Question

Elemental sulfur \((1.256 \mathrm{g})\) is combined with fluorine, \(\mathbf{F}_{2}\) to give a compound with the formula \(\mathrm{SF}_{x}\) a very stable, colorless gas. If you have isolated \(5.722 \mathrm{g}\) of \(\mathrm{SF}_{x}\), what is the value of \(x ?\)

Short answer

The value of \(x\) is 6.

Step-by-step solution

Calculate Moles of Sulfur

First, we need to determine the moles of sulfur present in the sample. The molar mass of sulfur (S) is approximately 32.07 g/mol. Use the formula: \[\text{moles of S} = \frac{\text{mass of S}}{\text{molar mass of S}} = \frac{1.256 \, \text{g}}{32.07 \, \text{g/mol}} \approx 0.03915 \, \text{mol}\]

Calculate Moles of SF_x

Now, calculate the moles of \(\text{SF}_x\) produced using its mass. First, however, calculate the molar mass of \(\text{SF}_x\) if possible. Note that the exact molar mass requires knowing \(x\), but we'll see later that knowing only \(5.722 \, \text{g}\) is critical. Here, without \(x\) known directly, moles can be considered conceptually as:\[\text{moles of } \text{SF}_x = \frac{5.722 \, \text{g}}{\text{molar mass of } \text{SF}_x}\]where, \(\text{molar mass of } \text{SF}_x = 32.07 + 18.998 \times x\).

Establish a Relationship between Moles

According to the chemical formula, 1 mole of S will react to form 1 mole of \(\text{SF}_x\). Therefore:\[\text{moles of } \text{SF}_x = \text{moles of S} \approx 0.03915 \, \text{mol}\]

Solve for x

Using the moles calculated and the mass of \(\text{SF}_x\) isolated, substitute into the molar mass equation:\[\text{molar mass of } \text{SF}_x \times 0.03915 \approx 5.722 \]Solving for \(\text{molar mass of } \text{SF}_x\) yields:\[\text{molar mass of } \text{SF}_x \approx \frac{5.722}{0.03915} \approx 146.1 \, \text{g/mol}\]Substituting back:\[32.07 + (18.998 \times x) \approx 146.1 \]Solving for \(x\):\[18.998 \times x \approx 146.1 - 32.07 \]\[18.998 \times x \approx 114.03\]\[x \approx \frac{114.03}{18.998} \approx 6\]

Conclusion

Based on the calculations, the value of \(x\) in the compound \(\text{SF}_x\) is determined to be 6.

Key concepts

Stoichiometry and Its Role in Chemical Reactions

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between the amounts of reactants and products in a chemical reaction. It allows us to use the balanced chemical equation to determine the proportions of reactants required and the amount of products that can be formed. Understanding stoichiometry is essential for calculating the composition of compounds and the yield of chemical reactions.

In our example, stoichiometry is crucial for determining the amount of fluorine needed to react with sulfur to form the compound \(\text{SF}_x\). By calculating the moles of sulfur and using stoichiometric coefficients from the balanced equation, we can understand the conversion between reactants and products. These calculations are fundamental in predicting how much of each substance is involved in and produced by the reaction.

• Helps in finding the right amounts of reactants to avoid excess waste.
• Predicts the formation quantities of products based on the reactants available.
• Guides in maintaining the cost-efficiency and efficacy of industrial chemical processes.

Stoichiometry is, therefore, the backbone for any practical application of chemistry, ensuring that reactions are approached with precision and accuracy, just as we did with the sulfur and fluorine example.

Understanding Chemical Formulas

A chemical formula represents the elements within a compound and the ratio of atoms of each element. It provides essential information about the molecular structure of the compound. In our example, the compound \(\text{SF}_x\) implies that sulfur (S) is combined with \(x\) atoms of fluorine (F).

The subscript \(x\) in \(\text{SF}_x\) is an unknown that we solved for using stoichiometric calculations. It signifies the number of fluorine atoms that combine with one sulfur atom to form a stable molecule. The chemical formula offers insights into:

• The atomic composition of a compound.
• How molecules interact with one another.
• The molecular weight, which aids in moles calculation.

Knowing the chemical formula is vital for predicting how a compound will behave in a reaction, its reactivity, and the properties it will exhibit. This makes it one of the most fundamental concepts in chemistry.

Mastering Moles Calculation

Moles calculation is a core concept in chemistry that involves converting mass into moles using the molar mass, providing a basis for further stoichiometric calculations. The mole is the standard unit for measuring quantity in chemistry and represents \(6.022 \times 10^{23}\) particles, whether they be atoms, molecules, or ions.

In the problem at hand, we calculated the moles of sulfur based on its mass and molar mass. This was the first step in analyzing the chemical reaction and determining the composition of \(\text{SF}_x\). We then related the moles of sulfur to \(\text{SF}_x\), using the reaction's stoichiometry.

• Important for determining how much of a substance is reacting.
• Facilitates conversions between grams and moles using the molar mass.
• Crucial for applying stoichiometric coefficients to calculate reactant-product relationships.

Moles calculation allows chemists to count particles in a substance through mass measurements, making it possible to predict and quantify the results of chemical reactions. As demonstrated, mastering this calculation leads to clearer insights into how substances interact and transform during reactions.