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 C, F, and B; it is 42.23% C and 55.66% F by mass.

a. What is the empirical formula of BARF?

b. A 2.251-g sample of BARF dissolved in 347.0 mL of solution produces a 0.01267 Msolution. What is the molecular formula of BARF?

Short answer

a. The empirical formula is C₁₈F₁₅B.

b. The molecular formula is C₁₈F₁₅B.

Step-by-step solution

Determination of molecular formula of BARF

The mass percent of B is

$\mathbf{100}\mathbf{\%}\mathbf{-}\mathbf{(}\mathbf{42}\mathbf{.}\mathbf{23}\mathbf{\%}\mathbf{+}\mathbf{55}\mathbf{.}\mathbf{66}\mathbf{\%}\mathbf{)}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{11}\mathbf{\%}$

A 100 g sample of BARF contains 42.23 g C, 55.66 g F, and 2.11 g B.

Now moles of C, F, and B are:

$$\begin{gathered} \mathbf{moles}\mathbf{}\mathbf{of}\mathbf{}\mathbf{C}\mathbf{}\mathbf{=}\frac{\mathbf{mass}}{\mathbf{molar}\mathbf{}\mathbf{mass}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{42}\mathbf{.}\mathbf{23}\mathbf{g}}{\mathbf{12}\mathbf{.}\mathbf{01}\mathbf{}\mathbf{g}\mathbf{/}\mathbf{mol}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{52}\mathbf{}\mathbf{mol} \end{gathered}$$

$$\begin{gathered} \mathbf{moles}\mathbf{}\mathbf{of}\mathbf{}\mathbf{F}\mathbf{}\mathbf{=}\frac{\mathbf{55}\mathbf{.}\mathbf{66}\mathbf{}\mathbf{g}}{\mathbf{18}\mathbf{.}\mathbf{99}\mathbf{}\mathbf{g}\mathbf{/}\mathbf{mol}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{93}\mathbf{mol} \end{gathered}$$

$$\begin{gathered} \mathbf{moles}\mathbf{}\mathbf{of}\mathbf{}\mathbf{B}\mathbf{}\mathbf{=}\frac{\mathbf{2}\mathbf{.}\mathbf{11}\mathbf{}\mathbf{g}}{\mathbf{10}\mathbf{.}\mathbf{81}\mathbf{}\mathbf{g}\mathbf{/}\mathbf{mol}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{195}\mathbf{mol} \end{gathered}$$

Divide all by the least number, that is, 0.195 mol.

$$\begin{gathered} \mathbf{C}\mathbf{}\mathbf{=}\frac{\mathbf{3}\mathbf{.}\mathbf{52}\mathbf{mol}}{\mathbf{0}\mathbf{.}\mathbf{195}\mathbf{mol}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{18} \end{gathered}$$

$$\begin{gathered} \mathbf{F}\mathbf{}\mathbf{=}\frac{\mathbf{2}\mathbf{.}\mathbf{93}\mathbf{mol}}{\mathbf{0}\mathbf{.}\mathbf{195}\mathbf{mol}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{15} \end{gathered}$$

$$\begin{gathered} \mathbf{B}\mathbf{}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{195}\mathbf{mol}}{\mathbf{0}\mathbf{.}\mathbf{195}\mathbf{mol}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{1} \end{gathered}$$

Thus, the empirical formula is C₁₈F₁₅B.

Determination of molecular formula

Moles of BARF are:

$$\begin{gathered} \mathbf{Moles}\mathbf{=}\mathbf{molarity}\mathbf{\times }\mathbf{volume} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{01267}\mathbf{\times }\mathbf{0}\mathbf{.}\mathbf{347}\mathbf{L} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0044}\mathbf{mol} \end{gathered}$$

Now, the molar mass of BARF is

$$\begin{gathered} \mathbf{molar}\mathbf{}\mathbf{mass}\mathbf{=}\frac{\mathbf{mass}}{\mathbf{moles}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{2}\mathbf{.}\mathbf{251}\mathbf{g}}{\mathbf{0}\mathbf{.}\mathbf{0044}\mathbf{mol}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{511}\mathbf{.}\mathbf{59}\mathbf{}\mathbf{g}\mathbf{/}\mathbf{mol} \end{gathered}$$

The empirical formula mass is:

$$\begin{gathered} \mathbf{embrical}\mathbf{}\mathbf{formula}\mathbf{}\mathbf{=}\left(18\times 12.01g/\mathrm{mol}\right)\mathbf{+}\left(15\times 18.99g/\mathrm{mol}\right)\mathbf{+}\mathbf{(}\mathbf{1}\mathbf{\times }\mathbf{10}\mathbf{.}\mathbf{81}\mathbf{}\mathbf{g}\mathbf{/}\mathbf{mol}\mathbf{)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\left(216.18+284.99+10.81\right)\mathbf{g}\mathbf{/}\mathbf{mol} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{511}\mathbf{.}\mathbf{98}\mathbf{}\mathbf{g}\mathbf{/}\mathbf{mol} \end{gathered}$$

Now divide the molar mass by the empirical formula mass.

$\frac{\mathbf{511}\mathbf{.}\mathbf{59}\mathbf{}\mathbf{g}\mathbf{/}\mathbf{mol}}{\mathbf{511}\mathbf{.}\mathbf{98}\mathbf{}\mathbf{g}\mathbf{/}\mathbf{mol}}\mathbf{=}\mathbf{1}$

Thus, the molecular formula is C₁₈F₁₅B.