Chapter 10, Problem 58QP

A compound of P and F was analyzed as follows: Heating 0.2324 g of the compound in a ${\text{378-cm}}^{\text{3}}$ container turned all of it to gas. which had a pressure of 97.3 mmHg at $\text{ 77°C}$. Then the gas was mixed with calcium chloride solution, which converted all the F to 0.2631 g of ${\text{CaF}}_{\text{2}}$. Determine the molecular formula of the compound.

Interpretation Introduction

Interpretation:

The molecular formula of the gas is to be calculated.

Concept introduction:

The ideal gas equation elaborates the physical properties of gases by relating the pressure, volume, temperature, and number of moles linked with each other with the help of four gas laws. This can be shown by:

$PV=nRT$

Here, $n$ is number of moles, $R$ is gas constant, $T$ is temperature, $V$ is volume, and $P$ is pressure.

The formula for conversion of temperature from degree Celsius to kelvin is represented as

$T\left(\text{K}\right)=T\left(°\text{C}\right)+273.15$

Answer to Problem 58QP

Solution: ${\text{P}}_{\text{2}}{\text{F}}_{\text{4}}$

Explanation of Solution

Given information:

Pressure $P=97.3\text{ mmHg}$

Temperature $T={77}^{\text{o}}\text{C}$

Volume is $V=378{\text{ cm}}^3$

The pressure is $97.3\text{ mmHg}$.

In 1mmHg, the pressure is $1/760\text{ atm}$.

Convert mmHg to atm as follows:

$\begin{array}{c}97.3\text{ mmHg}=\left(97.3\cancel{\text{mmHg}}\right)\left(\frac{1/760\text{ atm}}{1\cancel{\text{mmHg}}}\right)\\ =0.128\text{ atm}\end{array}$

The volume is $378{\text{ m}}^3$.

In 1cubic centimeter, there are ${10}^{-3}\text{ L}$.

Convert cubic centimeter to liter as follows:

$\begin{array}{c}378{\text{ cm}}^3=\left(378\cancel{{\text{cm}}^3}\right)\left(\frac{{10}^{-3}\text{ L}}{1\cancel{{\text{cm}}^3}}\right)\\ =0.378\text{ L}\end{array}$

The temperature is 77°C.

The conversion of temperature from degree Celsius to kelvin can be done by using the formula given below:

$\begin{array}{c}T\left(\text{K}\right)=T\left(°\text{C}\right)+273.15\\ =\left(77+273.15\right)\\ =350.15\text{ K}\end{array}$

The standard form of theideal gas equation is as follows:

$PV=nRT$

Rearrange the above equation for the number of moles as follows:

$n=\frac{RT}{PV}$

Substitute $0.128\text{ atm}$ for $P$, $350.15\text{ K}$ for $T$, $0.08206\text{ L}\text{.atm}/\text{K}\text{.mol}$ for $R$, and $0.378\text{ L}$ for $V$ in the above equation

$\begin{array}{c}n=\frac{\left(0.128\cancel{\text{atm}}\right)\left(\text{0}\text{.378 }\cancel{\text{L}}\right)}{\left(0.08206\cancel{\text{L}}\text{.}\cancel{\text{atm}}/\cancel{\text{K}}\text{.mol}\right)\left(350.15\cancel{\text{K}}\right)}\\ =\frac{0.048}{28.73}\text{ mol}\\ =0.00168\text{ mol}\\ =1.68\times {\text{10}}^{-3}\text{ mol}\end{array}$ $\begin{array}{c}n=\frac{\left(0.128\text{ atm}\right)\left(\text{0}\text{.378 L}\right)}{\left(0.08206\text{ L}\text{.atm}/\text{K}\text{.mol}\right)\left(350.15\text{ K}\right)}\\ =\frac{0.048}{28.73}\text{ mol}\\ =0.00168\text{ mol}\\ =1.68\times {\text{10}}^{-3}\text{ mol}\end{array}$

Calculate the molar mass as follows:

$M=\frac{m}{n}$

Substitute $1.68\times {\text{10}}^{-3}$ for $n$ and $0.2324\text{ g}$ for $m$ in the above equation

$\begin{array}{c}M=\frac{0.2324\text{ g}}{1.68\times {\text{10}}^{-3}\text{ mol}}\\ =138\text{ g}/\text{mol}\end{array}$

The molar mass of fluorine is $19\text{ g}/\text{mol}$.

Now, the mass of F in $0.2631\text{ g}$ ${\text{CaF}}_2$ is

$\begin{array}{c}{m}_{\text{F}}=\frac{0.2631{\text{ g CaF}}_2}{78.08\cancel{\text{g}/\text{mol}}{\text{ CaF}}_2}\times 2\times 19\cancel{\text{g}/\text{mol}}\text{ F}\\ =\text{0}\text{.0033}\times 38\text{ g}\\ =\text{0}.1280\text{ g}\end{array}$ $\begin{array}{c}{m}_{\text{F}}=\frac{0.2631{\text{ g CaF}}_2}{78.08\text{ g}/{\text{mol CaF}}_2}\times 2\text{ mol}\times 19\text{ g}/\text{mol F}\\ =\text{0}\text{.0033}\times 38\text{ g}\\ =\text{0}.1280\text{ g}\end{array}$

The compound has P and H in the $0.2324\text{ g}$ sample. So, the mass of P is

$\begin{array}{c}{m}_P=0.2321\text{ g}-0.1280\text{ g}\\ =\text{0}\text{.1044 g}\end{array}$

Calculate the number of moles of fluorine as follows:

$n_{\text{F}}=\frac{m_{\text{F}}}{M_{\text{F}}}$

Substitute $0.1280\text{ g}$ for $m_{\text{F}}$ and $19\text{ g}/\text{mol}$ for $M_{\text{F}}$ in the above equation

$\begin{array}{c}{n}_{\text{F}}=\frac{0.1280\cancel{\text{g}}}{19\text{ g}\cancel{/}\text{mol}}\\ =0.006737\text{ mol}\end{array}$ $\begin{array}{c}{n}_{\text{F}}=\frac{0.1280\text{ g}}{19\text{ g}/\text{mol}}\\ =0.006737\text{ mol}\end{array}$

The molar mass of phosphorous is $\text{31 g}/\text{mol}$.

Calculate the number of moles of phosphorous as follows:

$n_{\text{P}}=\frac{m_{\text{P}}}{M_{\text{P}}}$

Substitute $0.1044\text{ g}$ for $m_{\text{P}}$ and $\text{31 g}/\text{mol}$ for $M_{\text{P}}$ in the above equation

$\begin{array}{c}{n}_{\text{P}}=\frac{0.1044\cancel{\text{g}}}{\text{31 }\cancel{\text{g}}/\text{mol}}\\ =0.00337\text{ mol}\end{array}$ $\begin{array}{c}{n}_{\text{P}}=\frac{0.1044\text{ g}}{\text{31 g}/\text{mol}}\\ =0.00337\text{ mol}\end{array}$

So, the compound can be written as

$P_{0.00337}{F}_{0.006737}$

As the empirical formula is always an integer, the subscript can be simplified by dividing the formula with $0.00337$. On simplification, the empirical formula of the compound is ${\text{PF}}_2$.

Calculate the Empirical formula massas follows:

$31\text{ g}/\text{mol}+38\text{ g}/\text{mol}=69\text{ g}/\text{mol}$

Calculate the number of moles as follows:

$n=\frac{\text{Molar mass}}{\text{Empirical formula mass}}$

Substitute $\text{138 g}/\text{mol}$ for molar mass and $69\text{ g}/\text{mol}$ for empirical formula mass in the above equation

$\begin{array}{c}n=\frac{138\cancel{\text{g}/\text{mol}}}{69\cancel{\text{g}/\text{mol}}}\\ \approx 2\end{array}$ $\begin{array}{c}n=\frac{138\text{ g}/\text{mol}}{69\text{ g}/\text{mol}}\\ \approx 2\end{array}$

The molecular formula is always a whole number multiple.

So, the required molecular formula is

$\begin{array}{c}\text{Molecular formula}=2\times {\text{PF}}_2\\ ={\left({\text{PF}}_2\right)}_2\\ ={\text{P}}_2{\text{F}}_4\end{array}$

Therefore, the molecular formula of the gas is ${\text{P}}_{\text{2}}{\text{F}}_{\text{4}}$.

Conclusion

The molecular formula of the gas is ${\text{P}}_{\text{2}}{\text{F}}_{\text{4}}$.