Problem 76 - Chemistry and Chemical Reactivity

Question

Vitamin C is the simple compound \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{6}\). Besides being an acid, it is a reducing agent. One method for determining the amount of vitamin C in a sample is therefore to titrate it with a solution of bromine, \(\mathbf{B r}_{2},\) an oxidizing agent. \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{6}(\mathrm{aq})+\mathrm{Br}_{2}(\mathrm{aq}) \longrightarrow 2 \mathrm{HBr}(\mathrm{aq})+\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}_{6}(\mathrm{aq})\) A \(1.00-\mathrm{g}\) " chewable" vitamin C tablet requires \(27.85 \mathrm{mL}\) of \(0.102 \mathrm{M} \mathrm{Br}_{2}\) for titration to the equivalence point. What is the mass of vitamin C in the tablet?

Step-by-Step Solution

Answer

The mass of vitamin C in the tablet is 0.500 g.

Step 1: Write the Balanced Reaction Equation

The chemical equation for the reaction is already provided: \( \mathrm{C}_{6} \mathrm{H}_{8}\mathrm{O}_{6}(\mathrm{aq})+\mathrm{Br}_{2}(\mathrm{aq}) \longrightarrow 2\mathrm{HBr}(\mathrm{aq})+\mathrm{C}_{6} \mathrm{H}_{6}\mathrm{O}_{6}(\mathrm{aq}) \). This is balanced as written.

Step 2: Determine Moles of Bromine Used

Using the volume and molarity of \( \mathrm{Br}_2 \), calculate the moles of \( \mathrm{Br}_2 \) used. Number of moles \( = Molarity \times Volume = 0.102 \, \mathrm{mol/L} \times 0.02785 \, \mathrm{L} = 0.0028417 \, \mathrm{mol} \).

Step 3: Relate Moles of Bromine to Moles of Vitamin C

According to the reaction equation, 1 mole of \( \mathrm{Br}_2 \) reacts with 1 mole of \( \mathrm{C}_6 \mathrm{H}_8 \mathrm{O}_6 \). Therefore, the moles of vitamin C must also be 0.0028417.

Step 4: Calculate Mass of Vitamin C

The molecular mass of vitamin C (\( \mathrm{C}_6 \mathrm{H}_8 \mathrm{O}_6 \)) is calculated as follows: \( 6(12.01) + 8(1.01) + 6(16.00) = 176.12 \, \mathrm{g/mol} \). Multiply the number of moles by this molar mass: \( 0.0028417 \, \mathrm{mol} \times 176.12 \, \mathrm{g/mol} = 0.500 \, \mathrm{g} \).

Key Concepts

Redox reactions

Vitamin C titration is fascinating because it involves redox reactions. Redox reactions are chemical processes where the oxidation state of molecules changes. In simpler terms, one substance gets oxidized while another gets reduced, like a teeter-totter in balance. Here, Vitamin C (ascorbic acid) acts as a reducing agent, meaning it donates electrons. Bromine, the oxidizing agent, accepts these electrons, undergoing reduction in the process. In the reaction equation provided, ascorbic acid reduces bromine to hydrobromic acid (\(2 \ \text{HBr}\)). Understanding this electron transfer is key to grasping how the titration works. Each molecule plays its role in this electron-exchanging dance, driving the reaction to completion.

Molarity calculations

Molarity, a term you might often encounter, is a measure of concentration. It tells us how many moles of a solute are present in a liter of solution. In this exercise, we used the molarity of the bromine solution to calculate the amount used during the titration. By multiplying the molarity of bromine (0.102 M) by the volume used (0.02785 liters), we find the moles of bromine involved in the reaction to be 0.0028417 moles. This step is crucial because knowing the exact amount of bromine tells us how much Vitamin C it reacted with.

Stoichiometry

Stoichiometry is like the road map of chemistry. It lets us understand the quantitative relationships between reactants and products in a chemical reaction. For this vitamin C titration problem, stoichiometry tells us that one mole of bromine reacts with one mole of vitamin C. This one-to-one relationship is derived directly from the balanced chemical equation. Thus, the same amount of moles of vitamin C is used as the moles of bromine, simplifying our calculations. Stoichiometry helps in predicting quantities needed and produced, making our calculations precise and accurate.

Chemical equations

Chemical equations are more than just symbols and numbers; they tell a story of what's happening on a molecular level. Our equation: \(\text{C}_6\text{H}_8\text{O}_6(\text{aq}) + \text{Br}_2(\text{aq}) \longrightarrow 2\text{HBr}(\text{aq}) + \text{C}_6\text{H}_6\text{O}_6(\text{aq})\) illustrates how the reactants transform into products. A balanced equation is key, ensuring that matter is conserved, or in other words, the same amount of each element is present before and after the reaction. Balancing ensures that your equation follows the law of conservation of mass, forming the backbone for further calculations in the titration process.

Molecular mass determination

Determining molecular mass is crucial for converting moles to grams. In this vitamin C titration exercise, knowing the molecular mass of vitamin C (ascorbic acid) helped calculate its weight in the sample. The molecular formula \(\text{C}_6\text{H}_8\text{O}_6\) allows us to add up the atomic masses: 6 atoms of carbon (6 x 12.01 g/mol), 8 atoms of hydrogen (8 x 1.01 g/mol), and 6 atoms of oxygen (6 x 16.00 g/mol), totaling 176.12 g/mol. Using this, we convert the moles of vitamin C to grams, providing the mass present in the vitamin C tablet. This conversion is necessary for practical applications and understanding real-world proportions.