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: $$5 \mathrm{mL} 4.0 \mathrm{M} \text { acetone }+10 \mathrm{mL} 1.0 \mathrm{M} \mathrm{HCl}+10 \mathrm{mL} 0.0050 \mathrm{M} \mathrm{I}_{2}+25 \mathrm{mL} \mathrm{H}_{2} \mathrm{O}$$ a. How many moles of acetone were in the reaction mixture? Recall that, for a component \(A\), moles \(A=M_{A} \times V,\) where \(M_{A}\) is the molarity of \(A\) and \(V\) is the volume in liters of the solution of \(A\) that was used. ___________ moles acetone b. What was the molarity of acetone in the reaction mixture? The volume of the mixture was \(50 \mathrm{mL}\). 0.050 L, and the number of moles of acctone was found in Part (a). Again, \(M_{A}=\frac{\text { moles of } A}{V \text { of soln. in liters }}\) ___________________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. 0.02 moles acetone. b. 0.4 M acetone. c. Use 10 mL of 4.0 M acetone.

Step 1: Calculate Moles of Acetone

To find the moles of acetone, use the formula: \( \text{moles} = M \times V \). Here, the molarity \( M \) of acetone is \( 4.0 \, \text{M} \) and the volume \( V \) is in liters. Convert the given 5 mL to liters: \( 5 \times 10^{-3} \, \text{L} \). Now calculate the moles: \[ \text{moles of acetone} = 4.0 \, \text{M} \times 5 \times 10^{-3} \, \text{L} = 0.02 \, \text{moles}. \]

Step 2: Determine Molarity of Acetone in Reaction Mixture

Use the molarity formula \( M = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \). The total volume of the reaction mixture is given as 50 mL or 0.050 L. We already calculated 0.02 moles of acetone in Step 1. Use these values to find the molarity: \[ M = \frac{0.02 \, \text{moles}}{0.050 \, \text{L}} = 0.4 \, \text{M}. \]

Step 3: Double the Molarity of Acetone

To double the molarity to 0.8 M while maintaining the total volume at 50 mL, increase the volume of 4.0 M acetone used. Let \( V \) be the new volume of 4.0 M acetone needed: \( 4.0 \, \text{M} \times \frac{V}{\text{L}} = 0.8 \, \text{M} \). Therefore, \( V = 0.8 \times 0.050 \, \text{L} = 0.01 \, \text{L} = 10 \, \text{mL} \). Use 10 mL of 4.0 M acetone, adjust the amount of water, and keep the rest of the components constant.

Key Concepts

Molarity

Molarity is a fundamental concept in chemistry that helps in understanding how concentrated a solution is. It's defined as the number of moles of solute divided by the volume of the solution in liters.
To calculate molarity, one important step is determining the number of moles of your solute. For example, if you have a reaction mixture containing acetone, knowing its molarity helps in adjusting reaction conditions effectively.
When you dilute or concentrate a solution, the molarity can change. In the provided example, after calculating the moles of acetone, the total volume of the mixture was 50 mL (or 0.050 L). The molarity of acetone in this mixture can be found using the formula:

• \( M = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \)

The calculated molarity of acetone was found to be 0.4 M. Understanding this concept is crucial for solving chemical reaction problems efficiently.

Moles Calculation

Calculating moles is the first essential step in determining the amount of a substance in a given volume. Moles represent a unit of matter defined as 6.022 x 10^23 particles (atoms, molecules, etc.).
The formula to find moles from molarity and volume is:

• \( \text{moles} = M \times V \)

where \( M \) is molarity and \( V \) is volume in liters.
From the example, the initial volume given was 5 mL, converted to liters (\( 5 \times 10^{-3} \text{ L} \)), and the molarity was 4.0 M. Using these values, the moles of acetone were calculated as 0.02 moles. This simple conversion process is essential in any chemical calculation because it lays the groundwork for further calculations like molarity and reaction ratios.

Solution Preparation

Solution preparation involves creating a solution of a precise molarity by mixing a known amount of solute with a solvent. The challenge often lies in adjusting concentrations while keeping certain conditions consistent.
Your goal might be to adjust the molarity of a solution without altering the total volume, as seen in the exercise example.

• If the aim is to double the molarity while keeping the solution at the same volume, you'll need to alter the amount of the solute or adjust other components.

In this case, the solution involved doubling the molarity of acetone from 0.4 M to 0.8 M, which required using double the amount of the acetone (from 5 mL to 10 mL) but keeping total solution volume consistent at 50 mL by adjusting the volume of water accordingly.
This meticulous process allows for precise control over reaction conditions, ensuring consistent and accurate results in chemical reactions.