Question

When an evacuated 63.8 -mL glass bulb is filled with a gas at \(22^{\circ} \mathrm{C}\) and \(747 \mathrm{mmHg}\), the bulb gains \(0.103 \mathrm{~g}\) in mass. Is the gas \(\mathrm{N}_{2}, \mathrm{Ne},\) or \(\mathrm{Ar} ?\)

Short answer

The gas is argon (\(Ar\)).

Step-by-step solution

- Convert Measurements to Suitable Units

Convert the given volume to liters: \[ 63.8 \text{ mL} = 0.0638 \text{ L} \] Convert the pressure from mmHg to atm: \[ 747 \text{ mmHg} = \frac{747}{760} \text{ atm} = 0.983 \text{ atm} \] Convert the temperature from Celsius to Kelvin: \[ 22^{\bullet} \text{C} = 22 + 273 = 295 \text{ K} \]

- Calculate the Number of Moles Using the Ideal Gas Law

Use the ideal gas law equation: \[ PV = nRT \] Where, \(P\) is the pressure in atm, \(V\) is the volume in liters, \(n\) is the number of moles, \(R\) is the ideal gas constant \(0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1}\), and \(T\) is the temperature in Kelvin. Substitute the values: \[ (0.983 \text{ atm}) \cdot (0.0638 \text{ L}) = n \cdot (0.0821) \cdot (295 \text{ K}) \] Solving for \(n\): \[ n = \frac{0.983 \times 0.0638}{0.0821 \times 295} = 0.00262 \text{ mol} \]

- Calculate the Molar Mass of the Gas

The molar mass (M) can be calculated using the mass of the gas and the number of moles: \[ M = \frac{\text{mass}}{\text{moles}} \] Given \( \text{mass} = 0.103 \text{ g} \) and \( \text{moles} = 0.00262 \text{ mol} \), \[ M = \frac{0.103 \text{ g}}{0.00262 \text{ mol}} = 39.31 \text{ g/mol} \]

- Identify the Gas by Comparing Molar Masses

Compare the calculated molar mass with the known molar masses of nitrogen (\(N_2\)), neon (\(Ne\)), and argon (\(Ar\)): \(M_{N2} = 28.01 \text{ g/mol}\), \(M_{Ne} = 20.18 \text{ g/mol}\), \(M_{Ar} = 39.95 \text{ g/mol}\). The molar mass found is 39.31 g/mol, which is closest to the molar mass of argon (\(Ar\)).

Key concepts

Gas Laws

The gas laws describe the behavior of gases in relation to pressure, volume, and temperature. The most fundamental of these is the Ideal Gas Law represented by the equation:
\[PV = nRT\]Here, P stands for pressure, V for volume, n for the number of moles, R for the ideal gas constant, and T for temperature in Kelvin. By understanding and applying this equation, we can predict how a change in one of these variables affects the others. In this exercise, we use the Ideal Gas Law to determine the number of moles of the gas in the bulb by substituting the measured values of pressure, volume, and temperature.

Molar Mass Calculation

Molar mass is the mass of one mole of a substance (usually expressed in g/mol). Calculating the molar mass involves taking the given mass and dividing it by the number of moles. For this exercise, using the equation:
\[M = \frac{\text{mass}}{\text{moles}}\]we find the molar mass of the unknown gas. Given a mass of 0.103 g and 0.00262 moles, the calculation reveals a molar mass of approximately 39.31 g/mol. This molar mass can then be used to identify the gas by comparing it to known molar masses of elements such as nitrogen, neon, and argon.

Unit Conversion

Unit conversions are crucial in scientific calculations to ensure consistency. In this exercise, we convert volume from milliliters to liters, pressure from mmHg to atmospheres, and temperature from Celsius to Kelvin. Each of these conversions is important because the Ideal Gas Law requires specific units for each variable.
Following conversions were performed:

• Volume: 63.8 mL to 0.0638 L
• Pressure: 747 mmHg to 0.983 atm
• Temperature: 22°C to 295 K

These conversions allow us to correctly plug values into the Ideal Gas Law equation for accurate calculations.

Identifying Gases

Identifying an unknown gas involves comparing the calculated molar mass to known values. In this exercise, after determining the molar mass to be 39.31 g/mol, we compare it with common gases:

• Nitrogen (N_2): 28.01 g/mol
• Neon (Ne): 20.18 g/mol
• Argon (Ar): 39.95 g/mol

The calculated molar mass is closest to that of argon, suggesting that the unknown gas is most likely argon. This process of identification is essential in many scientific fields, especially chemistry and environmental science.