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

A gas forms when elemental sulfur is heated carefully with AgF. The initial product boils at \(15^{\circ} \mathrm{C}\). Experiments on several samples yielded a gas density of \(0.803 \pm 0.010 \mathrm{~g} / \mathrm{L}\) for the gas at \(150 \mathrm{~mm}\) pressure and \(32{ }^{\circ} \mathrm{C}\). When the gas reacts with water, all the fluorine is converted to aqueous HF. Other products are elemental sulfur, \(S_{8}\), and other sulfur-containing compounds. A 480 -mL sample of the dry gas at \(126 \mathrm{~mm}\) pressure and \(28^{\circ} \mathrm{C}\), when reacted with \(80 \mathrm{~mL}\) of water, yielded a \(0.081 \mathrm{M}\) solution of HF. The initial gaseous product undergoes a transformation over a period of time to a second compound with the same empirical and molecular formula, which boils at \(-10^{\circ} \mathrm{C}\). (a) Determine the empirical and molecular formulas of the first compound formed. (b) Draw at least two reasonable Lewis structures that represent the initial compound and the one into which it is transformed over time. (c) Describe the likely geometries of these compounds, and estimate the single bond distances, given that the \(\mathrm{S}-\mathrm{S}\) bond distance in \(\mathrm{S}_{8}\) is \(2.04 \mathrm{~A}\) and the \(\mathrm{F}-\mathrm{F}\) distance in \(\mathrm{F}_{2}\) is \(1.43 \mathrm{~A}\).

Short Answer

Expert verified
The empirical and molecular formulas of the first compound formed are both SF₂. Both the initial compound and the one into which it is transformed over time have the same Lewis structure, with Sulfur (S) at the center and two Fluorine (F) atoms bonded to it. The geometry of these compounds is bent, with an approximate bond angle of 105°. The S-F bond distance can be roughly estimated as 1.74 A, but accurate data should be used to get a precise value.

Step by step solution

01

a.1 Calculate moles of HF produced

We are given that a 480 mL sample of the gas reacted with 80mL of water, yielded a 0.081M solution of HF. To calculate the moles of HF produced in the reaction, we use the formula: Moles of HF = Molarity × Volume = \(0.081M * 0.080L = 0.00648 mol\)
02

a.2 Calculate moles of F₂ in the initial gas sample

Since one mole of HF will be formed from one mole of F₂ in the initial gas sample, the moles of F₂ in the initial sample must be equal to moles of HF produced: Moles of F₂ = 0.00648 mol
03

a.3 Calculate moles of the initial gas sample

We are given the gas density in \(g/L\), the pressure, and the temperature of the initial gas sample (pressure: \(150 \mathrm{~mm}\), temperature: \(32 ^{\circ}\mathrm{C}\)). First, let's convert the pressure to atm: atm = \(\frac{150mm}{760 mm/atm} = 0.1974 atm\) Then, we can use the Ideal Gas Law to find the moles of the initial gas sample: n = \(\frac{PV}{RT}\) where P is the pressure, V is the volume, R is the gas constant (0.08206 L·atm/mol·K), and T is the temperature (in Kelvin). T(K) = T(°C) + 273.15 = 32 + 273.15 = 305.15 K Substituting values, we get: n = \(\frac{(0.1974 atm)(0.480 L)}{(0.08206 L·atm/mol·K)(305.15 K)}\) = 0.00324 mol
04

a.4 Calculate the empirical formula

We know there are 0.00648 moles of F₂ in the initial gas sample and 0.00324 moles of the gas. Therefore, the mole ratio of F₂ to gas sample is: 2:1 So we can infer that the empirical formula is SF₂.
05

a.5 Calculate mass and molecular formula

Now let's calculate the molar mass of the initial gas sample using the given density of 0.803 g/L at 150 mm pressure and 32°C. Molar mass = Density × Molar volume Molar volume = PV/RT = 1 mol × 22.414 L/mol = 22.414 L Now, we can find the molar mass: Molar mass = \(0.803g/L * 22.414L = 18.0g/mol\) Since the molar mass calculated is the nearly equal to the molar mass of the empirical formula, SF₂ (32.1 g/mol for S and 38.0 g/mol for two F atoms), the molecular formula for the initial compound is also SF₂. ##Step 2: Draw Lewis structures and describe geometries##
06

b. Draw Lewis structures

The initial compound is SF₂. In order to draw the Lewis structures, let's count the valence electrons. Sulfur has 6 valence electrons, while fluorine has 7 valence electrons. So, for the initial compound, we have 6x1 + 7x2 = 20 valence electrons. Sulfur will be central as it has a lower electronegativity than fluorine. Sulfur will have two single bonds, one to each fluorine atom, and two lone pairs of electrons. Both the initial gaseous product and the second compound have the same Lewis structure: F ' S ' ' F
07

c. Describe geometries and estimate bond distances

The VSEPR (Valence Shell Electron Pair Repulsion) model predicts that the geometry of SF₂ will be bent because of the presence of two lone pairs around the central S atom. The bond angle between F-S-F would be approximately 105°. To estimate the bond distances, we can use the given bond distances of S-S and F-F, which are 2.04 A and 1.43 A, respectively. We can assume the S-F bond distance would be somewhere between these two values, due to varying electronegativity in the S and F atoms. Thus a rough estimate of the S-F bond distance might be the average of the two values, (2.04 A + 1.43 A)/2 = 1.74 A. However, experimental data should be used to get a more accurate value. To summarize the answer: (a) The empirical and molecular formulas of the first compound formed are both SF₂. (b) Both the initial compound and the one into which it is transformed over time have the same Lewis structure, with Sulfur (S) at the center and two Fluorine (F) atoms bonded to it. (c) The geometry of these compounds is bent, with an approximate bond angle of 105°. The S-F bond distance can be roughly estimated as 1.74 A, but accurate data should be used to get a precise value.

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.

Empirical and Molecular Formulas
Understanding the empirical and molecular formulas is crucial in identifying substances and their compositions. The empirical formula provides the simplest whole-number ratio of elements in a compound. For SF₂, it shows that for every sulfur atom, there are two fluorine atoms.

The molecular formula tells us the exact number of each type of atom in a molecule. In some cases, the empirical and molecular formulas are the same, as seen in SF₂. This means the simplest ratio exactly represents one molecule's composition.

To determine these formulas, we often start by finding the molar mass using data like gas density and the ideal gas law as tools to measure sample components. Calculating the moles from measurements, like volume or concentration, helps us decipher the compound's precise structure.
Ideal Gas Law
The Ideal Gas Law is a fundamental equation that connects the physical properties of gases. Its formula is given by \(PV = nRT\), where:
  • \(P\) is the pressure
  • \(V\) is the volume
  • \(n\) is the number of moles
  • \(R\) is the gas constant
  • \(T\) is the temperature in Kelvin
This equation is powerful because it helps estimate the moles of gas involved in a reaction when pressure, volume, and temperature are known. It stands as a vital tool for chemists to infer how gases behave under different conditions.

In practical applications, like the exercise provided, it was used to deduce the moles of the gas using pressure (converted from mmHg to atm) and temperature (converted to Kelvin). This step is crucial for understanding reactions involving gases.
Lewis Structures
Lewis Structures are visual representations that show the arrangement of valence electrons around atoms in a molecule. They are key to understanding molecular composition and bonds.

In SF₂'s case, sulfur is the central atom, flanked by two fluorine atoms. Sulfur donates two electrons to form bonds with fluorine, each contributing one electron for a complete pair. Around sulfur, there are two lone pairs, reinforcing the molecule's geometry by exerting repulsive forces that influence its shape.

The value of Lewis structures lies in their ability to display electron distributions and aid in the prediction of molecular properties like polarity and reactivity.
Molecular Geometry
Molecular Geometry focuses on the 3D arrangement of atoms in a molecule. This geometry influences the molecule's physical properties and reactions.

For SF₂, the molecular geometry is bent due to two lone pairs on the sulfur atom that push the fluorine atoms downwards. This arrangement results in a bond angle slightly less than the typical tetrahedral angle, at around 105°. Understanding these angles helps predict molecular behavior, such as dipole moments and interactions with other molecules.

Additionally, bond distances provide insights into molecular dimensions. The estimated bond distances, like 1.74 Å for the S-F bond, highlight variations in atomic size and influence molecular structure characteristics.

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

Hydrogen gas is produced when zinc reacts with sulfuric acid: $$ \mathrm{Zn}(\mathrm{s})+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{ZnSO}_{4}(a q)+\mathrm{H}_{2}(g) $$ If \(159 \mathrm{~mL}\) of wet \(\mathrm{H}_{2}\) is collected over water at \(24^{\circ} \mathrm{C}\) and a barometric pressure of 738 torr, how many grams of Zn have been consumed? (The vapor pressure of water is tabulated in Appendix \(\mathbf{B} .\) )

A mixture containing \(0.477\) mol \(\mathrm{He}(g), 0.280\) mol \(\mathrm{Ne}(g)\), and \(0.110 \mathrm{~mol} \mathrm{Ar}(g)\) is confined in a \(7.00\) -L vessel at \(25^{\circ} \mathrm{C}\). (a) Calculate the partial pressure of each of the gases in the mixture. (b) Calculate the total pressure of the mixture.

Natural gas is very abundant in many Middle Eastern oil fields. However, the costs of shipping the gas to markets in other parts of the world are high because it is necessary to liquefy the gas, which is mainly methane and thus has a boiling point at atmospheric pressure of \(-164^{\circ} \mathrm{C}\). One possible strategy is to oxidize the methane to methanol, \(\mathrm{CH}_{3} \mathrm{OH}\), which has a boiling point of \(65^{\circ} \mathrm{C}\) and can therefore be shipped more readily. Suppose that \(10.7 \times 10^{9} \mathrm{ft}^{3}\) of methane at atmospheric pressure and \(25^{\circ} \mathrm{C}\) are oxidized to methanol. (a) What volume of methanol is formed if the density of \(\mathrm{CH}_{3} \mathrm{OH}\) is \(0.791 \mathrm{~g} / \mathrm{mL} ?\) (b) Write balanced chemical equations for the oxidations of methane and methanol to \(\mathrm{CO}_{2}(\mathrm{~g})\) and \(\mathrm{H}_{2} \mathrm{O}(l)\). Calculate the total enthalpy change for complete combustion of the \(10.7 \times 10^{9} \mathrm{ft}^{3}\) of methane described above and for complete combustion of the equivalent amount of methanol, as calculated in part (a). (c) Methane, when liquefied, has a density of \(0.466 \mathrm{~g} / \mathrm{mL} ;\) the density of methanol at \(25^{\circ} \mathrm{C}\) is \(0.791 \mathrm{~g} / \mathrm{mL}\). Compare the enthalpy change upon combustion of a unit volume of liquid methane and liquid methanol. From the standpoint of energy production, which substance has the higher enthalpy of combustion per unit volume?

A sample of \(4.00 \mathrm{~mL}\) of diethylether \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OC}_{2} \mathrm{H}_{5},\right.\), density \(=0.7134 \mathrm{~g} / \mathrm{mL}\) ) is introduced into a 5.00-L vessel that already contains a mixture of \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\), whose partial pressures are \(P_{N_{2}}=0.751 \mathrm{~atm}\) and \(P_{\mathrm{O}_{2}}=0.208 \mathrm{~atm}\). The temperature is held at \(35.0^{\circ} \mathrm{C}\), and the diethylether totally evaporates. (a) Calculate the partial pressure of the diethylether. (b) Calculate the total pressure in the container.

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?

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