Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Effusion Rates
Effusion rates refer to the speed at which gas molecules escape through a tiny hole into a vacuum. According to Graham's Law of Effusion, these rates are inversely proportional to the square root of the molar mass of the gas molecules. This law helps us understand the movement of different gases under similar conditions of temperature and pressure.

In the context of arsenic(III) sulfide, effusion rates are crucial. They allowed us to compare its escape rate to that of xenon atoms. Given that arsenic(III) sulfide effuses at only 52% of the rate of xenon, we can use this proportional relationship to draw conclusions about the molar mass of arsenic(III) sulfide. By employing the formula \( \frac{Rate_1}{Rate_2} = \sqrt{\frac{Molar \ Mass_2}{Molar \ Mass_1}} \) we can find the necessary values for further calculation.

This understanding of effusion is not only fascinating in theoretical chemistry but is also widely applied in separating gases and studying new compounds.
Molar Mass Calculation
The molar mass calculation allows us to find the mass of a given substance's molecules, expressed in grams per mole. By utilizing Graham's Law of Effusion, as mentioned earlier, we can calculate the molar mass of an unknown gas when we know the effusion rates and the molar mass of a reference gas.

In our problem, we know the molar mass of xenon (131.29 g/mol) and its effusion data relative to arsenic(III) sulfide. By substituting the known values into the equation from Graham's Law,\( 0.52 = \sqrt{\frac{131.29 \ g/mol}{Molar \ Mass_1}} \), and solving for the molar mass of arsenic(III) sulfide (Molar Mass 1), we discover that it approximately equals 485.31 g/mol.

This ability to determine molar mass is essential for chemistry, as it lays the groundwork for deeper insights into molecular formulas and structures.
Molecular Formula Determination
Determining the molecular formula of a compound tells us the number and type of atoms present. For arsenic(III) sulfide, it is known to contain arsenic (As) and sulfur (S). The molar masses of As and S are useful here: 74.92 g/mol and 32.07 g/mol, respectively.

Using these values, we construct an equation that represents the sum of the individual elements' contributions to the total molar mass:\(74.92x + 32.07y \approx 485.31 \ g/mol\).Given knowledge about the typical structure of arsenic(III) sulfide, commonly found as As\(_2\)S\(_3\), we verify this configuration:
  • 2 As atoms at 74.92 g/mol = 149.84 g/mol
  • 3 S atoms at 32.07 g/mol = 96.21 g/mol
  • Total = 299.70 g/mol
When we take a whole number multiple of this standard formula to match the calculated molar mass, we find approximately 1.62, pointing to a rounding decision. By keeping the formula as As\(_2\)S\(_3\), it aligns with the given chemical characteristics.

Finding a molecular formula is like piecing together a puzzle. When the numbers add up, it facilitates the comprehension of the material's behavior in different states and environments.

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

Suppose you are given two 2 -L flasks and told that one contains a gas of molar mass 28 , the other a gas of molar mass 56 , both at the same temperature and pressure. The mass of gas in the flask \(A\) is \(1.0 \mathrm{~g}\) and the mass of gas in the flask \(\mathrm{B}\) is \(2.0 \mathrm{~g}\). Which flask contains the gas of molar mass 28 , and which contains the gas of molar mass 56 ?

Gaseous iodine pentafluoride, \(\mathrm{IF}_{5}\), can be prepared by the reaction of solid iodine and gaseous fluorine: $$\mathrm{I}_{2}(s)+5 \mathrm{~F}_{2}(g) \longrightarrow 2 \mathrm{IF}_{5}(g)$$ A \(5.00-\mathrm{L}\) flask containing \(10.0 \mathrm{~g}\) of \(\mathrm{I}_{2}\) is charged with \(10.0 \mathrm{~g}\) of \(\mathrm{F}_{2},\) and the reaction proceeds until one of the reagents is completely consumed. After the reaction is complete, the temperature in the flask is \(125^{\circ} \mathrm{C}\). (a) What is the partial pressure of \(\mathrm{IF}_{5}\) in the flask? (b) What is the mole fraction of \(\mathrm{IF}_{5}\) in the flask (c) Draw the Lewis structure of IF \(_{5}\). (d) What is the total mass of reactants and products in the flask?

Which of the following statements is false? (a) Gases are far less dense than liquids. (b) Gases are far more compressible than liquids. (c) Because liquid water and liquid carbon tetrachloride do not mix, neither do their vapors. (d) The volume occupied by a gas is determined by the volume of its container.

An 8.40 -g sample of argon and an unknown mass of \(\mathrm{H}_{2}\) are mixed in a flask at room temperature. The partial pressure of the argon is \(44.0 \mathrm{kPa},\) and that of the hydrogen is \(57.33 \mathrm{kPa} .\) What is the mass of the hydrogen?

Assume that a single cylinder of an automobile engine has a volume of \(524 \mathrm{~cm}^{3}\). (a) If the cylinder is full of air at \(74^{\circ} \mathrm{C}\) and \(99.3 \mathrm{kPa}\), how many moles of \(\mathrm{O}_{2}\) are present? (The mole fraction of \(\mathrm{O}_{2}\) in dry air is \(0.2095 .\) ) (b) How many grams of \(\mathrm{C}_{8} \mathrm{H}_{18}\) could be combusted by this quantity of \(\mathrm{O}_{2}\), assuming complete combustion with formation of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O} ?\)

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