Problem 142 - Chemistry An Atoms First Approach

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 \(42.23 \%\) C and \(55.66 \%\) F by mass. a. What is the empirical formula of BARF? b. A \(2.251-\mathrm{g}\) sample of BARF dissolved in \(347.0 \mathrm{mL}\) of solution produces a 0.01267-M solution. What is the molecular formula of BARF?

Step-by-Step Solution

Answer

The empirical formula of BARF is CF, and its molecular formula is C₁₆F₁₆B.

Step 1: Calculate the moles of each element from mass percentages

To find the empirical formula, assume that we have a 100g sample of BARF. In such a sample, we have 42.23g of C, and 55.66g of F. We can use their molar masses to find the moles of each element: Moles of C: \(\frac{42.23\,\mathrm{g}}{12.01\,\mathrm{g/mol}} \approx 3.517\,\mathrm{mol}\) Moles of F: \(\frac{55.66\,\mathrm{g}}{18.998\,\mathrm{g/mol}} \approx 2.929\,\mathrm{mol}\)

Step 2: Calculate the mole ratio

To find the empirical formula, we need to find the mole ratio between C and F. We achieve this by dividing each value by the smallest mole value: Ratio of moles of C: \(\frac{3.517}{2.929} \approx 1.2 : 1\) Ratio of moles of F: \(\frac{2.929}{2.929} \approx 1 : 1\) Since the ratio is approximately 1:1, the empirical formula is CF.

Step 3: Find the molar mass of BARF

We are given the information that a 2.251g sample of BARF dissolved in 347.0 mL of solution produces a 0.01267-M solution. Let's use these data to find the molar mass of BARF: First, calculate the moles of BARF in the solution: Moles of BARF = 0.01267 mol/L * 0.347 L = 0.00439609 mol Now, find the molar mass of BARF: Molar Mass of BARF = \(\frac{2.251\,\mathrm{g}}{0.00439609\,\mathrm{mol}} \approx 512\,\mathrm{g/mol}\)

Step 4: Find the molecular formula of BARF

Now that we have the molar mass of BARF, we can find the molecular formula. First, determine the mass of the empirical formula CF: Mass of CF = 12.01g/mol (C) + 18.998g/mol (F) = 31.008g/mol Now, divide the molar mass of BARF by the mass of the empirical formula to find the number of empirical formula units: Number of CF units in BARF = \(\frac{512\,\mathrm{g/mol}}{31.008\,\mathrm{g/mol}} \approx 16.5\) Since the number of units should be a whole number, we can round this value to 16 (being the closest integer), which indicates that the molecular formula of BARF is 16 CF units. Molecular Formula of BARF = C₁₆F₁₆B (we have one B atom due to the empirical formula) Thus, the empirical formula is CF and the molecular formula is C₁₆F₁₆B.

Key Concepts

Polymerization

Polymerization is a chemical process that joins monomer units together to form a polymer, which is a larger, repeating molecular structure. This process is fundamental in creating various synthetic materials such as plastics, resins, and fibers. To explain further, imagine many small building blocks (monomers) that are linked together to construct a large and complex structure (polymer).

In the context of our BARF (Tris(pentafluorophenyl)borane) exercise, BARF is used to initiate the polymerization of small molecules like ethylene or propylene. This initiation occurs in the presence of a catalytic transition metal compound and leads to the creation of large polymeric chains. The length and properties of the resulting polymers can greatly vary, depending on the kind of monomers used and the conditions under which the polymerization is carried out.

There are several types of polymerization, including addition (also known as chain-growth) and condensation (step-growth) polymerization. BARF is typically involved in addition polymerization where the growth of polymer chains proceeds through reactions between monomers that have unsaturated bonds.

Transition Metal Catalysts

Transition metal catalysts are substances that contain a transition metal and are capable of increasing the rate of chemical reactions without being consumed by the reaction itself. They play a pivotal role in the polymerization process because they can effectively lower the activation energy needed for the reaction to proceed, thus facilitating the bonding of monomers into long polymer chains.

In our exercise, a transition metal catalysts works in conjunction with BARF to initiate polymerization. Transition metals, which are elements found in the center of the periodic table, are uniquely suited for this role due to their ability to adopt multiple oxidation states and to form stable complexes with various organic compounds.

Such catalysts are designed to be efficient and specific, often leading to polymers with desired characteristics, such as tensile strength, flexibility, or resistance to chemicals. The use of transition metal catalysts in industrial processes contributes significantly to the production of a wide array of materials that we use in our daily lives, from packaging materials to clothing fibers.

Molar Mass Calculation

Molar mass calculation is a fundamental concept in chemistry that involves determining the mass of one mole of a substance. The molar mass is expressed in grams per mole (g/mol) and is calculated by summing the atomic masses of all atoms in a molecule as listed on the periodic table of elements.

For instance, to calculate the molar mass of BARF's empirical formula (CF), you add the molar mass of carbon (C), which is approximately 12.01 g/mol, to the molar mass of fluorine (F), roughly 18.998 g/mol, giving 31.008 g/mol. Understanding how to calculate molar mass is essential for converting between the mass of a sample and the amount of substance in moles, which is needed for stoichiometric calculations and determining empirical and molecular formulas.

In our exercise example, after establishing the empirical formula CF for BARF, we determined its molar mass from a dissolved sample to find the molecular formula. The molar mass calculation process, as shown in the step by step solution, is crucial in quantitatively describing the composition of compounds and plays a vital role in chemical analysis and synthesis.