Question

Chlorine dioxide, \(\mathrm{ClO}_{2}\), reacts with fluorine to give a new gas that contains \(\mathrm{Cl}, \mathrm{O},\) and \(\mathrm{F}\). In an experiment, you find that \(0.150 \mathrm{g}\) of this new gas has a pressure of \(17.2 \mathrm{mm}\) Hg in a 1850 -mL flask at \(21^{\circ} \mathrm{C} .\) What is the identity of the unknown gas?

Short answer

The gas is likely chlorine trifluoride (\(\text{ClF}_3\)).

Step-by-step solution

Convert Units of Pressure

First, convert the pressure from mm Hg to atmospheres (atm), which is more commonly used in gas calculations. Use the conversion factor: \(1\, \text{atm} = 760\, \text{mm Hg}\). \[ \text{Pressure in atm} = \frac{17.2\, \text{mm Hg}}{760\, \text{mm Hg/atm}} \approx 0.0226\, \text{atm} \]

Convert Volume to Liters

Next, convert the volume of the flask from mL to liters by dividing by 1000 because there are 1000 mL in a liter. \[ \text{Volume in L} = \frac{1850\, \text{mL}}{1000} = 1.850\, \text{L} \]

Convert Temperature to Kelvin

Convert the temperature from degrees Celsius to Kelvin by adding 273.15. This is necessary for using the ideal gas law. \[ T (K) = 21^{\circ}\text{C} + 273.15 = 294.15\, \text{K} \]

Use the Ideal Gas Law

Apply the ideal gas law equation \( PV = nRT \), where \( R = 0.0821\, \text{L atm/mol K} \). Plug in the known values for pressure \( P = 0.0226\, \text{atm} \), volume \( V = 1.850\, \text{L} \), and temperature \( T = 294.15\, \text{K} \), and solve for \( n \), the number of moles. \[ n = \frac{PV}{RT} = \frac{(0.0226)(1.850)}{(0.0821)(294.15)} \approx 0.00173\, \text{mol} \]

Calculate Molar Mass

Determine the molar mass of the unknown gas using the mass (0.150 g) and the number of moles calculated. \[ \text{Molar Mass} = \frac{\text{mass of gas}}{n} = \frac{0.150\, \text{g}}{0.00173\, \text{mol}} \approx 86.7\, \text{g/mol} \]

Identify the Gas

With a molar mass of approximately 86.7 g/mol, the gas is most likely chlorine trifluoride \( \text{ClF}_3 \), which has a molar mass calculated as follows: \( \text{Cl (35.5 g/mol)} + 3 \times \text{F (19 g/mol)} = 86.5 g/mol \). The molar mass fits the experimental value closely.

Key concepts

Ideal Gas Law

The ideal gas law is a fundamental equation used in chemistry to relate the physical properties of gases. It is expressed as \( PV = nRT \), where:

• \( P \) represents the pressure of the gas.
• \( V \) is the volume occupied by the gas.
• \( n \) indicates the number of moles of gas.
• \( R \) is the universal gas constant, which is \( 0.0821\, \text{L atm/mol K} \).
• \( T \) is the temperature in Kelvin.

By using the ideal gas law, you can solve for any one of these variables if the other three are known.
In the case of our example, we calculated the pressure, volume, and temperature to find the number of moles of the unknown gas using this equation.
This equation assumes that the gas being examined behaves ideally, meaning it perfectly follows the gas laws without any deviations due to interactions or volume of the gas particles themselves.

Molar Mass Determination

Molar mass is the mass of one mole of a substance and is usually expressed in grams per mole (g/mol). It is crucial for identifying unknown gases by comparing experimental values with those of known substances.
To determine molar mass, you need both the mass of the gas and the number of moles. From the ideal gas law calculation, the number of moles \( n \) was found to be \( 0.00173 \) mol.
Using the formula \( \text{Molar Mass} = \frac{\text{mass of the gas}}{n} \), the experimental molar mass of our unknown gas was calculated as approximately \( 86.7 \, \text{g/mol} \).
This value is then compared to known values of molar masses to identify the substance.

Unit Conversions

Unit conversions are an essential skill in chemistry, especially when dealing with gas laws.
For pressure, volume, and temperature, standard units typically include atmospheres for pressure, liters for volume, and Kelvin for temperature. Here are the steps we followed in our scenario:

• Pressure was initially provided in mm Hg and converted to atmospheres, using the conversion \( 1\, \text{atm} = 760\, \text{mm Hg} \).
• Volume was converted from milliliters to liters, where \( 1\, \text{L} = 1000\, \text{mL} \).
• Temperature was converted from degrees Celsius to Kelvin by adding 273.15.

Always ensure that all the units match those required for using equations like the ideal gas law.

Chemical Reactions

In this exercise, understanding chemical reactions is key to identifying the unknown gas.
Here, chlorine dioxide reacts with fluorine to form a new gas composed of chlorine, oxygen, and fluorine. To predict the identity of this gas, we use the chemical reaction information resulting in the production of \( \text{ClF}_3 \).
The stoichiometry of the reaction helps us hypothesize that combining chlorine and fluorine would likely form chlorine trifluoride, \( \text{ClF}_3 \), given its molar mass closely matches our calculation.
Recognizing changes in reactants and products aids in identifying the composition of the newly formed compounds, thereby supporting gas identification.