Problem 65 - Chemistry and Chemical Reactivity

Question

You have four gas samples: 1\. \(1.0 \mathrm{L}\) of \(\mathrm{H}_{2}\) at \(\mathrm{STP}\) 2\. \(1.0 \mathrm{L}\) of \(\mathrm{Ar}\) at \(\mathrm{STP}\) 3\. \(1.0 \mathrm{L}\) of \(\mathrm{H}_{2}\) at \(27^{\circ} \mathrm{C}\) and \(760 \mathrm{mm} \mathrm{Hg}\) 4\. \(1.0 \mathrm{L}\) of He at \(0^{\circ} \mathrm{C}\) and \(900 \mathrm{mm} \mathrm{Hg}\) (a) Which sample has the largest number of gas particles (atoms or molecules)? (b) Which sample contains the smallest number of particles? (c) Which sample represents the largest mass?

Step-by-Step Solution

Answer

(a) Sample 4, (b) Sample 3, (c) Sample 2.

Step 1: Understand Conditions

Standard Temperature and Pressure (STP) is defined as a temperature of \(0^{\circ}C\) and a pressure of \(1\,\text{atm}\) or \(760 \, \text{mm Hg}\). This means samples 1 and 2 are both at STP conditions.

Step 2: Calculate Moles at STP

At STP, 1 mole of an ideal gas occupies \(22.4\,\text{L}\). For samples 1 and 2: \( \text{moles} = \frac{1.0 \, \text{L}}{22.4 \, \text{L/mol}} \approx 0.0446 \,\text{mol} \).

Step 3: Calculate Moles for Sample 3 Using Ideal Gas Law

Sample 3 is 1.0 L of \( \mathrm{H}_{2} \) at \( 27^{\circ} \mathrm{C} \) and \( 760 \, \text{mm Hg} \). Use the ideal gas law: \( PV = nRT \). Convert temperature to Kelvin: \( 27^{\circ}C + 273.15 = 300.15\,\text{K} \). Use \(P = 1 \, \text{atm}\), \( R = 0.0821 \, \text{L atm/mol K} \), \(V = 1.0 \, \text{L} \). Solve for \(n\): \[ n = \frac{1.0 \, \text{atm} \times 1.0 \, \text{L}}{0.0821 \, \text{L atm/mol K} \times 300.15 \, \text{K}} \approx 0.0401 \, \text{mol} \]

Step 4: Calculate Moles for Sample 4 Using Ideal Gas Law

Sample 4 is 1.0 L of He at \(0^{\circ} \mathrm{C}\) and \(900 \, \text{mm Hg} \). Convert pressure to atm: \( \frac{900}{760} \approx 1.1842 \, \text{atm} \). Use the ideal gas law to find moles: \[ n = \frac{1.1842 \, \text{atm} \times 1.0 \, \text{L}}{0.0821 \, \text{L atm/mol K} \times 273.15 \, \text{K}} \approx 0.0526 \, \text{mol} \]

Step 5: Determine Largest Number of Particles

To find the largest number of particles, compare moles calculated. Sample 4 (He) has the highest number of moles (\(0.0526 \, \text{mol}\)), hence the largest number of particles.

Step 6: Determine Smallest Number of Particles

Compare moles calculated; Sample 3 (\( \mathrm{H}_{2} \) at \(27^{\circ} \mathrm{C}\)) has the smallest amount of moles (\(0.0401 \, \text{mol}\)), leading to the fewest particles.

Step 7: Calculate Mass for Each Sample

Use molar mass to determine mass:1. \( \mathrm{H}_{2} \): \(2.016 \, \text{g/mol} \) so \(0.0446 \, \text{mol} \times 2.016 = 0.0899 \, \text{g}\)2. \( \mathrm{Ar} \): \(39.948 \, \text{g/mol} \) so \(0.0446 \, \text{mol} \times 39.948 = 1.7837 \, \text{g}\)3. \( \mathrm{H}_{2} \): \(0.0401 \, \text{mol} \times 2.016 = 0.0809 \, \text{g}\)4. He: \(4.0026 \, \text{g/mol}\) so \(0.0526 \, \text{mol} \times 4.0026 = 0.2105 \, \text{g}\)

Step 8: Determine Largest Mass

From the calculated masses, Sample 2 (Ar) has the largest mass of \(1.7837 \, \text{g}\).

Key Concepts

STP (Standard Temperature and Pressure)

STP, which stands for Standard Temperature and Pressure, serves as a reference point in chemical calculations involving gases. It standardizes conditions to make comparisons more straightforward. At STP, the temperature is fixed at 0°C (or 273.15 K), and the pressure is fixed at 1 atm, which is equivalent to 760 mm Hg.
These conditions are particularly important when working with gas laws, such as the Ideal Gas Law, as they simplify calculations.
At STP, 1 mole of an ideal gas occupies a volume of 22.4 liters, making it easier to convert between the number of moles and volume with these fixed reference points.
When comparing gases like in our exercise, acknowledging STP helps us deduce that samples at these conditions contain the same number of particles or moles per volume and also allows for a direct comparison of samples with differing conditions using the Ideal Gas Law.

Molar Mass

Molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol). It's a crucial concept when converting between moles and grams, which is essential in chemical stoichiometry and practical applications like this exercise.
Each element has its own molar mass, determined by the atomic mass found on the periodic table. For instance, hydrogen (\( H_2 \)) has a molar mass of approximately 2.016 g/mol, whereas argon (\( Ar \)) has a significantly higher molar mass of 39.948 g/mol.
In our exercise, calculating each gas sample's mass involves multiplying the number of moles by the sample’s molar mass. This shows how mass varies with molar mass despite having the same volume and initial conditions like at STP. For instance, even at the same number of moles, argon has a larger mass due to its greater molar mass compared to hydrogen.

Number of Particles

Understanding the number of particles in a gas sample involves connecting molecular quantities and macroscopic measurements through concepts like moles. A mole, by definition, contains Avogadro's number (\(6.022 \times 10^{23}\)) of particles – be they atoms, molecules, or ions.
This connection allows us to predict the number of particles in any given sample size by first determining the number of moles using the Ideal Gas Law or measurements under specified conditions like STP.
For example, in our exercise, identifying the sample with the most particles comes down to comparing moles across different samples. The sample with the highest number of moles, such as helium in our scenario, results in the most significant number of particles due to the direct proportionality between moles and particles.
Thus, even under changing pressures and temperatures, we can estimate particle counts effectively by working through these concepts.