Question

The empirical formula of a gaseous fluorocarbon is${\mathbf{CF}}_{\mathbf{2}}$ . At a certain temperature and pressure, a 1-L volume holds 8.93 g of this fluorocarbon, whereas under the same conditions, the 1-L volume holds only 1.70 g gaseous fluorine$\left(F_2\right)$. Determine the molecular formula of this compound.

Short answer

The molecular formula of the compound is${\left({\text{CF}}_{\text{2}}\right)}_{\text{4}}$ or ${\text{C}}_{\text{4}}{\text{F}}_{\text{8}}$

Step-by-step solution

Molecular formula

Molecular formulas are connected with gram molecular masses. These are whole number multiples of the interrelated empirical formula mass.

Determining the molecular formula of the compounds

The empirical weight of fluorocarbon${\text{CF}}_{\text{2}}$is 50 g.

The molecular formula is${\left({\text{CF}}_{\text{2}}\right)}_{\text{n}}$.

Here n is the ratio of;

$\text{n =}\frac{\text{molecular weight}}{\text{empirical weight}}$

Under the same conditions of T and P, all the gases have the equal number of molecules.Hence, the number of moles of${\text{F}}_{\text{2}}$is equal to the number of moles of gaseous fluorocarbon.

The moles of fluorocarbon in its empirical formula;

$$\begin{gathered} {\text{Moles ofF}}_{\text{2}}\text{=}\frac{\text{1.70g}}{{\text{38gmol}}^{\text{- 1}}} \\ \text{= 0.045mol} \end{gathered}$$

Hence, the number of moles of fluorocarbon with the empirical formula of is 0.045 mol.

Now, themolar massof${\text{CF}}_{\text{2}}$:

$$\begin{gathered} {\text{Moles of CF}}_{\text{2}}\text{=}\frac{\text{Mass of the gas}}{{\text{Molar mass of CF}}_{\text{2}}} \\ \text{0.045mol =}\frac{\text{Mass of the gas}}{{\text{Molar mass of CF}}_{\text{2}}} \end{gathered}$$

The molar mass of${\text{CF}}_{\text{2}}$is ${\text{198.44gmol}}^{\text{- 1}}$.

The given empirical formula is${\text{CF}}_{\text{2}}$. The mass of${\text{CF}}_{\text{2}}$is 50 g.

Therefore,

$$\begin{gathered} \text{n =}\frac{\text{molecular weight}}{\text{empirical weight}} \\ \text{n =}\frac{\text{198.44g/mol}}{\text{50 g/mol}} \\ \text{n}\simeq 4 \end{gathered}$$

The molar mass,is ${\text{198.44gmol}}^{\text{- 1}}$approximately 4 times the empirical formula mass which is 50 g/mol.

Hence, there are four${\text{CF}}_{\text{2}}$units in the molecular formula of fluorocarbon.

The molecular formula of the compound is${\left({\text{CF}}_{\text{2}}\right)}_{\text{4}}$or${\text{C}}_{\text{4}}{\text{F}}_{\text{8}}$.