Problem 69 - Chemistry and Chemical Reactivity

Question

There are five compounds in the family of sulfur-fluorine compounds with the general formula \(\mathrm{S}_{x} \mathrm{F}_{y}\). One of these compounds is \(25.23 \%\) S. If you place 0.0955 g of the compound in a \(89-\mathrm{mL}\). flask at \(45^{\circ} \mathrm{C},\) the pressure of the gas is \(83.8 \mathrm{mm}\) Hg. What is the molecular formula of \(\mathrm{S}_{x} \mathrm{F}_{y}^{2}\)

Step-by-Step Solution

Answer

The molecular formula is \( \text{S}_2\text{F}_{10} \).

Step 1: Calculate Moles of Gas

First, we use the Ideal Gas Law equation, which is \( PV = nRT \), to find the number of moles \( n \). Convert the pressure from mm Hg to atm by dividing by 760. \[ P = \frac{83.8 \, \text{mm Hg}}{760} = 0.1103 \, \text{atm} \] Convert 45°C to Kelvin: \( T = 45 + 273.15 = 318.15 \, \text{K} \).The volume \( V \) in liters is: \( 89 \, \text{mL} = 0.089 \, \text{L} \). Using \( R = 0.0821 \, \text{L atm/mol K} \), we calculate \( n \): \[ n = \frac{PV}{RT} = \frac{0.1103 \, \text{atm} \times 0.089 \, \text{L}}{0.0821 \, \text{L atm/mol K} \times 318.15 \, \text{K}} = 3.801 \times 10^{-4} \, \text{mol} \]

Step 2: Determine Molar Mass

The mass of the compound is given as 0.0955 g. Using the moles calculated in Step 1: \[ \text{Molar Mass} = \frac{\text{Mass}}{\text{Moles}} = \frac{0.0955 \, \text{g}}{3.801 \times 10^{-4} \, \text{mol}} = 251.38 \, \text{g/mol} \]

Step 3: Calculate Moles of Sulfur in Compound

Given that the compound is 25.23% sulfur, we calculate the mass of sulfur in 1 mole of the compound. \[ \text{Mass of } S = 0.2523 \times 251.38 \, \text{g/mol} = 63.45 \, \text{g/mol} \] The atomic mass of sulfur is approximately 32.07 g/mol, so the moles of sulfur: \[ \text{Moles of } S = \frac{63.45 \, \text{g/mol}}{32.07 \, \text{g/mol}} \approx 1.98 \approx 2 \] Thus, there are 2 sulfur atoms.

Step 4: Determine Moles of Fluorine

Subtract the mass of sulfur from the total molar mass to find the mass of fluorine: \[ \text{Mass of F} = 251.38 \, \text{g/mol} - 63.45 \, \text{g/mol} = 187.93 \, \text{g/mol} \] The atomic mass of fluorine is approximately 19.00 g/mol, so the moles of fluorine: \[ \text{Moles of } F = \frac{187.93 \, \text{g/mol}}{19.00 \, \text{g/mol}} \approx 9.89 \approx 10 \] Thus, there are 10 fluorine atoms.

Step 5: Determine Molecular Formula

Using the calculated values for sulfur and fluorine, the molecular formula of the compound is \( \text{S}_2\text{F}_{10} \).

Key Concepts

Sulfur-Fluorine Compounds

Sulfur-fluorine compounds are a fascinating group of chemicals, primarily known for their intriguing bonding structures and unique properties. These compounds, with a general formula of \( \mathrm{S}_{x}\mathrm{F}_{y} \), consist of sulfur (S) and fluorine (F) atoms in various combinations. Sulfur, with its higher reactivity with fluorine, forms strong covalent bonds that make these compounds stable and useful in various industrial applications.
Due to the electronegativity of fluorine, sulfur-fluorine compounds often exhibit unique chemical behaviors, such as high thermal stability and chemical inertness. For students exploring chemistry, understanding these interactions helps in grasping concepts around molecular stability and bonding.

Ideal Gas Law

The Ideal Gas Law is a powerful equation used to relate the properties of gases, combining pressure (\(P\)), volume (\(V\)), temperature (\(T\)), and the number of moles (\(n\)). The equation is represented as \( PV = nRT \), where \( R \) is the ideal gas constant. This law assumes gases behave ideally, meaning they occupy no space and have no interaction forces.
In practice, the Ideal Gas Law is instrumental in calculating unknown properties when other variables are known. For instance, by rearranging the equation, students can solve for the number of moles \( n \), which is crucial for understanding gas behavior under different conditions.

• This enables determination of quantities like molar mass and molecular formulas from volume and pressure data.
• It also offers insight into real-world applications such as gas behavior prediction in various temperatures and pressures.
  

Understanding the Ideal Gas Law is essential for students aiming to master chemical principles involving gases.

Molar Mass Calculation

Molar mass is an important concept that represents the mass of one mole of a substance, typically expressed in grams per mole (g/mol). Calculating the molar mass begins by dividing the mass of a sample by the number of moles of the substance.
For example, if a compound's mass is 0.0955 g and contains \(3.801 \times 10^{-4}\) moles, then its molar mass is calculated as:
\[\text{Molar Mass} = \frac{\text{Mass}}{\text{Moles}} = \frac{0.0955 \text{ g}}{3.801 \times 10^{-4} \text{ mol}} = 251.38 \text{ g/mol}\]
This value helps in determining the molecular formula of a compound by giving insight into the quantity of each element present. Students should consistently apply this calculation process to identify molecular structures in unknown samples.
Moreover, knowing molar mass is essential for quantitative analyses and reactions, facilitating conversion between mass of a compound and number of moles.

Elemental Composition Analysis

Elemental composition analysis is a technique used to determine the percentage composition of each element in a compound. This type of analysis is crucial for understanding the actual molecular formula of a compound. For instance, if a compound is 25.23% sulfur, it indicates that in every 100 grams of the compound, there are 25.23 grams of sulfur.
To find out the number of atoms of each element in the compound, this mass percentage is helpful. First, calculate the mass of the specific element per mole, and divide it by the atomic mass of the element to determine the number of moles.

• For sulfur: \(\text{Mass of } S = 0.2523 \times \text{Molar Mass of Compound} \).
• The atomic mass is approximately \(32.07 \text{ g/mol} \), leading to the determination of moles for forming the empirical formula.

This analysis not only helps identify molecular formulas but also helps understand the proportions and relationships between different elements within a compound. It is a fundamental part of chemical analysis yielding detailed insights into chemical structures.