Question

Tris(pentafluorophenyl)borane, commonly known by its acronym BARF, is frequently used to initiate polymerization of ethylene or propylene in the presence of a catalytic transition metal compound. It is composed solely of \(\mathrm{C}, \mathrm{F}\), and \(\mathrm{B}\); it is \(42.23 \% \mathrm{C}\) by mass and \(55.66 \% \mathrm{~F}\) by mass. a. What is the empirical formula of BARF? b. A \(2.251-g\) sample of BARF dissolved in \(347.0 \mathrm{~mL}\) of solution produces a \(0.01267 M\) solution. What is the molecular formula of BARF?

Short answer

The empirical formula of BARF is \(\mathrm{C_{18}F_{15}B}\), and its molecular formula is also \(\mathrm{C_{18}F_{15}B}\).

Step-by-step solution

Determine empirical formula

First, we will assume a total mass of 100 grams for the compound. From the given mass percentages, we have 42.23 g of Carbon (C), 55.66 g of Fluorine (F), and the rest being Boron (B). We will calculate the mass of Boron first and then convert each element's mass to moles so we can compare their mole ratios. Mass of Boron = 100 - 42.23 - 55.66 = 2.11 g Now, convert to moles using the molar masses of the elements: Moles of Carbon = 42.23 g / 12.01 g/mol = 3.52 mol Moles of Fluorine = 55.66 g / 18.998 g/mol = 2.93 mol Moles of Boron = 2.11 g / 10.81 g/mol = 0.195 mol Next, we need to divide each mole value by the lowest mole value to get the mole ratio: Mole Ratio of Carbon = 3.52 / 0.195 = 18 Mole Ratio of Fluorine = 2.93 / 0.195 = 15 Mole Ratio of Boron = 0.195 / 0.195 = 1 The empirical formula is \(\mathrm{C_{18}F_{15}B}\).

Determine molecular formula

The molecular formula is a multiple of the empirical formula. From the given problem, we have the molarity (0.01267 M) and volume of the solution (347.0 mL). We can use these to calculate the number of moles for the given mass of the compound in the solution and then calculate the molar mass. First, convert volume to liters: Volume of solution = 347.0 mL x (1 L / 1000 mL) = 0.3470 L Now, calculate the number of moles of BARF in the given sample using the volume and molarity: Moles of BARF = 0.01267 M x 0.3470 L = 0.0043929 mol Next, calculate molar mass using the given mass of the sample and the total number of moles: Molar Mass of BARF = 2.251 g / 0.0043929 mol = 512.34 g/mol Now, we need the molar mass of the empirical formula from Step 1: Molar Mass of \(\mathrm{C_{18}F_{15}B}\) = (18 x 12.01) + (15 x 18.998) + (1 x 10.81)= 324.18 + 284.97 + 10.81 = 619.96 g/mol Divide the molar mass of the molecular formula by the molar mass of the empirical formula to find the multiple (n): n = 512.34 g/mol / 619.96 g/mol = 0.826 As n must be an integer, the empirical formula is also the molecular formula. Thus, the molecular formula of BARF is \(\mathrm{C_{18}F_{15}B}\).

Key concepts

Molar Mass Calculation

Understanding molar mass is crucial for determining various properties of chemical compounds. In chemistry, the molar mass of a substance is the mass of one mole of its atoms, molecules, or compounds. This value is usually expressed in grams per mole (g/mol).

To calculate the molar mass, you sum the atomic masses of all the atoms present in a molecule. Consider the compound \(\text{C}_{18}\text{F}_{15}\text{B}\). Here, carbon (C) has an atomic mass of roughly 12.01 g/mol, fluorine (F) is about 18.998 g/mol, and boron (B) is around 10.81 g/mol.

Here's how you calculate the molar mass step by step:

• Multiply the atomic mass of each element by the number of atoms present in the formula.
• Add these values together to get the total molar mass.

For \(\text{C}_{18}\text{F}_{15}\text{B}\):

• Carbon: \(18 \times 12.01 = 216.18\) g/mol
• Fluorine: \(15 \times 18.998 = 284.97\) g/mol
• Boron: \(1 \times 10.81 = 10.81\) g/mol

The sum is \(216.18 + 284.97 + 10.81 = 511.96\) g/mol. Calculating this correctly is essential in finding the molecular formula and other chemical properties.

Molecular Formula Determination

Determining the molecular formula of a compound involves figuring out the exact number of each type of atom in one molecule of the substance. Once you have the empirical formula, you can find the molecular formula by considering the compound's molar mass.

Start by using the molarity and volume data to find the number of moles of the solute. In the given problem, the molarity is 0.01267 M, and the volume is 347 mL. Convert the volume to liters by dividing by 1000, then multiply by the molarity to find the moles:

\(\text{Moles} = 0.01267 \, \text{M} \times 0.347 \, \text{L} = 0.0043929 \, \text{mol}\).

Next, use the given mass of the sample and the calculated moles to find the molar mass (512.34 g/mol in this instance). The molecular formula is a whole-number multiple of the empirical formula. Divide the molar mass by the empirical formula mass to find the multiple (n). If the number is not an integer, double-check calculations or experimental data.

In this exercise, the empirical formula mass matches the calculated molar mass of \(\text{C}_{18}\text{F}_{15}\text{B}\), confirming it as both the molecular and empirical formula.

Polymerization Initiators

Polymerization initiators like tris(pentafluorophenyl)borane (BARF) play a vital role in the formation of polymers. These substances start or "initiate" the polymerization process through chemical reactions with monomers like ethylene or propylene.

In polymerization, initiators decompose into reactive species that interact with monomers, creating long chains of repeating units. BARF is particularly useful in coordinating with transition metal catalysts to enhance the reaction efficiency.

Key aspects of polymerization initiators include:

• Reactivity: They must have the ability to react quickly and effectively with monomers.
• Stability: Initiators should remain stable until polymerization conditions are met.
• Compatibility: They need to work well with specific catalysts and conditions of the reaction.

Understanding initiators and their roles helps chemists design better polymerization processes, which lead to new materials with desirable properties, such as plastics or resins, enhancing numerous industrial applications.