Problem 22 - Chemistry and Chemical Reactivity

Question

Black smokers are found in the depths of the oceans (page 110 ). Thinking that the conditions in these smokers might be conducive to the formation of organic compounds, two chemists in Germany found the following reaction could occur in similar conditions. $$ 2 \mathrm{CH}_{3} \mathrm{SH}+\mathrm{CO} \rightarrow \mathrm{CH}_{3} \mathrm{COSCH}_{3}+\mathrm{H}_{2} \mathrm{S} $$ 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 \(8.65 \mathrm{g}\) of \(\mathrm{CH}_{3} \mathrm{COSCH}_{3}\) is isolated, what is its percent yield?

Step-by-Step Solution

Answer

(a) Theoretical yield: 9.39 g; (b) Percent yield: 92.09%.

Step 1: Calculate Molar Mass

First, calculate the molar mass of the compounds involved. For \(\mathrm{CH}_{3}\mathrm{SH}\): \(\mathrm{C} = 12.01\, \mathrm{g/mol}, \ \mathrm{H} = 1.01\, \mathrm{g/mol}\) will contribute \(3 \times 1.01\, \mathrm{g/mol}, \ \mathrm{S} = 32.07\, \mathrm{g/mol}\). The molar mass of \(\mathrm{CH}_{3}\mathrm{SH}\) is thus \(12.01 + 3.03 + 32.07 = 48.11\, \mathrm{g/mol}\). The molar mass for \(\mathrm{CH}_{3}\mathrm{COSCH}_{3}\) is calculated similarly: \(2 \times (12.01 + 3.03) + 32.07 + 16.00 = 90.20\, \mathrm{g/mol}\).

Step 2: Calculate Moles of Reactant

Determine the moles of \(\mathrm{CH}_{3}\mathrm{SH}\): \[\text{moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{10.0\, \mathrm{g}}{48.11\, \mathrm{g/mol}} = 0.208\, \mathrm{mol}\]

Step 3: Determine Theoretical Yield of Product

Using the balanced reaction, \(2 \mathrm{CH}_{3}\mathrm{SH}\) produces \(\mathrm{CH}_{3}\mathrm{COSCH}_{3}\), therefore half the amount of \(\mathrm{CH}_{3}\mathrm{COSCH}_{3}\) moles is produced: \[0.208\, \mathrm{mol} \times \frac{1}{2} = 0.104\, \mathrm{mol}\] Now, calculate the mass of \(\mathrm{CH}_{3}\mathrm{COSCH}_{3}\) produced theoretically: \[0.104\, \mathrm{mol} \times 90.20\, \mathrm{g/mol} = 9.39\, \mathrm{g}\]

Step 4: Calculate Percent Yield

Percent yield is calculated using \[\text{Percent Yield} = \left(\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\right) \times 100\%\] Substitute the known values: \[\frac{8.65\, \mathrm{g}}{9.39\, \mathrm{g}} \times 100 = 92.09\%\]

Key Concepts

Molar Mass

Understanding molar mass is crucial when performing chemical reactions involving specific amounts of substances. Molar mass, often expressed in grams per mole (g/mol), represents the mass of one mole of a substance. Each element has a specific atomic mass contributing to the total molar mass of a compound.
For example, in the given reaction, you need to know the molar mass of both reactants and products to determine yields.

• The molar mass of a compound is calculated by summing the atomic masses of all atoms present in its formula.
• For instance, to find the molar mass of \(\mathrm{CH}_3\mathrm{SH}\), you add the carbon (C), hydrogen (H), and sulfur (S) contributions. The calculation is \(12.01 + 3 \times 1.01 + 32.07 = 48.11\, \mathrm{g/mol}\).

Molar mass is essential for determining the number of moles, a critical step in calculating theoretical and percent yields.

Percent Yield

Percent yield is a measure of efficiency in a chemical reaction. It compares the actual yield—the amount of product actually obtained from a reaction—to the theoretical yield, which is the maximum amount possible under perfect conditions.

• To calculate percent yield, use the formula: \[\text{Percent Yield} = \left(\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\right) \times 100\%\]
• For example, if your theoretical yield of \(\mathrm{CH}_3\mathrm{COSCH}_3\) is \(9.39\, \mathrm{g}\) and the actual yield is \(8.65\, \mathrm{g}\), your percent yield would be \(\frac{8.65}{9.39} \times 100 = 92.09\%\).

Understanding percent yield helps identify how effective a reaction was, taking into account losses due to incomplete reactions or side reactions.

Chemical Reaction

A chemical reaction involves the transformation of reactants into products, typically represented by a balanced chemical equation. In these equations, substances (molecules, atoms, or ions) undergo bond formation or breakage, resulting in new substances.

• The equation provided, \(2 \mathrm{CH}_3\mathrm{SH} + \mathrm{CO} \rightarrow \mathrm{CH}_3\mathrm{COSCH}_3 + \mathrm{H}_2\mathrm{S}\), shows how two molecules of \(\mathrm{CH}_3\mathrm{SH}\) react with one molecule of carbon monoxide (CO) to produce one molecule of \(\mathrm{CH}_3\mathrm{COSCH}_3\) and hydrogen sulfide (\(\mathrm{H}_2\mathrm{S}\)).
• A balanced equation ensures that the number of each type of atom is equal on both sides, reflecting the law of conservation of mass.

Understanding the stoichiometry of a chemical reaction allows you to calculate the theoretical yield and assess how much of each reactant is needed.