Problem 87 - Chemistry: The Central Science in SI Units, Global Edition

Question

Arsenic(III) sulfide sublimes readily, even below its melting point of \(320^{\circ} \mathrm{C}\). The molecules of the vapor phase are found to effuse through a tiny hole at 0.52 times the rate of effusion of Xe atoms under the same conditions of temperature and pressure. What is the molecular formula of arsenic(III) sulfide in the gas phase?

Step-by-Step Solution

Answer

The molecular formula of arsenic(III) sulfide in the gas phase remains the same as in the solid phase, which is \(As_2S_3\).

Step 1: Understanding Graham's law of effusion

Graham's law of effusion states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass: \( \frac{Rate _1}{Rate _2} = \sqrt{\frac{Molar \ Mass_2}{Molar \ Mass_1}} \) In this case, Rate 1 corresponds to the rate of effusion of arsenic(III) sulfide, and Rate 2 corresponds to the rate of effusion of Xe atoms.

Step 2: Given information

We have the following given information: \( \frac{Rate _1}{Rate _2} = 0.52 \) (where Rate 1 is arsenic(III) sulfide and Rate 2 is Xe atoms) Molar mass of Xe = 131.29 g/mol

Step 3: Calculate the molar mass of arsenic(III) sulfide

Using Graham's law, we can find the molar mass of arsenic(III) sulfide: \( 0.52 = \sqrt{\frac{Molar \ Mass_2}{Molar \ Mass_1}} \) \( 0.52 = \sqrt{\frac{131.29 \ g/mol}{Molar \ Mass_1}} \) Now we can solve for the molar mass of arsenic(III) sulfide (Molar Mass 1): \( Molar \ Mass_1 = \frac{131.29 \ g/mol}{0.52^2} \) Molar Mass 1 ≈ 485.31 g/mol

Step 4: Determine the molecular formula

We know that the arsenic(III) sulfide molecule contains arsenic (As) and sulfur (S). The molar masses of As and S are: As = 74.92 g/mol S = 32.07 g/mol Let x be the number of arsenic (As) atoms and y be the number of sulfur (S) atoms in the molecule. Then we can write the equation for the molar mass of the molecule: \(74.92x + 32.07y ≈ 485.31 \ g/mol \) Since we know arsenic(III) is As_2S_3 (the prefix "III" tells us there are 3 sulfur atoms for every two arsenic atoms), we can use ratios to determine the possible molecular formula: (2 As atoms x 74.92 g/mol) + (3 S atoms x 32.07 g/mol) = 299.70 g/mol Now we need a multiple of 299.7 that will equal the molar mass of the gas phase molecule, 485.31 g/mol: \( \frac{485.31}{299.70} ≈ 1.62\) Since the multiple should be a whole number, we round down to 1 and infer that the molecular formula remains the same in the gas phase, which is: As_2S_3

Key Concepts

Effusion Rates

Effusion rates refer to the speed at which gas molecules escape through a tiny hole into a vacuum. According to Graham's Law of Effusion, these rates are inversely proportional to the square root of the molar mass of the gas molecules. This law helps us understand the movement of different gases under similar conditions of temperature and pressure.

In the context of arsenic(III) sulfide, effusion rates are crucial. They allowed us to compare its escape rate to that of xenon atoms. Given that arsenic(III) sulfide effuses at only 52% of the rate of xenon, we can use this proportional relationship to draw conclusions about the molar mass of arsenic(III) sulfide. By employing the formula \( \frac{Rate_1}{Rate_2} = \sqrt{\frac{Molar \ Mass_2}{Molar \ Mass_1}} \) we can find the necessary values for further calculation.

This understanding of effusion is not only fascinating in theoretical chemistry but is also widely applied in separating gases and studying new compounds.

Molar Mass Calculation

The molar mass calculation allows us to find the mass of a given substance's molecules, expressed in grams per mole. By utilizing Graham's Law of Effusion, as mentioned earlier, we can calculate the molar mass of an unknown gas when we know the effusion rates and the molar mass of a reference gas.

In our problem, we know the molar mass of xenon (131.29 g/mol) and its effusion data relative to arsenic(III) sulfide. By substituting the known values into the equation from Graham's Law,\( 0.52 = \sqrt{\frac{131.29 \ g/mol}{Molar \ Mass_1}} \), and solving for the molar mass of arsenic(III) sulfide (Molar Mass 1), we discover that it approximately equals 485.31 g/mol.

This ability to determine molar mass is essential for chemistry, as it lays the groundwork for deeper insights into molecular formulas and structures.

Molecular Formula Determination

Determining the molecular formula of a compound tells us the number and type of atoms present. For arsenic(III) sulfide, it is known to contain arsenic (As) and sulfur (S). The molar masses of As and S are useful here: 74.92 g/mol and 32.07 g/mol, respectively.

Using these values, we construct an equation that represents the sum of the individual elements' contributions to the total molar mass:\(74.92x + 32.07y \approx 485.31 \ g/mol\).Given knowledge about the typical structure of arsenic(III) sulfide, commonly found as As\(_2\)S\(_3\), we verify this configuration:

• 2 As atoms at 74.92 g/mol = 149.84 g/mol
• 3 S atoms at 32.07 g/mol = 96.21 g/mol
• Total = 299.70 g/mol

When we take a whole number multiple of this standard formula to match the calculated molar mass, we find approximately 1.62, pointing to a rounding decision. By keeping the formula as As\(_2\)S\(_3\), it aligns with the given chemical characteristics.

Finding a molecular formula is like piecing together a puzzle. When the numbers add up, it facilitates the comprehension of the material's behavior in different states and environments.