Problem 89 - Chemistry and Chemical Reactivity

Question

A newly synthesized compound containing boron and fluorine is \(22.1 \%\) boron. Dissolving \(0.146 \mathrm{g}\) of the compound in \(10.0 \mathrm{g}\) of benzene gives a solution with a vapor pressure of \(94.16 \mathrm{mm}\) Hg at \(25^{\circ} \mathrm{C} .\) (The vapor pressure of pure benzene at this temperature is 95.26 mm Hg.) In a separate experiment, it is found that the compound does not have a dipole moment. (a) What is the molecular formula for the compound? (b) Draw a Lewis structure for the molecule, and suggest a possible molecular structure. Give the bond angles in the molecule and the hybridization of the boron atom.

Step-by-Step Solution

Answer

(a) The compound is likely \( BF_3 \). (b) Draw the structure showing boron bonded to three fluorines in a trigonal planar geometry, with 120° bond angles and sp² hybridization.

Step 1: Calculate Molar Mass of the Compound

Determine the molar mass of the compound using the percentage of boron and the mass of boron in the compound. If 22.1% of the compound is boron, multiply 0.146 g by 0.221 to find the mass of boron and divide by its molar mass (approximately 10.81 g/mol for B) to find moles of boron, then use this to deduce the total molar mass of the compound.

Step 2: Calculate the Mole Fraction of Benzene

Use Raoult's Law to calculate the mole fraction, \( X \), of benzene: \( P = X P^0 \), where \( P \) is the vapor pressure of the solution (94.16 mm Hg) and \( P^0 \) is the vapor pressure of pure benzene (95.26 mm Hg). \[ X = \frac{P}{P^0} = \frac{94.16}{95.26} \approx 0.9884 \]

Step 3: Calculate the Number of Moles of the Compound

Using the mole fraction of benzene, calculate the moles of benzene: \( 10 \text{ g of benzene} / 78.11 \text{ g/mol} = 0.128 \text{ mol benzene} \). Substitute into the mole fraction equation: \[ 0.9884 = \frac{0.128}{0.128 + n} \] Solve for \( n \) to find the moles of the compound.

Step 4: Determine Molar Mass of the Compound

From the calculated moles of the compound (\( n \)), determine the molar mass using the given mass of the compound (0.146 g): \[ ext{Molar Mass} = \frac{0.146 ext{ g}}{n} \].

Step 5: Propose a Molecular Formula

With the molar mass and knowing that the compound is entirely composed of boron and fluorine, propose a formula. The potential structures could be \( BF_3 \), \( B_2F_4 \), and other lower molecular weight compounds.

Step 6: Draw the Lewis Structure

Draw the Lewis structure using the proposed molecular formula, considering boron often forms three bonds and no dipole moment indicates a symmetrical structure.

Step 7: Determine Molecular Shape and Bond Angles

Considering the symmetry and Lewis structure, determine the molecular shape (e.g., trigonal planar if \( BF_3 \)), note bond angles (120° in the case of trigonal planar), and the hybridization of boron (\( sp^2 \) for trigonal planar).

Key Concepts

Vapor Pressure

Vapor pressure is a critical concept in understanding how substances behave when dissolved. It measures the tendency of a substance to evaporate and is defined as the pressure exerted by a vapor in equilibrium with its liquid form. This concept can help explain how compounds like the one in our exercise affect solution properties.

• When a solute is dissolved in a solvent, the vapor pressure of the solvent decreases.
• The non-volatile solute component results in less volatile solvent molecules on the surface, thus reducing vapor pressure.
• Raoult’s Law, used here, provides a way to calculate vapor pressure of the solution: \[ \text{P}_{\text{solution}} = X_{\text{solvent}} \times \text{P}^{0}_{\text{solvent}} \]

Understanding vapor pressure helps pinpoint why the compound, when dissolved in benzene, resulted in a lower vapor pressure than that of pure benzene.
This change in pressure reveals vital information about the solute’s interaction with the solvent.

Molecular Formula

A molecular formula provides the exact number of each type of atom in a molecule. Determining this is crucial for understanding the chemical identity and properties of a compound. For the given boron-fluorine compound, the molecular formula reflects the composition:

• Percentage composition of elements is the starting point. Here, it involves calculating boron’s contribution using its percentage by mass.
• By adjusting known masses into moles, and utilizing mole ratio techniques, one can propose molecular formulas like \( BF_3 \) or \( B_2F_4 \).

The task becomes an exercise in balancing known values with hypothetical constructs to deduce the correct formula that fits the observed physical data: namely mass and vapor pressure changes.

Hybridization

Hybridization explains how atomic orbitals transform to form new orbitals, allowing atoms to form chemical bonds. Knowing the hybridization of an atom helps predict molecular geometry and bonding characteristics:

• For boron, a common hybridization state is \( sp^2 \). This corresponds to the formation of three equivalent orbitals arranged 120° apart.
• This hybridization is common in trigonal planar arrangements, like \( BF_3 \), where all boron-fluorine bonds are equivalent.

In this exercise, the lack of a dipole moment indicates a symmetrical distribution of charge, suggesting the hybridization leading to an equilateral shape.
Understanding hybridization provides insight into both the physical shape and the electronic structure of molecules.

Molecular Shape

Molecular shape dictates how a molecule interacts with its surroundings, influencing both chemical properties and reactivity. Knowing this helps in predicting behavior based on structure:

• The shape of a molecule is determined by the arrangement of atoms and electrons around the central atom.
• For example, in a compound like \( BF_3 \), trigonal planar geometry arises to minimize repulsion between electron pairs, with bond angles of 120°.
• Lack of a dipole moment further confirms a symmetrical shape, as seen in this exercise's compound.

Molecular shape is more than a physical description; it’s a tool for predicting molecular interactions and stability.
This information proves invaluable in fields like material science, pharmacology, and chemistry, making it crucial knowledge for students to master.