Problem 124 - Chemistry Principles and Practice

Question

The reaction of sulfur, \(\mathrm{S}_{8}\), with fluorine, \(\mathrm{F}_{2}\), yields a product with the general formula \(\mathrm{SF}_{x}\). If \(4.01 \mathrm{~g} \mathrm{~S}_{8}\) reacts with \(4.76 \mathrm{~g} \mathrm{~F}_{2}\) to yield only \(\mathrm{SF}_{x},\) what is the value of \(x ?\) Draw the Lewis structure of this compound, indicating the \(\mathrm{F}-\mathrm{S}-\mathrm{F}\) bond angles and the hybrid orbitals on sulfur.

Step-by-Step Solution

Answer

The value of \(x\) is 8 making the compound \(\mathrm{SF}_6\); it has octahedral geometry with \(90^\circ\) bond angles.

Step 1: Determine Molar Masses

To find the value of \(x\) in \(\mathrm{SF}_x\), we start by calculating the molar masses of \(\mathrm{S}_8\) and \(\mathrm{F}_2\). Molar mass of \(\mathrm{S}_8\) is approximately 256.48 g/mol (32.06 g/mol for sulfur times 8) and for \(\mathrm{F}_2\), it is approximately 38.00 g/mol (19.00 g/mol for fluorine times 2).

Step 2: Calculate Moles of Reactants

Using the molar masses, calculate moles of each reactant:Moles of \(\mathrm{S}_8 = \frac{4.01 \text{ g}}{256.48 \text{ g/mol}} \approx 0.0156 \text{ mol}\).Moles of \(\mathrm{F}_2 = \frac{4.76 \text{ g}}{38.00 \text{ g/mol}} \approx 0.1253 \text{ mol}\).

Step 3: Determine Limiting Reactant

The chemical formula \(\mathrm{SF}_x\) implies that sulfur and fluorine will react in a certain ratio. There is sufficient fluorine to react with all sulfur since \(\frac{0.1253}{0.0156} \approx 8\), but precise stoichiometry needs consideration for octatomic sulfur. Sulfur is limiting since \(\mathrm{S}_8\) molecule implicates a high fluorine need, suggesting that all sulfur is consumed first. Determine moles of \(\mathrm{SF}_x\) by sulfur moles, \(0.0156 \text{ mol}\), from the complete reaction of sulfur.

Step 4: Calculate Total Fluorine Used

To find \(x\) precisely, multiply moles of \(\mathrm{SF}_x\) formed by \(x\), and equate this to fluorine used:\(x \times 0.0156 \text{ mol (number for \( \mathrm{SF}_x\))} = 0.1253 \text{ mol (fluorine moles used) }\).Solving gives \(x = \frac{0.1253}{0.0156} \approx 8\).

Step 5: Draw Lewis Structure and Identify Geometry

The compound \(\mathrm{SF}_8\) is called sulfur hexafluoride. Drawing its Lewis structure implies having six bonds between the central sulfur atom and the fluorine atoms around it. Each fluorine makes a single bond with sulfur, and the geometry is octahedral.

Step 6: Specify Bond Angles and Hybridization

In an octahedral geometry, all \(\mathrm{F-S-F}\) bond angles are \(90^\circ\). Sulfur in \(\mathrm{SF}_6\) uses \(\mathrm{sp}^3\mathrm{d}^2\) hybrid orbitals to form bonds with fluorine atoms, accommodating the six bonds in its geometry.

Key Concepts

Molar Mass

Molar mass is an essential concept in stoichiometry, as it is the weight of one mole of a given substance, measured in grams per mole (g/mol). This is necessary for converting grams of a substance into moles, which is a fundamental step in solving stoichiometric problems.
To find the molar mass of a compound, you sum the atomic masses of all the atoms in a molecule. For instance, the molar mass of sulfur, \(\mathrm{S}_8\), is calculated by multiplying the atomic mass of sulfur (approximately 32.06 g/mol) by 8, which gives approximately 256.48 g/mol.
Fluorine, \(\mathrm{F}_2\), calculated similarly with an atomic mass of approximately 19.00 g/mol, results in a molar mass of approximately 38.00 g/mol.
Knowing these values, you can find the number of moles by dividing the given mass of the substance by its molar mass. This conversion is crucial when determining how much of each reactant participates in a chemical reaction.

Limiting Reactant

The limiting reactant is the compound in a chemical reaction that determines the maximum amount of product that can be formed. It is the reactant that is completely consumed first, stopping further reaction.
To identify the limiting reactant, compare the mole ratio of the reactants used in the reaction to the mole ratio required by the balanced chemical equation.
In our example, we calculated the moles of \(\mathrm{S}_8\) and \(\mathrm{F}_2\): \(0.0156\) mol and \(0.1253\) mol, respectively. The calculation \(\frac{0.1253}{0.0156}\) reveals a ratio of approximately 8:1, suggesting there is enough fluorine to react with sulfur. However, because the eight sulfur atoms in \(\mathrm{S}_8\) require a great deal of fluorine, sulfur is the limiting reactant.
This concept helps chemists determine which reactant will limit the formation of product, namely \(\mathrm{SF}_x\), ensuring accurate predictions and efficiency in chemical synthesis.

Lewis Structure

A Lewis structure is a visual representation of the arrangement of atoms within a molecule, showing how the atoms are bonded together with lines to represent shared electron pairs or bonds.
For \(\mathrm{SF}_6\), the Lewis structure involves placing the sulfur atom at the center and arranging the six fluorine atoms around it, each forming a single bond with sulfur. This configuration satisfies all the atoms' valence shell requirements, with sulfur accommodating six bonds directly.
The Lewis structure helps visualize the molecule's shape and predict physical properties like bond angles. Furthermore, it elucidates the electron arrangement within the molecule, aiding the understanding of its reactivity and interaction with other molecules.
In our molecule, \(\mathrm{SF}_6\), the Lewis structure plays a critical role in confirming that sulfur forms octahedral geometry, establishing the compound's structural integrity.

Hybrid Orbitals

Hybrid orbitals are formed when atomic orbitals on an atom combine to create new orbitals that align better with the geometry of a molecule, enabling stable bond formation.
In \(\mathrm{SF}_6\), sulfur forms \(\mathrm{sp}^3\mathrm{d}^2\) hybrid orbitals. Six electrons from sulfur hybridize by combining one s orbital, three p orbitals, and two d orbitals, accommodating the six fluorine atoms and their electron pairs.
This hybridization results in an octahedral geometry where all bond angles between the \(\mathrm{F-S-F}\) atoms are \(90^\circ\).
Understanding hybrid orbitals is crucial as it explains why molecules take on specific shapes, impacting their properties and functions. In essence, hybridization allows for the elaborate structural design of complex molecules.