Problem 90 - Chemistry Matter and Change

Question

Ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) is produced from the fermentation of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) in the presence of enzymes. \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightarrow 4 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(1)+4 \mathrm{CO}_{2}(\mathrm{g})\) Determine the theoretical yield and the percent yield of ethanol if 684 \(\mathrm{g}\) of sucrose undergoes fermentation and 349 \(\mathrm{g}\) of ethanol is obtained.

Step-by-Step Solution

Answer

The theoretical yield of ethanol is 368.544 g, with a percent yield of 94.69%.

Step 1: Calculate Molar Masses

To solve this problem, we first need to determine the molar masses of sucrose (C₁₂H₂₂O₁₁) and ethanol (C₂H₅OH). For sucrose: - Carbon (C): 12 atoms × 12.01 g/mol = 144.12 g/mol - Hydrogen (H): 22 atoms × 1.008 g/mol = 22.18 g/mol - Oxygen (O): 11 atoms × 16.00 g/mol = 176.00 g/mol The molar mass of sucrose = 144.12 + 22.18 + 176.00 = 342.30 g/mol. For ethanol: - Carbon (C): 2 atoms × 12.01 g/mol = 24.02 g/mol - Hydrogen (H): 6 atoms × 1.008 g/mol = 6.048 g/mol - Oxygen (O): 1 atom × 16.00 g/mol = 16.00 g/mol The molar mass of ethanol = 24.02 + 6.048 + 16.00 = 46.068 g/mol.

Step 2: Determine Moles of Sucrose

We need to convert the mass of sucrose to moles. The formula for moles is:\[ \text{moles} = \frac{\text{mass}}{\text{molar mass}} \]Given that the mass of sucrose is 684 g, we use:\[ \text{moles of sucrose} = \frac{684 \, \text{g}}{342.30 \, \text{g/mol}} = 2.00 \, \text{moles} \]

Step 3: Calculate Theoretical Yield of Ethanol

From the balanced equation, 1 mole of sucrose produces 4 moles of ethanol. Therefore, 2.00 moles of sucrose will produce:\[ 2.00 \, \text{moles of sucrose} \times 4 \, \text{moles of ethanol/mole of sucrose} = 8.00 \, \text{moles of ethanol} \]We convert moles of ethanol to grams using its molar mass, 46.068 g/mol:\[ \text{grams of ethanol} = 8.00 \, \text{moles} \times 46.068 \, \text{g/mol} = 368.544 \, \text{g} \]

Step 4: Calculate Percent Yield of Ethanol

The percent yield is calculated using the formula:\[ \text{Percent Yield} = \left( \frac{\text{actual yield}}{\text{theoretical yield}} \right) \times 100 \%\]Given that the actual yield of ethanol is 349 g, and the theoretical yield is 368.544 g:\[ \text{Percent Yield} = \left( \frac{349 \, \text{g}}{368.544 \, \text{g}} \right) \times 100 \% = 94.69 \% \]

Key Concepts

Percent Yield

Percent yield is a measure of how efficiently a chemical reaction produces the desired outcome. It compares the actual amount of product obtained from a reaction to the theoretical yield, which is the maximum possible amount of product predicted by stoichiometry. Calculating percent yield is essential in the laboratory to determine the effectiveness of a reaction.

To calculate percent yield, use the formula:

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

A percent yield value of 100% indicates that the reaction produced the exact amount predicted, while values below 100% suggest that not all reactants were converted to products.

Factors that can affect percent yield include incomplete reactions, side reactions, and losses during product recovery.

Molar Mass

Molar mass is a fundamental concept in chemistry, serving as the bridge from mass measurements to moles, which chemists use to count particles. A substance's molar mass is the mass in grams of one mole of that substance. Understanding molar mass is crucial when performing stoichiometry calculations in chemical reactions.

To find the molar mass of a compound, add up the atomic masses of all atoms in its chemical formula. For example, for ethanol

• Carbon (C): 2 atoms \(\times 12.01 \, \text{g/mol} = 24.02\, \text{g/mol}\)
• Hydrogen (H): 6 atoms \(\times 1.008 \, \text{g/mol} = 6.048\, \text{g/mol}\)
• Oxygen (O): 1 atom \(\times 16.00 \, \text{g/mol} = 16.00\, \text{g/mol}\)

The molar mass of ethanol is the sum of these values: 46.068 g/mol.

Molar mass allows you to convert weights into the number of moles, crucial for understanding how much of a substance reacts or is formed in a reaction.

Stoichiometry

Stoichiometry is the study of the quantitative relationships, or ratios, between the amounts of reactants and products in a chemical reaction. It is founded on the principle that matter is neither created nor destroyed in a chemical reaction.

The balanced chemical equation gives you the stoichiometric coefficients, which show the ratio of moles of each substance involved. For instance, in the fermentation of sucrose to ethanol, the equation \[ \text{C}_{12}\text{H}_{22}\text{O}_{11}(\text{aq})+\text{H}_{2}\text{O}(\text{g}) \rightarrow 4\text{C}_{2}\text{H}_{5}\text{OH}(1)+4\text{CO}_{2}(\text{g}) \] indicates that 1 mole of sucrose results in 4 moles of ethanol.

By understanding stoichiometry, you can predict the amount of products formed from a given amount of reactants and also determine how much reactant is needed to fully convert into products. This calculation was crucial in determining the theoretical yield of ethanol in the given exercise.

Fermentation

Fermentation is a metabolic process where organisms convert sugars into acids, gases, or alcohol in the absence of oxygen. It is an essential chemical reaction in various biological systems and industrial applications.

In the context of this exercise, fermentation involves the biochemical conversion of sucrose into ethanol and carbon dioxide, facilitated by enzymes. The equation showcases this transformation:

• \( \text{C}_{12}\text{H}_{22}\text{O}_{11}(\text{aq}) + \text{H}_{2}\text{O}(\text{g}) \rightarrow 4\text{C}_{2}\text{H}_{5}\text{OH}(1) + 4\text{CO}_{2}(\text{g}) \)

Fermentation is not only important in food and beverage production but also in biofuel production, such as ethanol for energy.

Understanding fermentation helps in optimizing conditions for desired product yields and recognizing the role of enzymes as biological catalysts in these transformations.