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?

Short Answer

Expert verified
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 by step solution

01

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.
02

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
03

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
04

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

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Most popular questions from this chapter

A gas of unknown molecular mass was allowed to effuse through a small opening under constant-pressure conditions. It required \(105 \mathrm{~s}\) for $1.0 \mathrm{~L}$ of the gas to effuse. Under identical experimental conditions it required \(31 \mathrm{~s}\) for \(1.0 \mathrm{~L}\) of \(\mathrm{O}_{2}\) gas to effuse. Calculate the molar mass of the unknown gas. (Remember that the faster the rate of effusion, the shorter the time required for effusion of $1.0 \mathrm{~L} ;$ in other words, rate is the amount that diffuses over the time it takes to diffuse.)

A deep-sea diver uses a gas cylinder with a volume of \(10.0 \mathrm{~L}\) and a content of \(51.2 \mathrm{~g}\) of \(\mathrm{O}_{2}\) and \(32.6 \mathrm{~g}\) of He. Calculate the partial pressure of each gas and the total pressure if the temperature of the gas is \(19^{\circ} \mathrm{C}\).

A sample of \(3.00 \mathrm{~g}\) of \(\mathrm{SO}_{2}(g)\) originally in a \(5.00-\mathrm{L}\) vessel at \(21{ }^{\circ} \mathrm{C}\) is transferred to a \(10.0-\mathrm{L}\) vessel at \(26^{\circ} \mathrm{C}\). A sample of $2.35 \mathrm{~g}\( of \)\mathrm{N}_{2}(g)\( originally in a \)2.50-\mathrm{L}$ vessel at \(20^{\circ} \mathrm{C}\) is transferred to this same \(10.0-\mathrm{L}\) vessel. \((\mathbf{a})\) What is the partial pressure of \(\mathrm{SO}_{2}(g)\) in the larger container? (b) What is the partial pressure of \(\mathrm{N}_{2}(g)\) in this vessel? (c) What is the total pressure in the vessel?

In the Dumas-bulb technique for determining the molar mass of an unknown liquid, you vaporize the sample of a liquid that boils below $100^{\circ} \mathrm{C}$ in a boiling-water bath and determine the mass of vapor required to fill the bulb. From the following data, calculate the molar mass of the unknown liquid: mass of unknown vapor, \(1.012 \mathrm{~g}\); volume of bulb, \(354 \mathrm{~cm}^{3}\); pressure, \(98.93 \mathrm{kPa} ;\) temperature, $99{ }^{\circ} \mathrm{C}$.

A glass vessel fitted with a stopcock valve has a mass of $337.428 \mathrm{~g}\( when evacuated. When filled with \)\mathrm{Ar}$, it has a mass of \(339.854 \mathrm{~g}\). When evacuated and refilled with a mixture of Ne and Ar, under the same conditions of temperature and pressure, it has a mass of \(339.076 \mathrm{~g} .\) What is the mole percent of Ne in the gas mixture?

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