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. $$\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightarrow 4 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{l})+4 \mathrm{CO}_{2}(\mathrm{g})$$ Determine the theoretical yield and the percent yield of ethanol if 684 g of sucrose undergoes fermentation and 349 g of ethanol is obtained.

Short answer

The theoretical yield of ethanol is 368.56 g, and the percent yield is 94.69%.

Step-by-step solution

Find the Molar Masses of the Compounds

To calculate the molar mass of sucrose and ethanol, we simply add up the atomic masses of all the atoms in their respective chemical formulas: Molar mass of sucrose, \(C_{12}H_{22}O_{11}\): (12 atoms of C x 12.01 g/mol) + (22 atoms of H x 1.01 g/mol) + (11 atoms of O x 16.00 g/mol) = 342.30 g/mol Molar mass of ethanol, \(C_{2}H_{5}OH\): (2 atoms of C x 12.01 g/mol) + (6 atoms of H x 1.01 g/mol) + (1 atom of O x 16.00 g/mol) = 46.07 g/mol

Find Moles of Sucrose

Now that we have the molar mass of sucrose, we can find the number of moles of sucrose used in the reaction: Moles of sucrose = \(\cfrac{\text{mass of sucrose}}{\text{molar mass of sucrose}}\) Moles of sucrose = \(\cfrac{684~\text{g}}{342.30~\text{g/mol}} \approx 2.00~\text{moles}\)

Use Stoichiometry to Find Theoretical Yield of Ethanol

According to the balanced chemical equation, 1 mole of sucrose reacts with 1 mole of water to produce 4 moles of ethanol and 4 moles of carbon dioxide: $$\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightarrow 4 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{l})+4 \mathrm{CO}_{2}(\mathrm{g})$$ Using the stoichiometric ratio, we find the moles of ethanol that can be formed by the reaction: Moles of ethanol = (2.00 moles of sucrose) x (4 moles of ethanol / 1 mole of sucrose) = 8.00 moles of ethanol Now, we can calculate the theoretical yield of ethanol in grams: Theoretical yield of ethanol = (8.00 moles of ethanol) x (46.07 g/mol) = 368.56 g

Calculate Percent Yield

To find the percent yield, we'll use the formula: Percent yield = \(\cfrac{\text{actual yield}}{\text{theoretical yield}}\) x 100 In this problem, the actual yield of ethanol is given as 349 g. Plugging the values into the formula, we get: Percent yield = \(\cfrac{349~\text{g}}{368.56~\text{g}}\) x 100 = 94.69% Thus, the theoretical yield of ethanol is 368.56 g, and the percent yield is 94.69%.

Key concepts

Stoichiometry

Stoichiometry is like a recipe for a chemical reaction, telling you how much of each substance is needed to make a certain amount of a product. Let's imagine making a sandwich—you need a certain number of slices of bread and pieces of cheese to make a certain number of sandwiches. Similarly, stoichiometry uses a balanced chemical equation to relate the amounts of reactants (starting materials) to the amounts of products (what we end up with).

In our exercise, we use stoichiometry to determine how many moles of ethanol we can get from fermenting sucrose. The balanced equation tells us that four moles of ethanol are produced for every mole of sucrose. By knowing this ratio, we can calculate the theoretical yield—the maximum amount of ethanol that can be made if everything goes perfectly. But in real life, reactions don't always go perfectly, which is why we also look at the percent yield, indicating how much ethanol was actually obtained compared to the theoretical yield.

Molar Mass

The molar mass is the weight of one mole of a substance (like the weight of a dozen eggs). One mole is like a universal number of molecules, just as a dozen is a count of 12. In chemistry, the molar mass allows us to convert between the weight of a substance and the number of moles. This is crucial because chemical reactions happen between molecules, not grams.

So, when we calculated the molar masses of sucrose and ethanol, we added up the atomic weights of each element in their formulas. Once we had the molar mass, we converted the weight of sucrose used in the experiment (684 g) to moles, because the balanced chemical equation speaks in terms of moles, not grams.

Chemical Reaction

A chemical reaction is like a dance where partners (elements or molecules) leave one group to join another. In the equation we have, sucrose and water are our initial dancers. When the music starts (the reaction happens), these dancers break apart and reform into new groups—ethanol and carbon dioxide.

The balanced chemical equation is our choreography, showing exactly how the dancers swap partners. It's key to making sure that we don't lose any dancers along the way because in chemical reactions, just like in a dance, no one disappears—they just change partners. All of this partnering and swapping is why stoichiometry and molar masses are so important. We need to know the details of the dance to predict its outcome.

Fermentation of Sucrose

The fermentation of sucrose is a natural process, somewhat like brewing beer. Yeast acts like a tiny factory, changing sucrose into ethanol and carbon dioxide—this is the same process used to make alcoholic drinks.

In our exercise, we look at how this process can be written down in the form of a chemical equation. The equation \(C_{12}H_{22}O_{11} + H_2O \rightarrow 4 C_{2}H_{5}OH + 4 CO_{2}\) tells us that sucrose and water transform into ethanol and carbon dioxide when the reaction occurs. Understanding this process helps us predict how much ethanol we can make from a certain amount of sucrose—which is what we needed to determine the theoretical yield in our problem.