Problem 91 - 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_6, determined by applying Graham's Law of effusion and using the provided rate of effusion compared to Argon atoms. Using the calculated molar mass of arsenic(III) sulfide (approximately 510.2 g/mol), the molecular formula As_2S_6 yields the appropriate molar mass.

Step 1: Recall 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. Mathematically, this can be represented as: \( \frac{Rate_1}{Rate_2} = \sqrt{ \frac{Molar \, Mass_2}{Molar \, Mass_1 }} \) Where Rate_1 and Rate_2 are the rates of effusion of gases 1 and 2, and Molar Mass_1 and Molar Mass_2 are the molar masses of the gases, respectively.

Step 2: Calculate the molar mass of Argon

We are given the rate of effusion of Argon relative to arsenic(III) sulfide, so we will need the molar mass of Argon (Ar). The molar mass of Argon is 39.95 g/mol.

Step 3: Set up the formula with known values and solve for the molar mass of arsenic(III) sulfide

We know that the rate of effusion of arsenic(III) sulfide is 0.28 times that of the rate of effusion of Argon atoms. Using Graham's Law of effusion: \( \frac{Rate_{As_2S_3}}{Rate_{Ar}} = \sqrt{ \frac{Molar \, Mass_{Ar}}{Molar \, Mass_{As_2S_3}}} \) Plugging in the known values: \( 0.28 = \sqrt{ \frac{39.95 \, g/mol}{Molar \, Mass_{As_2S_3}}} \)

Step 4: Solve for the molar mass of arsenic(III) sulfide

To solve for the molar mass of arsenic(III) sulfide, first square both sides of the equation: \( 0.28^2 = \frac{39.95 \, g/mol}{Molar \, Mass_{As_2S_3}} \) Next, solve for the molar mass of arsenic(III) sulfide in the gas phase: \( Molar \, Mass_{As_2S_3} = \frac{39.95 \, g/mol}{0.28^2} \) \( Molar \, Mass_{As_2S_3} ≈ 510.2 \, g/mol \)

Step 5: Determine the molecular formula of arsenic(III) sulfide based on the molar mass

The molecular formula for arsenic(III) sulfide is As_nS_m, where n and m are integers. Using the periodic table, the molar mass of As is 74.92 g/mol, and the molar mass of S is 32.07 g/mol. We can express the total molar mass of the formula as: \( 510.2 \, g/mol ≈ 74.92n + 32.07m \) In this case, trying the simplest ratios (n = 1, m = 3) won't result in the desired molar mass; however, doubling the ratio results in: \( 510.2 \, g/mol ≈ 74.92(2) + 32.07(6) \) Thus, the molecular formula of arsenic(III) sulfide in the gas phase is As_2S_6.

Key Concepts

Molecular Formula

When we talk about the molecular formula, we're referring to a notation that tells us the number of atoms of each element in a molecule. For students encountering these formulas, remember they give direct insight into the composition of a compound.
To determine the molecular formula, we often start by understanding the empirical formula, which is the simplest whole-number ratio. However, in this exercise, we use the molar mass to directly derive the molecular formula.
Given the problem above with arsenic(III) sulfide in the vapor phase, calculating its molecular formula depends heavily on knowing the combined weight of the atoms involved. The periodic table is a useful tool here as it provides atomic weights, like 74.92 g/mol for arsenic (As) and 32.07 g/mol for sulfur (S). From there, we estimate how many atoms of each exist in a molecule based on these weights and the determined molar mass. In our example, this involves the clever use of simple mathematical ratios. Indeed, even chemistry can be like solving a puzzle!

Molar Mass Calculation

Understanding how to calculate molar mass is essential for mastering chemistry, especially when dealing with gases and their behaviors. The molar mass of a compound is the sum of the masses of its constituent atoms, each multiplied by the number of times they appear in the formula. This concept is crucial for connecting the microscopic world of atoms to macroscopic measurements.
Let's break it down with arsenic(III) sulfide in the exercise. We calculated its molar mass using the molar masses of its elements: arsenic (As) and sulfur (S). Arsenic's molar mass is 74.92 g/mol, and sulfur's is 32.07 g/mol. For the formula As_2S_6, the calculation is straightforward:

• Arsenic: 2 atoms of As, so the total mass is \(2 \times 74.92\) g/mol
• Sulfur: 6 atoms of S, so the total mass is \(6 \times 32.07\) g/mol

Add these together for the molar mass of the compound: approximately 510.2 g/mol.
This calculation is central to understanding how molecules behave in different states and conditions, such as in vapor phase chemistry.

Vapor Phase Chemistry

Vapor phase chemistry can initially seem daunting, but it's all about understanding how substances behave and interact when they are in gaseous form.
The term 'vapor phase' refers to a gas state of a substance that is generally a solid or liquid at room temperature. In our exercise, arsenic(III) sulfide sublimes, meaning it transitions directly from a solid to a gas without becoming a liquid. This property can be quite fascinating and highlights how temperature influences the state of matter.
In this vapour state, molecules are far apart and move freely, which ensures that the principles of gas laws, like Graham's Law of Effusion, apply. This law tells us about the rate at which different gases escape through a tiny hole and is inversely related to the molar mass of the gas. Here, it allows us to connect the physical observation (the rate of effusion) with the molecular weight to rightfully ascertain the molecular formula of gases like arsenic(III) sulfide.
Through studying vapor phase chemistry, we learn critical insights not just about how individual molecules behave, but also about how we can make use of these behaviors in practical applications like purification and separation techniques in industry.