Problem 1 - Chemical Principles in the Laboratory

Question

In a reaction involving the iodination of acetone, the following volumes were used to make up the reaction mixture: $$\text { 10. mL } 4.0 \mathrm{M} \text { acetone }+10 . \mathrm{mL} 1.0 \mathrm{M} \mathrm{HCl}+10 . \mathrm{mL} 0.0050 \mathrm{M} \mathrm{I}_{2}+20 . \mathrm{mL} \mathrm{H}_{2} \mathrm{O}$$ a. How many moles of acetone were in the reaction mixture? Recall that, for a component A, moles of \(\mathrm{A}=[\mathrm{A}] \times V,\) where \([\mathrm{A}]\) is the molarity of \(\mathrm{A}\) and \(V\) is the volume in liters of the solution of \(\mathrm{A}\) that was used. _____moles acetone b. What was the molarity of acetone in the reaction mixture? The volume of the mixture was \(50 . \mathrm{mL}\), or \(0.050 \mathrm{L},\) and the number of moles of acetone was found in Part (a). Again, $$[\mathrm{A}]=\frac{\text { moles of } \mathrm{A}}{V \text { of soln. in liters }}$$ _____\(\mathrm{M}\) acetone c. How could you double the molarity of the acetone in the reaction mixture, keeping the total volume at \(50 . \mathrm{mL}\) and keeping the same concentrations of \(\mathrm{H}^{+}\) ion and \(\mathrm{I}_{2}\) as in the original mixture?

Step-by-Step Solution

Answer

a) 0.040 moles acetone; b) 0.80 M acetone; c) Use 20 mL acetone, 10 mL HCl, 10 mL I2, and no H2O.

Step 1: Calculate Volume in Liters

First, convert the volume of acetone from milliliters to liters. Given: \(\text{Volume} = 10 \text{ mL} = 0.010 \text{ L}\).

Step 2: Calculate Moles of Acetone

Use the formula for moles: \(\text{Moles of A} = [A] \times V\). With \([\text{Acetone}] = 4.0\, \text{M}\) and \(V = 0.010\, \text{L}\), calculate \(\text{Moles of Acetone} = 4.0 \times 0.010 = 0.040\, \text{moles}\).

Step 3: Calculate Molarity of Acetone in Mixture

The total volume of the reaction mixture is given as \(0.050\, \text{L}\). The molarity is calculated as \([\text{Acetone}] = \frac{\text{moles of Acetone}}{V} = \frac{0.040}{0.050} = 0.80\, \text{M}\).

Step 4: Modify Molarity While Keeping Total Volume Constant

To double the molarity to \(1.60\, \text{M}\) without changing the volume, the moles of acetone must also double. You can achieve this by using \(20\, \text{mL}\) of \(4.0\, \text{M}\) acetone and substituting the extra acetone volume for \(10\, \text{mL}\) of \(\text{H}_2\text{O}\) while making sure the original concentrations of \(\text{H}^+\) ion and \(\text{I}_2\) are maintained with \(10\, \text{mL}\) of \(1.0\, \text{M HCl}\) and \(10\, \text{mL}\) of \(0.0050\, \text{M } \text{I}_2\).

Key Concepts

Molarity

Molarity is a fundamental concept in chemistry, which describes the concentration of a solute within a solution. It is expressed as the number of moles of solute per liter of solution. In the context of the iodination of acetone, molarity helps to determine the concentration of different reactants involved in the reaction.

To calculate molarity, the formula \[ [ ext{A}] = \frac{ ext{moles of A}}{V ext{ of soln. in liters}} \] is used. In this formula,

• \([ ext{A}]\) signifies the molarity of component A.
• The numerator represents the moles of the solute.
• The denominator describes the volume of the solution in liters.

This approach allows us to understand not just the amount, but the concentration of acetone in a given reaction mixture. For example, knowing the molarity of acetone will enable chemists to control the rate of the reaction by altering the concentration.

Reaction Mixture

A reaction mixture encompasses all the reactants involved in a chemical reaction alongside any solvent present. In the iodination of acetone exercise, the reaction mixture includes acetone, hydrochloric acid (HCl), iodine (\( ext{I}_2\)), and water.

When preparing a reaction mixture, each component is added in specific volumes to achieve the desired concentration and to facilitate the progression of the reaction. In this case:

• \(10\) mL of \(4.0\) M acetone ensures enough reactant is present for the iodination process.
• \(10\) mL of \(1.0\) M HCl acts as a catalyst.
• \(10\) mL of \(0.0050\) M \(\text{I}_2\) provides the necessary iodine for the reaction.
• \(20\) mL of water complements the mixture, ensuring the right reaction volume.

Understanding the composition and purpose of each reactant within the reaction mixture is crucial for achieving optimal yield and efficiency.

Moles Calculation

Calculating moles is an essential step in determining how much of a substance is present in a reaction. The formula for calculating moles involves multiplying the molarity by the volume in liters. Specifically, \[ ext{Moles of A} = [A] imes V \]

In the iodination of acetone, for instance, the moles of acetone can be calculated using its molarity and volume in the reaction mixture:

• Given \, [ ext{Acetone}] = 4.0 \, ext{M}
• Volume \, (V) = 0.010 \, ext{L}
• Moles of acetone = 4.0 \, ext{M} \times 0.010 \, ext{L} = 0.040 \, ext{moles}

The ability to calculate moles helps chemists to understand the quantities of reactants they'll have available as the reaction proceeds, thus allowing for proper stoichiometric calculations and adjustments if necessary to achieve the desired outcome.