Problem 105 - Chemistry The Central Science

Question

A lithium salt used in lubricating grease has the formula \(\mathrm{LiC}_{n} \mathrm{H}_{2 n+1} \mathrm{O}_{2} .\) The salt is soluble in water to the extent of 0.036 \(\mathrm{g}\) per 100 \(\mathrm{g}\) of water at \(25^{\circ} \mathrm{C}\) . The osmotic pressure of this solution is found to be 57.1 torr. Assuming that molality and molarity in such a dilute solution are the same and that the lithium salt is completely dissociated in the solution, determine an appropriate value of \(n\) in the formula for the salt.

Step-by-Step Solution

Answer

The value of n in the formula for the lithium salt \(\mathrm{LiC}_{n}\mathrm{H}_{2n+1}\mathrm{O}_{2}\) is approximately 1, giving the molecular formula \(\mathrm{LiCH}_{3}\mathrm{O}_{2}\).

Step 1: Write down the osmotic pressure formula and given data

The osmotic pressure formula is given by: \(π = M * R * T\) where: π = osmotic pressure M = molarity R = gas constant (0.0821 L·atm/mol·K) T = temperature in Kelvin We are given: - solubility of lithium salt = 0.036 g per 100 g water at \(25^{\circ}\mathrm{C}\) - osmotic pressure (π) = 57.1 torr First, we need to convert the osmotic pressure from torr to atm and the temperature to Kelvin: - π = 57.1 torr × (1 atm / 760 torr) ≈ 0.0751 atm - T = \(25^{\circ}\mathrm{C} + 273\) = 298 K

Step 2: Calculate the molality and molarity of the solution

We need to calculate the molality (which is assumed to be equal to molarity), given the solubility of lithium salt: molality = (mass of solute) / (1000 g solvent / molecular weight) Let's denote the molality as M and the molecular weight as MW, we can write: M = (0.036 g) / (1000 g solvent / MW)

Step 3: Use osmotic pressure formula to determine molecular weight

Now, we can use the osmotic pressure formula to find the molecular weight of the lithium salt: π = M * R * T Substituting the values, 0.0751 atm = (0.036 g / (1000 g solvent / MW)) * 0.0821 L·atm/mol·K * 298 K Solving for MW, we get: MW ≈ 41.6 g/mol

Step 4: Calculate the value of n using molecular weight

The molecular formula of the lithium salt is \(\mathrm{LiC}_{n}\mathrm{H}_{2n+1}\mathrm{O}_{2}\). Let's find the total molecular weight in terms of n: MW = 6.94 + 12.01n + 2.02(2n+1) + 2(16) Substitute the value of MW: 41.6 ≈ 6.94 + 12.01n + 4.04n + 4.04 + 32 Now, solve for n: n ≈ 1

Step 5: Write down the final formula for the lithium salt

We found that the value of n equals 1. So, the molecular formula of the lithium salt is: \(\mathrm{LiC}_{1}\mathrm{H}_{2(1)+1}\mathrm{O}_{2}\) or \(\mathrm{LiCH}_{3}\mathrm{O}_{2}\)

Key Concepts

Lithium Salt

Lithium salts are compounds composed of the metal lithium (Li) combined with an anion, which provides stability and functionality to the compound. These salts are widely used in various applications, from industrial to medicinal purposes. One common use is in the production of lubricating greases.

Lithium's properties lend themselves well to this application because it forms stable compounds that can efficiently reduce friction. In our problem, the lithium salt has the molecular formula \(\mathrm{LiC}_{n}\mathrm{H}_{2n+1}\mathrm{O}_{2}\), which implies it is a type of lithium carboxylate, a compound known for its use in greases. These greases perform excellently under high temperatures and can effectively protect mechanical parts, highlighting the practical use of lithium salts in everyday applications.

Molecular Weight

Molecular weight, or molar mass, is the measure of the mass of a molecule. It is expressed in grams per mole (g/mol), and this value is crucial for determining the amount of a substance needed to achieve a desired concentration or effect in a solution.

In our problem, calculating the molecular weight was key to concluding the exact structure of the lithium salt. By rearranging the osmotic pressure equation, the molecular weight was found to be approximately 41.6 g/mol. This information allows us to infer the precise number of carbon, hydrogen, and oxygen atoms in the compound, eventually helping to identify the correct formula, \(\mathrm{LiCH}_{3}\mathrm{O}_{2}\), for the lithium salt in the solution.

Solubility

Solubility refers to the ability of a substance to dissolve in a solvent, creating a homogeneous solution. The measure of solubility is typically given in terms of concentration, such as grams of solute per 100 grams of solvent.

In the context of the problem, the solubility of the lithium salt is 0.036 g per 100 g of water at \(25^{\circ}\mathrm{C}\). This relatively low solubility indicates that the substance moderately dissolves in water, forming a dilute solution. Despite the low concentration, this solubility level was significant enough to calculate the osmotic pressure and determine the molecular weight, ultimately leading to calculations that revealed the specific molecular structure of the compound.

Molecular Formula

The molecular formula of a compound tells us exactly what atoms make up a molecule and the number of each type of atom present in the molecule. Understanding the molecular formula is crucial in chemistry as it determines both the properties and behaviors of the substance.

For the lithium salt in question, the formula \(\mathrm{LiC}_{n}\mathrm{H}_{2n+1}\mathrm{O}_{2}\) was given, where \(n\) represents the number of carbon atoms bonded in the chain. Using the calculated molecular weight and the assumption of complete dissociation in water, the specific formula \(\mathrm{LiCH}_{3}\mathrm{O}_{2}\) was derived, corresponding to when \(n = 1\). This complete identification aids in predicting how the lithium salt will behave in lubricating applications and other chemical processes.