Problem 81 - Chemistry 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 \(As_{3}S_{9}\), based on the given effusion rate and applying Graham's law of effusion.

Step 1: Using Graham's law of effusion

According to Graham's law of effusion, the rate of effusion of gas 1 divided by the rate of effusion of gas 2 is equal to the square root of the molar mass of gas 2 divided by the molar mass of gas 1. Mathematically it can be represented as: \( \dfrac{rate_{1}}{rate_{2}} = \sqrt{\dfrac{M_{2}}{M_{1}}} \) In our exercise, gas 1 is arsenic(III) sulfide, and gas 2 is Ar. We are given that rate of effusion of arsenic(III) sulfide is 0.28 times the rate of effusion of Ar, so we can write: \( \dfrac{0.28}{1} = \sqrt{\dfrac{M_{Ar}}{M_{As_{x}S_{y}}}} \) where \(M_{Ar}\) is the molar mass of Ar, and \(M_{As_{x}S_{y}}\) is the molar mass of arsenic(III) sulfide with x and y representing the subscripts in the molecular formula.

Step 2: Solving for the molar mass of arsenic(III) sulfide

Now, let's solve for \(M_{As_{x}S_{y}}\). First, square both sides of the equation: \( \dfrac{(0.28)^2}{1^2} = \dfrac{M_{Ar}}{M_{As_{x}S_{y}}} \) Next, multiply both sides of the equation by \(M_{As_{x}S_{y}}\): \( (0.28)^2 \times M_{As_{x}S_{y}} = M_{Ar} \) We know the molar mass of Ar is around 39.95 g/mol, so we can substitute this value and solve for \(M_{As_{x}S_{y}}\): \( M_{As_{x}S_{y}} = \dfrac{M_{Ar}}{(0.28)^2} = \dfrac{39.95}{0.0784} \) \( M_{As_{x}S_{y}} = 509.6 \)

Step 3: Finding the molecular formula

Now that we have found the molar mass of arsenic(III) sulfide in the gas phase, we can use this information to find the molecular formula. The molar mass of arsenic (As) is around 74.92 g/mol, and the molar mass of sulfur (S) is around 32.07 g/mol. Let the molecular formula be As_xS_y. Then, the molar mass of As_xS_y is: \( M_{As_{x}S_{y}} = 74.92x + 32.07y \) We have found in the previous step that \( M_{As_{x}S_{y}} = 509.6 \), so we can write: \( 74.92x + 32.07y = 509.6 \) Since this compound is arsenic(III) sulfide, the ratio between As and S is 1:3. Therefore, x = y/3. We can substitute this into our equation and solve for y: \( 74.92(\dfrac{y}{3}) + 32.07y = 509.6 \) \( 24.97y + 32.07y = 509.6 \) \( 57.04y = 509.6 \) \( y = \dfrac{509.6}{57.04} \) \( y ≈ 8.94 \) Since we cannot have fractional subscripts in the molecular formula, we can round it off to the nearest whole number, which is 9. Substituting this back into the x = y/3 equation, we get: \( x = \dfrac{9}{3} = 3 \) Therefore, the molecular formula of arsenic(III) sulfide in the gas phase is: \( As_{3}S_{9} \)