Chapter 5, Problem 5.90PAE

Interpretation Introduction

Interpretation:

You work in a semiconductor production plant that relies on several chlorofluorocarbons in its manufacturing process. One day, you find an unlabeled gas cylinder, and you are assigned to figure out what is in the tank. First, you fill a 1.000-L flask with the gas. At a pressure of 250.0 torr and a temperature of 25.00⁰C, you determine that the mass of the gas in the flask is 2.2980 g. Then, you send the flask to an outside lab for elemental analysis, and they report that the gas contains 14.05% C, 44.46% F, and 41.48% Cl by mass. The molecular formula of this gas should be identified.

Concept introduction:

An ideal gas which is known as the perfect gas is a gas whose volume V, Pressure P and temperature T are related through the ideal gas laws: -

$\text{PV = nRT}$

Here,

n = Number of moles of the gas

R = Universal gas constant

T = Temperature

P = Pressure

Ideal gases are described as the molecules which have negligible size but have an average molar kinetic energy which is dependent on the temperature. When temperature is low most of the gases behave like ideal gases and the ideal gas law might be applied to them.

Answer to Problem 5.90PAE

Solution:

${\text{C}}_{\text{2}}{\text{F}}_{\text{4}}{\text{Cl}}_{\text{2}}$

Given:

Volume of gas taken= 1.000L

Pressure = 250 torr = 0.329 atm

Temperature = (25+273) K = 298K

Mass of the gas = 2.298g

Percentage of elements present in gas:

$\begin{array}{l}\text{C}=\text{14}.0\text{5}\%\\ \text{F}=\text{44}.\text{46}\%\\ \text{Cl}=\text{41}.\text{48}\%\end{array}$

Explanation of Solution

The ideal gas equation is as follows:

$\text{PV}=\text{nRT}$

Or,

$\text{PV}=\left(\frac{\text{m}}{\text{M}}\right)\text{ RT}$

Or,

$\text{M}=\text{mRT}/\text{PV}$

$\begin{array}{l}=\frac{\left(2.298\text{ g}\right)\left(0.0821\text{ atm L/mol K}\right)\left(298\text{ K}\right)}{\left(\text{0}\text{.329 atm}\right)\left(\text{1 L}\right)}\\ =170.88\text{ g/mol}\end{array}$

Now, the percent composition of the given elements are:

$\begin{array}{l}\text{C}=\text{14}.0\text{5}\%\\ \text{F}=\text{44}.\text{46}\%\\ \text{Cl}=\text{41}.\text{48}\%\end{array}$

Divide all with their molar masses to get the number of atoms of each element as follows

$\begin{array}{l}\text{C = }\frac{\text{14}\text{.02}}{\text{12}}\text{ = 1}\text{.17}\hfill \\ \text{F = }\frac{\text{44}\text{.46}}{\text{18}\text{.9}}\text{ = 2}\text{.35}\hfill \\ \text{Cl = }\frac{\text{41}\text{.48}}{\text{35}\text{.5}}\text{ = 1}\text{.17}\hfill \end{array}$

Now, divide with the smallest number to calculate the empirical formula:

$\begin{array}{l}\text{C = }\frac{\text{1}\text{.17}}{\text{1}\text{.17}}\text{ =1 }\hfill \\ \text{F = }\frac{\text{2}\text{.35}}{\text{1}\text{.17}}\text{ = 2}\hfill \\ \text{Cl = }\frac{\text{1}\text{.17}}{\text{1}\text{.17}}\text{ = 1}\hfill \end{array}$

So, the empirical formula is ${\text{C}}_{\text{1}}{\text{F}}_{\text{2}}{\text{Cl}}_{\text{1}}$.

Calculate empirical formula mass as follows:

$\begin{array}{c}{\text{M}}_{\text{EF}}=\left(1\right)\left(12\text{ g/mol}\right)+\left(2\right)\left(18.9\text{ g/mol}\right)+\left(1\right)\left(35.5\text{ g/mol}\right)\\ =85.3\text{ g/mol}\end{array}$

Now,

$\text{n}\left({\text{M}}_{\text{EF}}\right){\text{=M}}_{\text{MF}}$

Or,

$\text{n=}\frac{{\text{M}}_{\text{MF}}}{{\text{M}}_{\text{EF}}}$

Where, ${\text{M}}_{\text{EF}}$ is empirical formula mass and ${\text{M}}_{\text{MF}}$ is molecular formula mass.

Thus,

$\begin{array}{c}\text{n}=\frac{\text{170}\text{.8 g/mol}}{\text{85}\text{.3 g/mol}}\\ \text{=2}\text{.002}\\ \approx \text{2}\end{array}$

Therefore, the molecular formula is ${\text{C}}_{\text{2}}{\text{F}}_{\text{4}}{\text{Cl}}_{\text{2}}$.

Conclusion

Ideal gases are described as the molecules which have negligible size but have an average molar kinetic energy which is dependent on the temperature. Using the ideal gas equation, the calculated molecular formula of gas is ${\text{C}}_{\text{2}}{\text{F}}_{\text{4}}{\text{Cl}}_{\text{2}}$.