Problem 74 - General Chemistry: Principles and Modern Applications

Question

In the reaction shown, \(100.0 \mathrm{g} \mathrm{C}_{6} \mathrm{H}_{11} \mathrm{OH}\) yielded \(64.0 \mathrm{g}\) \(\mathrm{C}_{6} \mathrm{H}_{10} .\) (a) What is the theoretical yield of the reaction? (b) What is the percent yield? (c) What mass of \(\mathrm{C}_{6} \mathrm{H}_{11} \mathrm{OH}\) would produce \(100.0 \mathrm{g} \mathrm{C}_{6} \mathrm{H}_{10}\) if the percent yield is that determined in part (b)? $$\mathrm{C}_{6} \mathrm{H}_{11} \mathrm{OH} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{10}+\mathrm{H}_{2} \mathrm{O}$$

Step-by-Step Solution

Answer

The theoretical yield of the reaction is approximately 95.4g. The actual yield being 64.0g yields a percent yield of roughly 67.0%. This means that to produce 100.0g of \(\mathrm{C}_{6}\mathrm{H}_{10}\), it would require about 144.7g of \(\mathrm{C}_{6} \mathrm{H}_{11} \mathrm{OH}\).

Step 1: Calculate Theoretical Yield

To calculate the theoretical yield, find the number of moles of the reactant \(\mathrm{C}_{6} \mathrm{H}_{11} \mathrm{OH}\) by using its molar mass (86.15 g/mol). Then, relate this to the moles of the product \(\mathrm{C}_{6}\mathrm{H}_{10}\) using the stoichiometric ratio of 1 to 1 given by the balanced equation. Further, convert the moles of the product back to grams using its molar mass (82.15 g/mol). The formula would look like: \[ \text{No. of moles of } \mathrm{C}_{6} \mathrm{H}_{11} \mathrm{OH}= \frac{100.0g}{86.15 g/mol} \] \[ \text{Theoretical Yield of } \mathrm{C}_{6}\mathrm{H}_{10}=\text{No. of moles of } \mathrm{C}_{6} \mathrm{H}_{11} \mathrm{OH} \times 82.15 g/mol \]

Step 2: Calculate Percent Yield

To calculate the percent yield, divide the actual yield by the theoretical yield and multiply the quotient by 100%. The percent yield formula would look like: \[ \text{Percent yield} = \frac{64.0g}{\text{Theoretical Yield}} \times 100\% \]

Step 3: Determine Initial Mass

To calculate the initial mass of the reactant that will yield 100.0g of the product given the calculated percent yield, first, figure out the theoretical yield of reactant needed to produce 100.0g of the product. Then, divide by the percent yield to account for the reaction's efficiency. The formula would look like: \[ \text{Theoretical yield of } \mathrm{C}_{6} \mathrm{H}_{11} \mathrm{OH} = \frac{100.0g}{82.15 g/mol} \] \[ \text{Initial mass of } \mathrm{C}_{6} \mathrm{H}_{11} \mathrm{OH} = \frac{\text{Theoretical yield of } \mathrm{C}_{6} \mathrm{H}_{11} \mathrm{OH}}{\text{Percent yield}} \times 100\% \]

Key Concepts

Theoretical Yield

In the realm of chemical reactions, understanding theoretical yield is crucial. The theoretical yield represents the maximum amount of product that can be generated in a perfect scenario, where the reaction occurs with complete efficiency and no loss. It's a calculated value that is based on the stoichiometry of the balanced chemical equation. For instance, in our exercise:

• The balanced equation is: \(\mathrm{C}_{6} \mathrm{H}_{11} \mathrm{OH} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{10}+\mathrm{H}_{2} \mathrm{O}\)
• For every mole of \(\mathrm{C}_{6} \mathrm{H}_{11} \mathrm{OH}\) consumed, one mole of \(\mathrm{C}_{6} \mathrm{H}_{10}\) is produced.
• Using the molar masses (\(86.15 \text{ g/mol}\) for \(\mathrm{C}_{6} \mathrm{H}_{11} \mathrm{OH}\) and \(82.15 \text{ g/mol}\) for \(\mathrm{C}_{6} \mathrm{H}_{10}\)), we calculate the theoretical yield of \(\mathrm{C}_{6} \mathrm{H}_{10}\) from 100.0 g of \(\mathrm{C}_{6} \mathrm{H}_{11} \mathrm{OH}\).

This involves converting grams of reactant to moles, using stoichiometry to find moles of product, and converting back to grams.

Percent Yield

Once we've established the theoretical yield, the next step in evaluating a chemical reaction's efficiency is determining the percent yield. Percent yield compares the actual amount of product obtained from the reaction to the theoretical yield, giving insight into how well the reaction performed. In practical situations, the actual yield is often less than the theoretical yield due to various factors like incomplete reactions, side reactions, and loss of product during processing.

• The formula for percent yield is:\[\text{Percent yield} = \left(\frac{\text{Actual yield}}{\text{Theoretical yield}}\right) \times 100\%\]
• For the given problem, the actual yield is 64.0 g of \(\mathrm{C}_{6} \mathrm{H}_{10}\).
• Calculate the percent yield by substituting the actual yield and the previously calculated theoretical yield into the formula.

Understanding percent yield helps us gauge the efficiency of a chemical process and identify potential areas for optimization.

Stoichiometry

Stoichiometry is a foundational concept in understanding chemical reactions at a deeper level. It involves the quantitative relationships between reactants and products in a chemical reaction. Simply put, stoichiometry allows us to predict how much of a substance is consumed or produced in a given reaction.In our given reaction:

• The stoichiometric ratio is 1:1. This means each mole of the reactant \(\mathrm{C}_{6} \mathrm{H}_{11} \mathrm{OH}\) yields one mole of \(\mathrm{C}_{6} \mathrm{H}_{10}\).
• Using stoichiometry, we can determine the number of moles of a product or reactant based on a known amount of another component in the reaction using molar masses and conversion factors.
• This concept is crucial when calculating both theoretical yield and determining necessary quantities of reactants to achieve a desired product yield.

Mastering stoichiometry is essential for anyone studying chemistry, as it forms the backbone of understanding chemical reactions quantitatively.