Chapter 5, Problem 5.54QP

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 5.54QP

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

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 $\text{0}\text{.00168}\text{mol}$

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 $\text{138}\text{g/mol}$

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.