Problem 46 - Chemistry Central Science

Question

Rank the following gases from least dense at \(1.00\) atm and \(298 \mathrm{~K}\) to most dense under these same conditions: \(\mathrm{SO}_{2}, \mathrm{HBr}, \mathrm{CO}_{2}\). Explain.

Step-by-Step Solution

Answer

The gases ranked from least dense to most dense at \(1.00\) atm and \(298 \ \mathrm{K}\) are \(\mathrm{CO}_{2}\) (\(1.81 \ \frac{\mathrm{g}}{\mathrm{L}}\)), \(\mathrm{SO}_{2}\) (\(2.63 \ \frac{\mathrm{g}}{\mathrm{L}}\)), and \(\mathrm{HBr}\) (\(3.32 \ \frac{\mathrm{g}}{\mathrm{L}}\)). This ranking was determined using the Ideal Gas Law and the molecular weights of each gas.

Step 1: Recall the Ideal Gas Law

The Ideal Gas Law is given by the formula \(PV = nRT\), where - \(P\) is the pressure given in atmospheres (atm), - \(V\) is the volume (in liters), - \(n\) is the number of moles, - \(R\) is the Ideal Gas Constant (\(0.0821 \frac{\mathrm{L} \cdot \mathrm{atm}}{\mathrm{mol} \cdot \mathrm{K}}\)), - and \(T\) is the temperature given in Kelvin (K).

Step 2: Determine the molecular weight for each gas

The molecular weights of the given gases are: - \(\mathrm{SO}_{2}\): \(S = 32.00 \ \mathrm{g/mol}\), \(O = 16.00 \ \mathrm{g/mol}\); \(\Rightarrow (1 \times 32.00) + (2 \times 16.00) = 64.00 \ \mathrm{g/mol}\) - \(\mathrm{HBr}\): \(H = 1.01 \ \mathrm{g/mol}\), \(Br = 79.90 \ \mathrm{g/mol}\); \(\Rightarrow (1 \times 1.01) + (1 \times 79.90) = 80.91 \ \mathrm{g/mol}\) - \(\mathrm{CO}_{2}\): \(C = 12.01 \ \mathrm{g/mol}\), \(O = 16.00 \ \mathrm{g/mol}\); \(\Rightarrow (1 \times 12.01) + (2 \times 16.00) = 44.01 \ \mathrm{g/mol}\)

Step 3: Determine the density of each gas

To find the density, we need to modify the Ideal Gas Law formula to: \(\rho = \frac{n}{V} \times \frac{m}{n} = \frac{m}{V} = \frac{M \times P}{R \times T}\), where - \(\rho\) is the density, - \(M\) is molecular weight, - \(m\) is mass, - \(n\) is moles. Using the given pressure (\(1.00 \mathrm{\ atm}\)) and temperature (\(298 \mathrm{\ K}\)), we can calculate the densities of each gas by plugging in the molecular weights found in Step 2: - Density of \(\mathrm{SO}_{2}\): \(\rho_{\mathrm{SO}_{2}} = \frac{64.00 \ \mathrm{g/mol} \times 1.00 \mathrm{\ atm}}{0.0821 \frac{\mathrm{L\cdot atm}}{\mathrm{mol\cdot K}} \times 298 \mathrm{K}} = 2.63 \ \frac{\mathrm{g}}{\mathrm{L}}\) - Density of \(\mathrm{HBr}\): \(\rho_{\mathrm{HBr}} = \frac{80.91 \ \mathrm{g/mol} \times 1.00 \mathrm{\ atm}}{0.0821 \frac{\mathrm{L\cdot atm}}{\mathrm{mol\cdot K}} \times 298 \mathrm{K}} = 3.32 \ \frac{\mathrm{g}}{\mathrm{L}}\) - Density of \(\mathrm{CO}_{2}\): \(\rho_{\mathrm{CO}_{2}} = \frac{44.01 \ \mathrm{g/mol} \times 1.00 \mathrm{\ atm}}{0.0821 \frac{\mathrm{L\cdot atm}}{\mathrm{mol\cdot K}} \times 298 \mathrm{K}} = 1.81 \ \frac{\mathrm{g}}{\mathrm{L}}\)

Step 4: Compare the densities

Comparing the three densities calculated in Step 3: - \(\rho_{\mathrm{SO}_{2}} = 2.63 \ \frac{\mathrm{g}}{\mathrm{L}}\), - \(\rho_{\mathrm{HBr}} = 3.32 \ \frac{\mathrm{g}}{\mathrm{L}}\), - \(\rho_{\mathrm{CO}_{2}} = 1.81 \ \frac{\mathrm{g}}{\mathrm{L}}\).

Step 5: Rank the gases by density

Ranking the gases from least dense to most dense: 1. \(\mathrm{CO}_{2}\): \(1.81 \ \frac{\mathrm{g}}{\mathrm{L}}\), 2. \(\mathrm{SO}_{2}\): \(2.63 \ \frac{\mathrm{g}}{\mathrm{L}}\), 3. \(\mathrm{HBr}\): \(3.32 \ \frac{\mathrm{g}}{\mathrm{L}}\).

Key Concepts

Molecular Weight

Molecular weight, also referred to as molecular mass, is a critical concept in chemistry that correlates with the sum of the atomic weights of all atoms present in a molecule. It is expressed in atomic mass units (amu) or grams per mole (g/mol), which indicates the mass of one mole of a given substance. To calculate the molecular weight of a compound, we sum up the atomic weights of all the atoms within a molecule. For example, sulfur dioxide (\text{SO}_2) has a molecular weight calculated as:
\[ (1 \times 32.00) + (2 \times 16.00) = 64.00 \, \text{g/mol} \]
Understanding molecular weight is a fundamental step in predicting the behavior of gases using the Ideal Gas Law, as it impacts the calculation of the density of gases.

Gas Density

Gas density is a measure of the mass per unit volume of a gas and is expressed in grams per liter (g/L). It directly affects how a gas will behave under certain conditions, such as pressure and temperature. Densities of gases are important in various scientific and industrial processes, including the calculation of buoyancy and the design of chemical reactors.

To determine the density of a gas under the Ideal Gas Law, we modify the equation to:
\[ \rho = \frac{M \times P}{R \times T} \]
where \( \rho \) represents the density, \( M \) the molecular weight, \( P \) the pressure, \( R \) the Ideal Gas Constant, and \( T \) the temperature in Kelvin. This relationship shows us that at constant temperature and pressure, the density of a gas is directly proportional to its molecular weight. Lighter gases will have lower densities and therefore tend to rise, unlike heavier gases.

Molar Mass

Molar mass is often considered synonymous with molecular weight — it signifies the mass of one mole of a substance and is expressed in grams per mole (g/mol). It is pivotal when converting between grams and moles, which is a common procedure in stoichiometry and chemical equations. For instance, in our Ideal Gas Law density calculation, molar mass (represented by \( M \)) plays a key role because it links the mass of the gas to its mole quantity, ultimately determining the density of the gas when considered with volume, temperature, and pressure. A higher molar mass signifies a denser gas, assuming that conditions of temperature and pressure remain equal. Hence, in the ranking exercise for the gases \text{SO}_2, HBr, and \text{CO}_2, knowing the molar masses helps us establish which gas is the densest under specified conditions.

In summary, by mastering the concepts of molecular weight, gas density, and molar mass, students can proficiently navigate problems involving the physical properties of gases, including their densities, and correctly apply the Ideal Gas Law.