Question

The brewing industry uses yeast to convert glucose to ethanol. The baking industry uses the carbon dioxide produced in the same reaction to make bread rise: $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s) \stackrel{\text { yeast }}{\longrightarrow} 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+2 \mathrm{CO}_{2}(g) $$ How many grams of ethanol can be produced from \(100 . \mathrm{g}\) of glucose? What volume of \(\mathrm{CO}_{2}\) is produced? (Assume \(1 \mathrm{~mol}\) of gas occupies \(22.4 \mathrm{~L}\) at the conditions used. \()\)

Short answer

51.15 g of ethanol and 24.86 L of CO2 are produced.

Step-by-step solution

- Write the Balanced Chemical Equation

Starting with the balanced chemical reaction: \[ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s) \stackrel{\text { yeast }}{\longrightarrow} 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+2 \mathrm{CO}_{2}(g) \]

- Calculate Molar Masses

Calculate the molar mass of glucose (\( \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \)): \[ 6(12.01) + 12(1.01) + 6(16.00) = 180.16 \text{g/mol} \] Calculate the molar mass of ethanol (\( \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} \)): \[ 2(12.01) + 6(1.01) + 16.00 = 46.08 \text{g/mol} \]

- Convert Glucose Mass to Moles

Number of moles of glucose in 100 grams: \[ \frac{100 \text{ g}}{180.16 \text{ g/mol}} = 0.555 \text{ mol} \]

- Use Stoichiometry to Find Moles of Ethanol

From the balanced equation, 1 mol of glucose produces 2 mol of ethanol. So, \[ 0.555 \text{ mol} \text{ glucose} \times \frac{2 \text{ mol} \text{ ethanol}}{1 \text{ mol} \text{ glucose}} = 1.11 \text{ mol ethanol} \]

- Convert Moles of Ethanol to Grams

Number of grams of ethanol produced: \[ 1.11 \text{ mol} \times 46.08 \text{ g/mol} = 51.15 \text{ g ethanol} \]

- Calculate Volume of CO2 Produced

From the balanced equation, 1 mol of glucose produces 2 mol of \( \mathrm{CO}_{2} \). Therefore, \[ 0.555 \text{ mol} \text{ glucose} \times \frac{2 \text{ mol} \mathrm{CO}_{2}}{1 \text{ mol} \text{ glucose}} = 1.11 \text{ mol} \mathrm{CO}_{2} \] Volume of \( \mathrm{CO}_{2} \) produced: \[ 1.11 \text{ mol} \times 22.4 \text{ L/mol} = 24.86 \text{ L} \]

Key concepts

Glucose Fermentation

In fermentation, glucose undergoes a chemical transformation facilitated by yeast. The key reaction for fermenting glucose to ethanol and carbon dioxide is as follows:
\(\text{C}_{6}\text{H}_{12}\text{O}_{6}(s) \rightarrow 2\text{C}_{2}\text{H}_{5}\text{OH}(l) + 2\text{CO}_{2}(g)\). This process is vital in industries such as brewing, where ethanol is the primary product, and baking, where carbon dioxide helps dough rise.

During the reaction, one glucose molecule produces two ethanol molecules and two carbon dioxide molecules. Understanding this balanced equation is crucial for solving stoichiometry problems related to fermentation.

The balanced equation shows a 1:2:2 ratio between glucose, ethanol, and carbon dioxide, respectively. This balance allows us to calculate the amounts of products formed from a given amount of glucose. Recognizing the stoichiometry of the reaction aids in solving related calculations efficiently and accurately.

Yeast Metabolism

Yeast metabolism is a fascinating biological process where yeast cells convert sugars, like glucose, into energy, ethanol, and carbon dioxide. This metabolic process is known as ethanol fermentation and occurs anaerobically, meaning it does not require oxygen to proceed.

Yeast enzymes break down glucose in a series of steps, resulting in the production of ethanol and carbon dioxide, along with ATP (adenosine triphosphate) for cellular energy. Here's the simplified overall reaction:

\(\text{C}_{6}\text{H}_{12}\text{O}_{6} \rightarrow 2\text{C}_{2}\text{H}_{5}\text{OH} + 2\text{CO}_{2}\)

The conversion process involves glycolysis, where glucose is split into pyruvate, and subsequent reactions where pyruvate is further converted to ethanol and carbon dioxide. Yeast's unique ability to perform this conversion is harnessed in various industries for product creation.

• In brewing, ethanol is the desired end product.
• In baking, the carbon dioxide produced forms air pockets in the dough, making it rise.

Understanding yeast metabolism's biochemical pathway is essential for appreciating how and why these industries utilize yeast.

Molar Mass Calculations

Calculating molar mass is fundamental in stoichiometry problems. It involves summing the masses of all the atoms present in a molecule, using their atomic weights from the periodic table. For our example in the glucose fermentation exercise, here’s how we calculate the molar masses:

For glucose (\(\text{C}_{6}\text{H}_{12}\text{O}_{6}\)), we sum the masses of the constituent atoms:

\[6(12.01) + 12(1.01) + 6(16.00) = 180.16 \text{ g/mol}\]

For ethanol (\(\text{C}_{2}\text{H}_{5}\text{OH}\)), we do a similar summation:

\[2(12.01) + 6(1.01) + 16.00 = 46.08 \text{ g/mol}\]

With these molar masses, we can convert between mass and moles, enabling us to solve various stoichiometry problems. For instance, to find how many grams of ethanol can be produced from 100 grams of glucose, we:

• Convert glucose mass to moles: \[ \frac{100 \text{ g}}{180.16 \text{ g/mol}} = 0.555 \text{ mol}\]
• Use the stoichiometric ratio from the balanced equation (1 mol glucose: 2 mol ethanol) to determine moles of ethanol: \[0.555 \text{ mol} \times 2 = 1.11 \text{ mol} \text{ ethanol}\]
• Convert moles of ethanol back to grams: \[1.11 \text{ mol} \times 46.08 \text{ g/mol} = 51.15 \text{ g} \text{ ethanol}\]

Mastering molar mass calculations is crucial for solving various chemical problems accurately.