Chapter 11, Problem 11.84QP

Interpretation Introduction

Interpretation:

The molecular formula of the given compound has to be calculated.

Concept Introduction:

Ideal gas is the most usually used form of the ideal gas equation, which describes the relationship among the four variables P, V, n, and T.  An ideal gas is a hypothetical sample of gas whose pressure-volume-temperature behavior is predicted accurately by the ideal gas equation.

$\text{PV = nRT}$

The molar mass formula

$\text{M}\text{=}\frac{\text{mass}\text{(in}\text{g)}}{\text{mol}}$

Answer to Problem 11.84QP

Hence, the molecular formula is ${\left({\text{PF}}_{\text{2}}\right)}_{\text{2}}\text{ or}{\text{P}}_{\text{2}}{\text{F}}_{\text{4}}$

If you determine the molar mass of the gas, you will be able to determine the molecular formula from the empirical formula.

Explanation of Solution

To calculate the moles of the compound.

$\begin{array}{l}\text{n}\text{=}\frac{\text{PV}}{\text{RT}}\\ \text{n}\text{=}\frac{\left(\text{97}\text{.3}\text{mmHg}\text{×}\frac{\text{1}\text{atm}}{\text{760}\text{mmHg}}\right)\text{(0}\text{.378}\text{L)}}{\left(\text{0}\text{.0821}\frac{\text{L}\text{.}\text{atm}}{\text{K}\text{.}\text{mol}}\right)\text{(77+}\text{273)}\text{K}}\text{=}\text{0}\text{.00168}\text{mol}\end{array}$

The mole of the compound is calculated by plugging in the values of the given volume,

Pressure and temperature.  The moles of the compound was found to be $\text{0}\text{.00168}\text{mol}$

To calculate the molar mass

$\text{M}\text{=}\frac{\text{mass}}{\text{mol}}\text{=}\frac{\text{0}\text{.2324}\text{g}}{\text{0}\text{.00168}\text{mol}}=\text{138}\text{g/mol}$

The molar mass is calculated by plugging in the given weight of compound and moles of compound.  The molar mass was found to be $\text{138}\text{g/mol}$

To calculate the empirical formula, first we need to find the mass of $\text{F in 0}\text{.2631 g of CaF}\text{.}$

$\text{0}\text{.2631}\text{g}{\text{CaF}}_{\text{2}}\text{×}\frac{\text{1}\text{mol}{\text{CaF}}_{\text{2}}}{\text{78}\text{.08}\text{g}{\text{CaF}}_{\text{2}}}\text{×}\frac{\text{2}\text{mol}\text{F}}{\text{1}\text{mol}{\text{CaF}}_{\text{2}}}\text{=}\frac{\text{19}\text{.00}\text{g}\text{F}}{\text{1}\text{mol}\text{F}}\text{=}\text{0}\text{.1280}\text{g}\text{F}$

Since the compound only contains P and F, the mass of P in the 0.2324 g sample is:

$\text{0}\text{.2324 g}\text{-}\text{0}\text{.1280 g = 0}\text{.1044 g P}$

The mass of $\text{F in 0}\text{.2631 g of CaF}$ is calculated by plugging in the given weight of compound and molecular weight of the compound.  The molar mass was found to be $\text{0}\text{.1280}\text{g}\text{F}$ and $\text{0}\text{.1044 g P}$

We can convert masses of P and F to moles of each substance.

$\text{?}\text{mol}\text{P}\text{=}\text{0}\text{.1044}\text{g}\text{P}\text{×}\frac{\text{1}\text{mol}\text{P}}{\text{30}\text{.97}\text{g}\text{P}}\text{=}\text{0}\text{.003371}\text{mol}\text{P}$

$\text{?}\text{mol}\text{F}\text{=}\text{0}\text{.1280}\text{g}\text{F}\text{×}\frac{\text{1}\text{mol}\text{F}}{\text{19}\text{.00}\text{g}\text{F}}\text{=}\text{0}\text{.006737}\text{mol}\text{F}$

The moles of each substance are calculated by plugging in the given weight of compound and molecular weight of compound.  The moles of each substance was found to be $\text{0}\text{.003371}\text{mol}\text{P}$ $\text{and}\text{0}\text{.006737}\text{mol}\text{F}$

Thus, we arrive at the formula $\text{P0}\text{.003371F0}\text{.006737}$. Dividing by the smallest number of moles (0.003371mol) gives the empirical formula ${\text{PF}}_{\text{2}}\text{.}$

To determine the molecular formula, divide the molar mass by the empirical mass.

$\frac{\text{molar}\text{mass}}{\text{empirical}\text{molar}\text{mass}}\text{=}\frac{\text{138}\text{g}}{\text{68}\text{.97}\text{g}}\approx \text{2}$

Hence, the molecular formula is ${\left({\text{PF}}_{\text{2}}\right)}_{\text{2}}\text{ or}{\text{P}}_{\text{2}}{\text{F}}_{\text{4}}$

Conclusion

The molecular formula of the given compound was calculated.