Question

What is the molecular formula of a gaseous fluoride of sulfur containing \(70.4 \%\) F and having a density of approximately \(4.5 \mathrm{g} / \mathrm{L}\) at \(20^{\circ} \mathrm{C}\) and \(1 \mathrm{atm} ?\)

Short answer

The molecular formula of the gaseous fluoride of sulfur is SF4.

Step-by-step solution

Calculate mass of sulfur

Given that the gaseous fluoride contains 70.4% fluorine (F), it means that it must contain 100% - 70.4% = 29.6% sulfur (S). If we consider 100g as a total, there are 70.4g of fluorine and 29.6g of sulfur.

Calculate moles of F and S

We calculate the moles by dividing the mass of each element by their respective atomic masses. The atomic mass of F = 19.00 g/mol and of S = 32.07 g/mol. Thus, the moles of F = 70.4g / 19.00 g/mol = 3.7 mol and the moles of S = 29.6g / 32.07 g/mol = 0.92 mol.

Determine the ratio of moles

To find the simplest whole-number ratio, divide each number of moles by the smallest number of moles calculated. In this case, divide both F and S moles by 0.92. So, ratio of moles of F and S becomes F:S = 3.7 / 0.92 : 0.92 / 0.92 = 4 : 1. This implies the empirical formula is SF4.

Calculate molecular formula

Given that the molecular mass of a gaseous fluoride of sulfur (M) can be calculated by using the ideal gas law equation \(M = dRT / P\), where d is density (4.5 g/L), R = 0.0821 L.atm/K.mol (ideal gas constant), T is temperature in Kelvin (20°C = 293.15 K), P is the pressure in atm (1 atm). Substituting these values into the ideal gas law equation gives M = 4.5g/L * 0.0821 L.atm/K.mol * 293.15 K / 1 atm = 108.15 g/mol.

Identify the molecular formula

The empirical formula SF4 has the molar mass as 32.07g + 4x19.00g = 108 g/mol, which is exactly the molar mass calculated in previous step. Therefore, the molecular formula is the same as the empirical formula - SF4.

Key concepts

Understanding Gaseous Fluoride of Sulfur

When we talk about a gaseous fluoride of sulfur, we're discussing a chemical compound where sulfur (S) is bonded to fluorine (F). In this exercise, the compound's composition is given as 70.4% fluorine. This means that the remaining portion, 29.6%, is sulfur. This kind of compound is essential in chemistry due to its unique properties and applications in different reactions. Gaseous fluorides of sulfur conduct research and industrial importance due to their oxidative and catalytic properties.

• Composition: Knowing the percentages of fluorine and sulfur helps determine the elemental makeup of the compound.
• Chemical Bonding: The properties and reactivity of gaseous fluorides depend on these elements being bonded together.

Understanding these core properties allows us to analyze and predict the behavior of the compound in various scenarios. Always remember that the percentages relate to the mass contribution of each element in the compound, which is crucial for further calculations.

Empirical Formula Calculation Made Simple

The empirical formula of a compound gives the simplest ratio of the elements within it. To determine this, we begin by using the percentages provided, assuming a hypothetical 100g sample. This simplifies the calculation without altering the results.

• Mole Calculation: We calculate the moles for each element by dividing the mass of each (given in percentages) by its atomic mass. For fluorine: \( \text{moles of F} = \frac{70.4}{19} \approx 3.7 \), and for sulfur: \( \text{moles of S} = \frac{29.6}{32.07} \approx 0.92 \).
• Ratio Determination: Next, find the lowest whole-number ratio by dividing all mole quantities by the smallest amount calculated. This yields the simplest ratio, \( F:S = 4:1 \).
• Empirical Formula: Thus, the empirical formula derived tells us the elemental relationship within the compound, which is SF₄ for this exercise.

By understanding each step along the way, determining the empirical formula becomes a straightforward task, even for more complex compounds.

The Role of Ideal Gas Law Application

The ideal gas law, a fundamental principle in chemistry, provides a relationship between pressure (P), volume (V), temperature (T), and the number of moles of a gas. It is crucial for calculating the molecular formula of gases, especially when their density is known.

• Formula Involved: \( PV = nRT \) which rearranges to find molecular mass as \( M = \frac{dRT}{P} \). This formula allows us to calculate the molar mass from given conditions.
• Application: In our problem, substituting density (\(4.5 \text{ g/L}\)), temperature (\(293.15 \text{ K}\)), and pressure (\(1 \text{ atm}\)) with the gas constant (\(0.0821 \text{ L.atm/K.mol}\)), provides the molar mass of \(108.15 \text{ g/mol}\).
• Connection to Molecular Formula: Comparing this molar mass with the empirical formula mass confirms they are equivalent, solidifying that the molecular formula is SF₄.

The ideal gas law is a powerful tool in chemistry, transforming our understanding of gas-based reactions and behaviors when we connect it to empirical and molecular formula calculations.