Problem 76 - Chemistry The Molecular Science

Question

Disulfur dichloride, which has a revolting smell, can be prepared by directly combining \(\mathrm{S}_{8}\) and \(\mathrm{Cl}_{2}\), but it can also be made by this reaction: \(3 \mathrm{SCl}_{2}(\ell)+4 \mathrm{NaF}(\mathrm{s}) \longrightarrow \mathrm{SF}_{4}(\mathrm{~g})+\mathrm{S}_{2} \mathrm{Cl}_{2}(\ell)+4 \mathrm{NaCl}(\mathrm{s})\) Calculate the mass of \(\mathrm{SCl}_{2}\) needed to react with excess NaF to prepare \(1.19 \mathrm{~g} \mathrm{~S}_{2} \mathrm{Cl}_{2}\) if the expected yield is \(51 \%\).

Step-by-Step Solution

Answer

5.34 g of \(\mathrm{SCl}_2\) is needed.

Step 1: Understand the Reaction

The balanced chemical equation provided is: \(3 \mathrm{SCl}_{2}(\ell)+4 \mathrm{NaF}(\mathrm{s}) \longrightarrow \mathrm{SF}_{4}(\mathrm{~g})+\mathrm{S}_{2} \mathrm{Cl}_{2}(\ell)+4 \mathrm{NaCl}(\mathrm{s})\). This equation shows that 3 moles of \(\mathrm{SCl}_2\) react to produce 1 mole of \(\mathrm{S}_{2} \mathrm{Cl}_{2}\).

Step 2: Relate Mass and Moles

Determine the molar mass of \(\mathrm{S}_{2} \mathrm{Cl}_{2}\): sulfur (S) has a molar mass of 32.07 g/mol and chlorine (Cl) has a molar mass of 35.45 g/mol. Thus, the molar mass of \(\mathrm{S}_{2} \mathrm{Cl}_{2}\) is \(2 \times 32.07 + 2 \times 35.45 = 135.04\) g/mol. Calculate the moles of \(\mathrm{S}_{2} \mathrm{Cl}_{2}\) in 1.19 g: \(\text{moles of } \mathrm{S}_{2} \mathrm{Cl}_{2} = \frac{1.19 \text{ g}}{135.04 \text{ g/mol}} \approx 0.00881 \text{ mol}\).

Step 3: Calculate Theoretical Yield of \(\mathrm{SCl}_2\)

Using the mole ratio from the equation (3 moles of \(\mathrm{SCl}_2\) produce 1 mole of \(\mathrm{S}_{2} \mathrm{Cl}_{2}\)), determine the moles of \(\mathrm{SCl}_2\) needed: \(0.00881 \text{ mol } \mathrm{S}_{2} \mathrm{Cl}_{2} \times \frac{3 \text{ mol } \mathrm{SCl}_2}{1 \text{ mol } \mathrm{S}_{2} \mathrm{Cl}_{2}} = 0.02643 \text{ mol } \mathrm{SCl}_2\).

Step 4: Adjust for Percentage Yield

Since the percent yield is 51%, the theoretical calculation must be adjusted: required moles of \(\mathrm{SCl}_2\) are actually \(\frac{0.02643 \text{ mol}}{0.51} \approx 0.0518 \text{ mol}\).

Step 5: Convert Moles of \(\mathrm{SCl}_2\) to Mass

Find the molar mass of \(\mathrm{SCl}_2\): \(32.07 + 35.45 \times 2 = 102.97\) g/mol. Convert moles of \(\mathrm{SCl}_2\) to grams: \(0.0518 \text{ mol} \times 102.97 \text{ g/mol} \approx 5.34 \text{ g}\). \(5.34 \text{ g}\) of \(\mathrm{SCl}_2\) is needed.

Key Concepts

Disulfur Dichloride

Disulfur dichloride, with the chemical formula \( \mathrm{S}_2\mathrm{Cl}_2 \), is a compound that can be produced from sulfur and chlorine under certain conditions. It's known for its pungent and unpleasant smell. In chemical reactions, disulfur dichloride can serve as a source of sulfur and chlorine when broken down or reacted with other substances. In the specific reaction provided, it is synthesized indirectly from sulfur dichloride (\( \mathrm{SCl}_2 \)) and sodium fluoride (\( \mathrm{NaF} \)).
This interesting compound plays a role in organic syntheses and in the vulcanization of rubber. Its preparation through such reactions underscores the ability to derive valuable products through carefully controlled chemical processes.

Molar Mass Calculations

Molar mass is key to converting between grams and moles, which are a chemist's bridge between the macroscopic and molecular worlds. Calculating the molar mass of disulfur dichloride involves adding the molar masses of its constituent elements. Sulfur (\( S \)) has a molar mass of \( 32.07 \text{ g/mol} \), while chlorine (\( \text{Cl} \)) has \( 35.45 \text{ g/mol} \). For \( \mathrm{S}_2\mathrm{Cl}_2 \), the molar mass is computed as follows:

• Molar mass of sulfur: \( 2 \times 32.07 = 64.14 \text{ g/mol} \)
• Molar mass of chlorine: \( 2 \times 35.45 = 70.9 \text{ g/mol} \)
• Total molar mass of \( \mathrm{S}_2\mathrm{Cl}_2 \): \( 64.14 + 70.9 = 135.04 \text{ g/mol} \)

This calculation is essential for determining how much of a compound is present in any given reaction, as it helps convert mass to moles.

Percentage Yield

In chemical reactions, the percentage yield indicates the efficiency of the reaction: it compares the actual yield (what you got) to the theoretical yield (what you expected). It's calculated as follows:
\[\text{Percentage Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\]
In the context of producing \( \mathrm{S}_2\mathrm{Cl}_2 \) from \( \mathrm{SCl}_2 \), if the expected or theoretical yield is known,—assuming 100% efficient reaction—and the actual yield is around 51%, the theoretical calculation must be adjusted.
This means that more \( \mathrm{SCl}_2 \) is needed to achieve the desired mass of \( \mathrm{S}_2\mathrm{Cl}_2 \). The correction needed for percentage yield ensures that actual results in the lab align more closely with theoretical expectations, taking into consideration practical limitations like side reactions and incomplete reactions.