glycol as an additive to the solvent, water.