Problem 30 - Chemistry and Chemical Reactivity

Question

A reaction studied by Wächtershäuser and Huber (see "Black Smokers and the Origins of Life") is $$ 2 \mathrm{CH}_{3} \mathrm{SH}+\mathrm{CO} \longrightarrow \mathrm{CH}_{3} \mathrm{COSCH}_{3}+\mathrm{H}_{2} \mathrm{S} $$ If you begin with \(10.0 \mathrm{g}\) of \(\mathrm{CH}_{3} \mathrm{SH}\), and excess \(\mathrm{CO}\), (a) What is the theoretical yield of \(\mathrm{CH}_{3} \mathrm{COSCH}_{3} ?\) (b) If 8.65 g of \(\mathrm{CH}_{3} \mathrm{COSCH}_{3}\) is isolated, what is its percent yield?

Step-by-Step Solution

Answer

(a) The theoretical yield is 9.39 g. (b) The percent yield is 92.1%.

Step 1: Determine Molar Masses

First, let's calculate the molar masses of the compounds involved. The molar mass of \(\mathrm{CH}_3\mathrm{SH}\) is calculated as: \(12.01 + 3(1.01) + 32.07 + 1.01 = 48.11 \, \text{g/mol}\). The molar mass of \(\mathrm{CH}_3\mathrm{COSCH}_3\) is calculated as: \(2(12.01) + 6(1.01) + 16.00 + 32.07 = 90.19 \, \text{g/mol}\).

Step 2: Calculate Moles of Initial Reactant

Convert the mass of \(\mathrm{CH}_3\mathrm{SH}\) into moles using its molar mass. \(\text{Moles of } \mathrm{CH}_3\mathrm{SH} = \frac{10.0 \, \text{g}}{48.11 \, \text{g/mol}} \approx 0.208 \, \text{mol}\).

Step 3: Use Reaction Stoichiometry

From the balanced equation, the molar ratio of \(\mathrm{CH}_3\mathrm{SH}\) to \(\mathrm{CH}_3\mathrm{COSCH}_3\) is 2:1. Therefore, \(0.208 \, \text{mol of } \mathrm{CH}_3\mathrm{SH}\) corresponds to \(0.104 \, \text{mol of } \mathrm{CH}_3\mathrm{COSCH}_3\).

Step 4: Calculate Theoretical Yield

Calculate the theoretical yield of \(\mathrm{CH}_3\mathrm{COSCH}_3\) using its molar mass. \(\text{Mass of } \mathrm{CH}_3\mathrm{COSCH}_3 = 0.104 \, \text{mol} \times 90.19 \, \text{g/mol} \approx 9.39 \, \text{g}\).

Step 5: Calculate Percent Yield

The percent yield is calculated by comparing the actual yield to the theoretical yield. \(\text{Percent Yield} = \left( \frac{8.65 \, \text{g}}{9.39 \, \text{g}} \right) \times 100 \approx 92.1\%\).

Key Concepts

Understanding Theoretical Yield

Theoretical yield is the maximum amount of product that can be formed in a chemical reaction. It's based on the starting amounts of reactants and is calculated assuming perfect conditions with no loss of material.
In the example equation, starting with 10.0 g of \( \mathrm{CH}_{3} \mathrm{SH} \) and using it fully with excess \( \mathrm{CO} \), allows us to predict the maximum amount of \( \mathrm{CH}_{3} \mathrm{COSCH}_{3} \) that could be produced. This requires calculating moles of reactants and using the balance of the chemical equation for stoichiometry.
This approach helps chemists plan and optimize reactions by knowing the potential yield—in essence, the best possible scenario where all reactants convert to products without any by-products or leftovers.

Calculating Percent Yield

Percent yield tells us how efficient a reaction is compared to its theoretical potential. It's the ratio of the actual yield (what you got) to the theoretical yield (what you could get), expressed as a percentage.
The formula is:

• \( \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100 \)

In our example, isolating 8.65 g of \( \mathrm{CH}_{3} \mathrm{COSCH}_{3} \) from a theoretical 9.39 g gives a yield of around 92.1%.
This indicates how much product was successfully made and helps identify any issues such as incomplete reactions or material loss.

The Role of Reaction Stoichiometry

Reaction stoichiometry involves the calculation of the quantitative relationships between reactants and products in a chemical reaction.
By examining the balanced equation:

• \( 2 \mathrm{CH}_{3} \mathrm{SH} + \mathrm{CO} \rightarrow \mathrm{CH}_{3} \mathrm{COSCH}_{3} + \mathrm{H}_{2} \mathrm{S} \)

We see that 2 moles of \( \mathrm{CH}_{3} \mathrm{SH} \) react with 1 mole of \( \mathrm{CO} \) to produce 1 mole of \( \mathrm{CH}_{3} \mathrm{COSCH}_{3} \).
This ratio is vital for determining how much product can be synthesized from a given amount of reactant, thus allowing us to solve for the theoretical yield using conversion factors between moles.

Exploring Molar Mass Calculations

Molar mass is the mass of one mole of a substance, defined in grams per mole. It helps convert between grams of a substance and moles, which is essential for stoichiometry.
To calculate the molar mass of \( \mathrm{CH}_{3} \mathrm{SH} \), we add the atomic weights of all atoms in the compound:

• Carbon (C): 12.01
• Hydrogen (H): 1.01 (there are 4)
• Sulfur (S): 32.07

Giving us: \( 12.01 + 4(1.01) + 32.07 = 48.11 \ \text{g/mol} \).
For \( \mathrm{CH}_{3} \mathrm{COSCH}_{3} \), repeat the process, incorporating all the atomic masses.
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