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?

Short Answer

Expert verified
The molecular formula of arsenic (III) sulfide in the gas phase is As2S3.

Step by step solution

01

Write Graham's law of effusion formula

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, it can be expressed as: \( \frac{Rate_1}{Rate_2} = \sqrt{\frac{M_2}{M_1}} \) Here, Rate1 and Rate2 are the rates of effusion for gas1 and gas2, respectively. M1 and M2 are their respective molar masses.
02

Substitute the given values into the formula

We are given that the rate of effusion of arsenic (III) sulfide (AS) to that of the argon (Ar) is 0.28 (or 28%). The molar mass of argon gas is 39.95 g/mol. We can rearrange the formula to find the molar mass of arsenic (III) sulfide. \( \frac{Rate_{AS}}{Rate_{Ar}} = 0.28 = \sqrt{\frac{M_{Ar}}{M_{AS}}} \)
03

Calculate the molar mass of arsenic (III) sulfide

Now, we can calculate the molar mass of arsenic (III) sulfide by squaring both sides of the equation and rearranging it: \( M_{AS} = \frac{M_{Ar}}{0.28^2} \) \( M_{AS} = \frac{39.95\,g/mol}{0.0784} \) \( M_{AS} ≈ 509.49\,g/mol \)
04

Determine the molecular formula of arsenic (III) sulfide

Arsenic has a molar mass of approximately 74.92 g/mol and sulfur has a molar mass of approximately 32.07 g/mol. Let's assume the molecular formula to be AsxSy, where x and y are the numbers of arsenic and sulfur atoms, respectively. We can express the molar mass of arsenic (III) sulfide as: \( 509.49 \approx 74.92x + 32.07y \) Since this is arsenic (III) sulfide, the oxidation state of arsenic is +3 and each sulfur atom has an oxidation state of -2. This means that for each arsenic atom, there will be 3/2 sulfur atoms in the formula. Therefore, we get: \( y = \frac{3}{2} x \) Now we can substitute this expression into the molar mass equation: \( 509.49 \approx 74.92x + 32.07\left(\frac{3}{2}x\right) \) Solving for x, we find: \( x \approx 2 \) Since y = 3/2x: \( y \approx 3 \)
05

Write the molecular formula

Based on our calculations, the molecular formula of arsenic (III) sulfide in the gas phase is: As2S3

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Calcium hydride, CaH \(_{2},\) reacts with water to form hydrogen gas: $$\mathrm{CaH}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q)+2 \mathrm{H}_{2}(g)$$ This reaction is sometimes used to inflate life rafts, weather balloons, and the like, when a simple, compact means of generating \(\mathrm{H}_{2}\) is desired. How many grams of \(\mathrm{CaH}_{2}\) are needed to generate 145 \(\mathrm{L}\) of \(\mathrm{H}_{2}\) gas if the pressure of \(\mathrm{H}_{2}\) is 825 torr at \(21^{\circ} \mathrm{C} ?\)

Assume that an exhaled breath of air consists of \(74.8 \% \mathrm{N}_{2}, 15.3 \% \mathrm{O}_{2}, 3.7 \% \mathrm{CO}_{2},\) and 6.2\(\%\) water vapor. (a) If the total pressure of the gases is 0.985 atm, calculate the partial pressure of each component of the mixture. (b) If the volume of the exhaled gas is 455 \(\mathrm{mL}\) and its temperature is \(37^{\circ} \mathrm{C}\) , calculate the number of moles of \(\mathrm{CO}_{2}\) exhaled. (c) How many grams of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) would need to be metabolized to produce this quantity of \(\mathrm{CO}_{2} ?\) (The chemical reaction is the same as that for combustion of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} .\) See Section 3.2 and Problem 10.57 )

(a) The compound 1-iodododecane is a nonvolatile liquid with a density of 1.20 \(\mathrm{g} / \mathrm{mL}\) . The density of mercury is 13.6 \(\mathrm{g} / \mathrm{mL} .\) What do you predict for the height of a barometer column based on 1 -iodododecane, when the atmospheric pressure is 749 torr? (b) What is the pressure, in atmospheres, on the body of a diver if he is 21 ft below the surface of the water when the atmospheric pressure is 742 torr?

A 1.42 -g sample of helium and an unknown mass of \(\mathrm{O}_{2}\) are mixed in a flask at room temperature. The partial pressure of the helium is 42.5 torr, and that of the oxygen is 158 torr. What is the mass of the oxygen?

You have a gas at \(25^{\circ} \mathrm{C}\) confined to a cylinder with a movable piston. Which of the following actions would double the gas pressure? \((\mathbf{a})\) Lifting up on the piston to double the volume while keeping the temperature constant; \((\mathbf{b})\) Heating the gas so that its temperature rises from \(25^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\) , while keeping the volume constant; \((\mathbf{c})\) Pushing down on the piston to halve the volume while keeping the temperature constant.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free