Question

When yeast is added to a solution of glucose or fructose, the sugars are said to undergo fermentation, and ethyl alcohol is produced. $${\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6(\mathrm{aq})\to 2{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}(\mathrm{aq})+2{\mathrm{CO}}_2(g)$$ This is the reaction by which wines are produced from grape juice, Calculate the mass of ethyl alcohol, ${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}$, produced when $5.25\mathrm{g}$ of glucose, ${\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_0,$ undergoes this reaction.

Short answer

When 5.25 g of glucose undergoes fermentation, 2.68 g of ethyl alcohol is produced.

Step-by-step solution

Calculate moles of glucose provided

First, we need to determine the number of moles of glucose. We can do this using the molar mass of glucose which we can calculate from the Periodic Table: Molar mass of Glucose: $C_6{H}_{12}{O}_6=6\times 12.01g/mol+12\times 1.01g/mol+6\times 16.00g/mol=180.18g/mol$ Now, we can calculate the moles of glucose: Moles of Glucose: $\frac{5.25g}{180.18g/mol}=0.0291mol$

Determine moles of ethyl alcohol produced using stoichiometry

According to the balanced equation, 1 mole of glucose produces 2 moles of ethyl alcohol: C6H12O6 (aq) → 2 C2H5OH (aq) + 2 CO2 (g) So, we can set up the stoichiometric relationship as follows: Moles of Ethyl Alcohol: $0.0291mol\times \frac{2mol{C}_2{H}_{5}OH}{1mol{C}_6{H}_{12}{O}_6}=0.0582mol$

Calculate the mass of ethyl alcohol produced

Now, we need to calculate the mass of ethyl alcohol produced using the molar mass of ethyl alcohol: Molar mass of Ethyl Alcohol: $C_2{H}_{5}OH=2\times 12.01g/mol+6\times 1.01g/mol+1\times 16.00g/mol=46.07g/mol$ Now we can calculate the mass of ethyl alcohol: Mass of Ethyl Alcohol: $0.0582mol\times 46.07g/mol=2.68g$ Thus, when 5.25 g of glucose undergoes fermentation, 2.68 g of ethyl alcohol is produced.

Key concepts

Stoichiometry

Stoichiometry is a branch of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. It is based on the conservation of mass, which states that in an isolated system, mass cannot be created or destroyed during a chemical reaction. Understanding stoichiometry is vital when performing chemical calculations because it helps you relate the quantities of one substance to those of another.

For example, in the fermentation reaction from the exercise, stoichiometry allows us to determine how much ethyl alcohol will be produced from a given amount of glucose. The balanced chemical equation indicates that for every molecule of glucose (\r$C_6{H}_{12}{O}_6$), two molecules of ethyl alcohol (\r$C_2{H}_{5}OH$) are produced. This ratio (1:2) is used to convert moles of glucose to moles of ethyl alcohol. Once you have the moles of the desired product, you're a step closer to finding out the actual mass of the product formed in the reaction.

Molar Mass Calculation

The molar mass of a substance is the weight of one mole of that substance; essentially, it's the sum of the atomic masses of all the atoms in a molecule. Molar mass is measured in grams per mole (\r$g/mol$), and it's a crucial concept for converting between the mass of a substance and the amount of substance in moles.

Let's break it down using the fermentation example. To find the molar mass of glucose (\r$C_6{H}_{12}{O}_6$), we calculate the sum of the atomic masses of six carbon atoms, twelve hydrogen atoms, and six oxygen atoms. Similarly, to find the molar mass of ethyl alcohol (\r$C_2{H}_{5}OH$), we add the atomic masses of two carbon atoms, six hydrogen atoms, and one oxygen atom. Knowing the molar masses allows us to convert the mass of glucose given in the problem to moles, and later the moles of ethyl alcohol to grams, which is the mass we're trying to find.

Chemical Equation Balancing

Balancing chemical equations is fundamental in stoichiometry. It ensures that the law of conservation of mass is satisfied, meaning that there is an equal number of each type of atom on both the reactant and product sides of a chemical equation.

In the fermentation chemical reaction provided in the exercise, the chemical equation is already balanced for us: \r$C_6{H}_{12}{O}_6(aq)\to 2C_2{H}_{5}OH(aq)+2C{O}_2(g)$. It shows that one glucose molecule reacts to form two molecules of ethyl alcohol and two molecules of carbon dioxide gas. This balance signifies that if we start with a certain number of moles of glucose, we can predict the number of moles of ethyl alcohol produced. Having a correctly balanced equation is crucial because it provides the exact stoichiometric ratios necessary to perform accurate calculations in any chemical reaction.