Question

A 100.0 mL flask contains 0.193 g of a volatile oxide of nitrogen. The pressure in the flask is \(760 \mathrm{mmHg}\) at \(17^{\circ} \mathrm{C}\) a. Calculate the density of the gas. b. Is the gas \(\mathrm{NO}, \mathrm{NO}_{2},\) or \(\mathrm{N}_{2} \mathrm{O}_{5} ?\)

Short answer

Question: Calculate the density of an unknown nitrogen oxide gas and identify which nitrogen oxide gas is present in a flask containing the following conditions: a mass of 0.193 g, a volume of 100.0 mL, a pressure of 760 mmHg, and a temperature of 17°C. Answer: a. The density of the gas is approximately 0.0273 g/mL. b. The gas in the flask is NO2.

Step-by-step solution

Write the ideal gas equation with density

Recall that the ideal gas equation can be written as follows: \(PV=nRT\), where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is the temperature. Now, let's write the equation in terms of density. The density of a gas, denoted as \(\rho\), is the mass (\(m\)) divided by the volume (\(V\)). Also, we know that the number of moles can be written as \(n=\frac{m}{M}\), where \(M\) is the molar mass of the gas. Substituting these expressions into the ideal gas equation, we get: \(P(\frac{m}{\rho}) = (\frac{m}{M})RT\) After rearranging the equation for \(\rho\) (density), we have: \(\rho = \frac{m}{V} = \frac{PM}{RT}\)

Convert the given values to appropriate units

Before plugging the values into the equation, let's ensure that they are in appropriate units. The pressure and the temperature should be converted to SI units: Pressure: \(760\, \mathrm{mmHg} \times \frac{101.325\, \mathrm{kPa}}{760\, \mathrm{mmHg}} = 101.325\, \mathrm{kPa}\) Temperature: \(17^{\circ} \mathrm{C} + 273.15 = 290.15\, \mathrm{K}\) We can use these values in the density equation.

Calculate the density of the gas

Using the density equation derived in step 1 and the values from step 2, we can calculate the density of the gas: \(\rho = \frac{PM}{RT} = \frac{101.325\, \mathrm{kPa} \times 0.193\, \mathrm{g}}{8.314\, \mathrm{\frac{J}{mol\,K}} \times 290.15\, \mathrm{K}\times 100.0\, \mathrm{mL} \times \frac{1\, \mathrm{L}}{1000\, \mathrm{mL}}}\) Upon calculation, we find the density of the gas to be: \(\rho ≈ 0.0273\, \frac{\mathrm{g}}{\mathrm{mL}}\)

Compare the calculated density to the known densities of \(\mathrm{NO}\), \(\mathrm{NO}_{2}\), and \(\mathrm{N}_{2}\mathrm{O}_{5}\)

Now we need to identify the nitrogen oxide by comparing the calculated density with the known densities of \(\mathrm{NO}\), \(\mathrm{NO}_{2}\), and \(\mathrm{N}_{2}\mathrm{O}_{5}\). First, let's calculate the molar mass for each of these: \(\mathrm{NO}\): \(M_{\mathrm{NO}} = (14.0067+15.9994)\, \mathrm{\frac{g}{mol}} = 30.0061\, \mathrm{\frac{g}{mol}}\) \(\mathrm{NO}_{2}\): \(M_{\mathrm{NO}_{2}} = (14.0067+2\times15.9994)\, \mathrm{\frac{g}{mol}} = 46.0055\, \mathrm{\frac{g}{mol}}\) \(\mathrm{N}_{2}\mathrm{O}_{5}\): \(M_{\mathrm{N}_{2}\mathrm{O}_{5}} = (2\times14.0067+5\times15.9994)\, \mathrm{\frac{g}{mol}} = 108.0104\, \mathrm{\frac{g}{mol}}\) Now, we can calculate their theoretical densities at the given pressure and temperature using the ideal gas equation and the calculated molar masses: \(\rho_{\mathrm{NO}} = \frac{101.325\, \mathrm{kPa} \times 30.0061\, \mathrm{\frac{g}{mol}}}{8.314\, \mathrm{\frac{J}{mol\,K}} \times 290.15\, \mathrm{K}} = 0.0128\, \frac{\mathrm{g}}{\mathrm{mL}}\) \(\rho_{\mathrm{NO}_{2}} = \frac{101.325\, \mathrm{kPa} \times 46.0055\, \mathrm{\frac{g}{mol}}}{8.314\, \mathrm{\frac{J}{mol\,K}} \times 290.15\, \mathrm{K}} = 0.0197\, \frac{\mathrm{g}}{\mathrm{mL}}\) \(\rho_{\mathrm{N}_{2}\mathrm{O}_{5}} = \frac{101.325\, \mathrm{kPa} \times 108.0104\, \mathrm{\frac{g}{mol}}}{8.314\, \mathrm{\frac{J}{mol\,K}} \times 290.15\, \mathrm{K}} = 0.0442\, \frac{\mathrm{g}}{\mathrm{mL}}\) Comparing these values to the calculated density, we find that the density of the gas in the flask is closest to the density of \(\mathrm{NO}_{2}\) (0.0197 g/mL). Therefore, the gas in the flask is \(\mathrm{NO}_{2}\). #Results# a. The density of the gas is approximately \(0.0273\, \frac{\mathrm{g}}{\mathrm{mL}}\). b. The gas in the flask is \(\mathrm{NO}_{2}\).

Key concepts

Gas Density Calculation

The concept of gas density involves determining the mass a specific volume of gas occupies. Unlike solids or liquids, gases spread out to fill their containers completely, making density calculations a bit trickier. Density is defined as mass divided by volume, often represented as \( \rho = \frac{m}{V} \). To calculate the gas density using the ideal gas law, you can use the formula \( \rho = \frac{PM}{RT} \) where:

• \( P \) is the pressure of the gas.
• \( V \) is the volume.
• \( M \) is the molar mass.
• \( R \) is the ideal gas constant \(8.314 \frac{J}{mol\cdot K}\).
• \( T \) is the temperature in Kelvin.

This adaptation of the ideal gas law links density directly with measurable factors like pressure and temperature, as well as the innate chemical property of molar mass connected to the gas itself. Knowing this equation helps understand how gases behave under different conditions and allows for the identification of unknown gases based on their physical properties.

Molar Mass Determination

The molar mass of a gas is fundamental in linking its microscopic and macroscopic properties. It's the mass of a given substance (element or compound) divided by the amount of substance, measured in moles. The process to determine it often involves rearranging the ideal gas law and making use of the calculated densities and experimental conditions.
For this, we utilize the equation: \( \frac{PM}{RT} = \rho \). From this, you can rearrange to solve for molar mass: \( M = \frac{\rho RT}{P} \).
By measuring densities at controlled temperatures and pressures, one can use known gases to calibrate methods for often more complicated or unisolable compounds. For example, knowing the densities and other conditions allows for deductive reasoning to determine which compound matches the measured properties, as seen when comparing potential candidates like \( \mathrm{NO} \), \( \mathrm{NO}_2 \), and \( \mathrm{N}_2\mathrm{O}_5 \).
Understanding and utilizing the molar mass aids in various scientific and industry applications—ranging from laboratory identification procedures to analyzing environmental contaminants.

Volatile Oxide Identification

Identifying a volatile oxide involves using calculated physical properties, such as density and molar mass, to conclude which chemical compound is present. In the problem observed, different nitrogen oxides (\( \mathrm{NO} \), \( \mathrm{NO}_2 \), and \( \mathrm{N}_2\mathrm{O}_5 \)) are candidates due to the nitrogen content. Bridging gas behavior with chemical identity, one compares measured values of gas density with theoretical densities calculated using known molar mass and conditions.
Using the theoretical density calculation formula \( \rho = \frac{PM}{RT} \), determined from the known properties of potential gas candidates, helps indicate the correct oxide. In the specific task, omparing the given density of the volatile oxide—it aligns most closely with \( \mathrm{NO}_2 \), indicating its identity.
This process sharpens skills in analytical chemistry, promoting a systematic approach to solving problems related to gas-phase reactions, mixture analyses, and understanding compound behaviors in gaseous environments.