Question

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

Short answer

The molecular formula is \(\mathrm{S}_{1} \mathrm{F}_{5}\).

Step-by-step solution

Calculate Moles of the Compound

First, convert the given pressure from mm Hg to atm since standard units for gas laws are in atm. Use the conversion factor: \(1 \, \text{atm} = 760 \, \text{mm Hg}\). So, \(83.8 \, \text{mm Hg} = \frac{83.8}{760} \, \text{atm} = 0.11026 \, \text{atm}\) (approx.). Then, convert temperature to Kelvin: \(T = 45^{\circ}C + 273.15 = 318.15 \, K\). Use the ideal gas law equation \(PV = nRT\) to find moles (n), where \(R = 0.0821 \, L \, atm \/ (mol \, K)\). \(V = 89 \, \text{mL} = 0.089 \, L\). Substitute these values into the equation: \(n = \frac{PV}{RT} = \frac{(0.11026 \, \text{atm})\times(0.089 \, L)}{0.0821 \, \text{L atm /(mol K)} \times 318.15 \, K}\approx 0.0000381 \, mol\).

Calculate Molar Mass of the Compound

The given mass of the compound is \(0.0955 \, g\). Using the number of moles found in Step 1, calculate the molar mass (M) of the compound: \(M = \frac{\text{Mass}}{\text{Moles}} = \frac{0.0955 \, g}{0.0000381 \, mol}\approx 2506.56 \, g/mol\).

Determine Empirical Formula

Given that sulfur constitutes \(25.23\%\) of the compound, calculate the mass of sulfur in \(1 \, mol\) of compound: \(M_S = 0.2523 \times 2506.56 \, g/mol = 632.28 \, g/mol\). Then calculate moles of sulfur: \(n_S = \frac{632.28 \, g/mol}{32.07 \, g/mol} \approx 19.72\), indicating 19 or 20 atoms of sulfur (depending on significant figure rounding). Also, calculate mass of fluorine: \(M_F = 2506.56\, g/mol - 632.28\, g/mol= 1874.28\, g/mol\), where \(M_F = 1874.28\, g/mol\) and molar mass of fluorine (F) is \(19.00\, g/mol\), giving \(n_F = \frac{1874.28}{19.00} = 98.65\) moles of fluorine. So, approximate the empirical formula to \(\mathrm{S}_{20} \mathrm{F}_{100}\).

Simplify to Molecular Formula

The empirical formula suggests \(\frac{n_F}{n_S} \approx 5\), therefore take the smallest whole number ratio, \(\mathrm{S}_{1} \mathrm{F}_{5}\). Hence, \(\mathrm{S}_{20} \mathrm{F}_{100}\) simplifies to \(\mathrm{S}_{1} \mathrm{F}_{5}\) as the molecular formula reflecting the actual molar mass divided by empirical formula mass match.

Key concepts

Gas Laws

Understanding the behavior of gases is essential in chemistry, especially when dealing with compounds in their gaseous form. One of the core equations in this domain is the Ideal Gas Law, represented as \(PV = nRT\). Here, \(P\) stands for pressure, \(V\) for volume, \(n\) for moles of the gas, \(R\) for the gas constant, and \(T\) for temperature in Kelvin. In our exercise, we began by converting measurements to standard units necessary for calculations. Pressure was converted from mm Hg to atm, as 1 atm equals 760 mm Hg. Thus, the compound's pressure was transformed to approximately 0.11026 atm. Similarly, Celsius temperature was converted to Kelvin by adding 273.15, resulting in 318.15 K.Using these converted values in the Ideal Gas Law allows us to calculate the moles of gas present. For example, substituting into \(PV = nRT\): \[n = \frac{PV}{RT}\]This step is crucial because it leads to the accurate calculation of gas moles, which is foundational for both empirical and molecular formula determination.

Empirical Formula

The empirical formula of a compound gives the simplest ratio of the elements present. In the case of sulfur fluorine compounds like \(\mathrm{S}_x \mathrm{F}_y\), calculating the empirical formula starts with knowing the percentage composition of elements. For our compound, sulfur constitutes 25.23% of the weight. By determining the total molar mass and applying these percentages, sulfur's contribution to the compound's molar mass is computed. This process involves multiplying the percentage by the total molar mass, providing us with the sulfur mass per mole of the compound. Next, to find the number of sulfur atoms, this mass is divided by sulfur's atomic weight (32.07 g/mol), leading to approximately 19.72, suggesting 19 or 20 sulfur atoms. Similarly, by subtracting the sulfur mass from the total molar mass, we find the fluorine mass and calculate atoms of fluorine in the compound. This key step establishes the ratio of elements, guiding us to an empirical formula such as \(\mathrm{S}_{20} \mathrm{F}_{100}\).

Molar Mass Calculation

The molar mass of a compound is an essential value indicating the mass of one mole of that compound, often identified through experiment. In this scenario, the molar mass was found using both the mass of the compound sample and the calculated number of moles. To calculate molar mass, we use the formula:\[\text{Molar Mass (M)} = \frac{\text{Mass of compound (g)}}{\text{Moles of compound}}\]Here, the sample provided a molar mass of about 2506.56 g/mol. This figure is essential for determining an accurate molecular formula. Confirming that the empirical formula's calculated mass aligns with this experimental value ensures that it reflects the compound's true structure. Finally, adjusting the empirical formula to its simplest ratio, based on this molar mass, reveals the molecular formula, providing further precision in understanding the compound's composition and structure.