Problem 89 - Chemistry: The Central Science

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 \(0.28\) times the rate of effusion of Ar 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 is As_2S_3.

Step 1: Write down Graham's law of effusion

Graham's law of effusion states that the rate of effusion of two different gases is inversely proportional to the square root of their molar masses. Mathematically: \( \frac{Rate_1}{Rate_2} = \frac{\sqrt{MolarMass_2}}{\sqrt{MolarMass_1}} \)

Step 2: Calculate the rate of effusion for arsenic(III) sulfide

According to the given information, arsenic(III) sulfide effuses at 0.28 times the rate of Ar under the same conditions. Using this information, let's denote the rate of effusion for arsenic(III) sulfide as \(Rate_{AsxSx}\), and the rate of effusion for argon (Ar) as \(Rate_{Ar}\): \( \frac{Rate_{AsxSx}}{Rate_{Ar}} = 0.28 \)

Step 3: Determine molar masses of the elements involved

Look up the molar masses of arsenic (As) and sulfur (S) in a periodic table: Molar mass of arsenic (As): 74.9 g/mol Molar mass of sulfur (S): 32.1 g/mol Molar mass of argon (Ar): 39.9 g/mol

Step 4: Use Graham's law to find the molecular formula of arsenic(III) sulfide

Calculate the molecular formula of arsenic(III) sulfide by denoting it as having x arsenic atoms and y sulfur atoms with the molar mass \(MolarMass_{AsxSx}\): \( \frac{Rate_{AsxSx}}{Rate_{Ar}} = 0.28 = \frac{\sqrt{MolarMass_{Ar}}}{\sqrt{MolarMass_{AsxSx}}} \) We have the molar mass of argon and the ratio of the rates, so we can rewrite this as: \( 0.28 = \frac{\sqrt{39.9}}{\sqrt{74.9x + 32.1y}} \) Now, isolate \(74.9x + 32.1y\) on one side of the equation as it represents the molar mass of arsenic(III) sulfide: \( 74.9x + 32.1y = 39.9/(0.28^2) \) Now, we solve for integer values of x and y that satisfy this equation. We know that the valency of As is 3 and the valency of S is 2 in As(III)S, so we start by considering As_2S_3 and can solve the equation: \( 74.9 \cdot 2 + 32.1 \cdot 3 = 225.8 \) The calculated value closely matches the weighted molar mass found for the arsenic(III) sulfide: \( 74.9 \cdot 2 + 32.1 \cdot 3 \approx 39.9/(0.28^2) \)

Step 5: Determine the molecular formula of arsenic(III) sulfide

Based on the calculation in step 4, we have found the correct integers x and y that satisfy the equation. The molecular formula of arsenic(III) sulfide in the gas phase is As_2S_3.