Question

In a reaction involving the iodination of acetone, the following volumes were used to make up the reaction mixture: $$10 \mathrm{mL} 4.0 \mathrm{M} \text { acetone }+10 \mathrm{mL} 1.0 \mathrm{M} \mathrm{HCl}+10 \mathrm{mL} 0.0050 \mathrm{M} \mathrm{I}_{2}+20 \mathrm{mL} \mathrm{H}_{2} \mathrm{O}$$ a. How many moles of acetone were in the reaction mixture? Recall that, for a component \(A\), no. moles \(A=M_{A} \times V,\) where \(M_{A}\) is the molarity of \(A\) and \(V\) the volume in liters of the solution of \(A\) that was used. __________________moles acetone. What was the molarity of acetone in the reaction mixture? The volume of the mixture was \(50 \mathrm{mL}\), 0.050 liter, and the number of moles of acetone was found in Part a. Again, \(M_{A}=\frac{\text { no. moles } A}{V \text { of soln. in liters }}\) __________________M acetone. 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?

Short answer

0.040 moles of acetone were used in the original reaction mixture, and the molarity of acetone in the reaction mixture was 0.80 M. To double the molarity of acetone in the reaction mixture while keeping the total volume at 50 mL and the same concentrations of H+ ion and I2, the new reaction mixture should consist of 20 mL of 4.0 M acetone, 10 mL of 1.0 M HCl, 10 mL of 0.0050 M I2, and only 10 mL of H2O, instead of 20 mL.

Step-by-step solution

Calculate the moles of acetone in the reaction mixture

We are given the volume and molarity of acetone in the mixture: V_acetone = 10 mL and M_acetone = 4.0 M. Since 1 L = 1000 mL, we convert the volume to liters: V_acetone = 10 mL × (1 L / 1000 mL) = 0.010 L. Then, the formula to calculate moles A = M_A × V helps us find the moles of acetone: moles_acetone = M_acetone × V_acetone = 4.0 M × 0.010 L = 0.040 moles. Answer: 0.040 moles acetone.

Calculate the molarity of acetone in the reaction mixture

We found the number of moles of acetone in part (a) to be 0.040 moles. To find the molarity of acetone, we need to divide the number of moles of acetone by the total volume of the reaction mixture in liters. The total volume of the reaction mixture is 50 mL, so in liters it is 0.050 L. Therefore, the molarity of acetone in the reaction mixture is M_acetone_mixture = (0.040 moles) / (0.050 L) = 0.80 M. Answer: 0.80 M acetone.

Double the molarity of acetone in the reaction mixture

To double the molarity of acetone in the reaction mixture while keeping the total volume at 50 mL and the concentrations of H+ ion and I2 constant, we can first calculate the new volume of 4.0 M acetone solution needed to achieve this. Let V_new_acetone be the volume in liters of the new 4.0 M acetone solution needed. Then, the total number of moles of acetone in the new mixture will be twice the moles found in Step 1, which is 2 × 0.040 moles = 0.080 moles. Using the formula for moles, A =M_A × V, we can solve for the new volume of acetone solution needed: V_new_acetone = (0.080 moles) / (4.0 M) = 0.020 L, or 20 mL. Next, we need to adjust the volume of water in the mixture to maintain a total volume of 50 mL. Since we are now using 20 mL of acetone, we will need to decrease the volume of water to 30 mL. Therefore, the new reaction mixture would consist of: - 20 mL of 4.0 M acetone - 10 mL of 1.0 M HCl - 10 mL of 0.0050 M I2 - 10 mL of H2O, instead of 20 mL This will ensure we double the molarity of acetone while keeping the total volume at 50 mL and the same concentrations of H+ ion and I2.

Key concepts

Molarity Calculation

Understanding the concept of molarity is essential in chemistry, especially when mixing solutions to react, as with iodination of acetone. Molarity, denoted as M, is a measure of the concentration of a solution. It is defined as the number of moles of solute (which is the substance being dissolved) per liter of solution. The equation for calculating molarity is:
\[ M = \frac{\text{number of moles of solute}}{\text{volume of solution in liters}} \]
When you're performing experiments or working with chemical reactions, being able to calculate the molarity of a solution allows you to determine how much of each reactant is present. In our exercise, we used the volumes and molarities of different reactants to find the moles of acetone present. For example, with 10 mL of a 4.0 M acetone, we first converted milliliters to liters, because molarity requires the volume in liters. We then used the formula \( \text{moles} = M \times V \) to calculate the number of moles of acetone, which was 0.040 moles in this case.

Why is Molarity Important?

Molarity is important as it provides a straightforward way to discuss the concentration of solutions in a uniform way, making it simpler to follow recipes in synthetic chemistry, adjust concentrations for experiments, and predict the outcomes of reactions.

Chemical Reaction Stoichiometry

Chemical reaction stoichiometry is about the quantitative relationship between the reactants and products in a chemical reaction. It forms the basis for the molarity calculations, as seen in the iodination of acetone. The stoichiometry allows us to predict how much product will form from given amounts of reactants and is vital for making sure reactions go to completion.
In practice, stoichiometry requires a balanced chemical equation where the number of atoms for each element is the same on both sides of the reaction. Once the equation is balanced, the coefficients can be used to determine the molar ratios between the reactants and the products. This helps in calculating how much of each reactant is needed and in assessing the efficiency of the reaction.

Practical Application of Stoichiometry

In the iodination of acetone, if we wanted to double the molarity of acetone, we would use stoichiometry to ensure that the amounts of the other reactants are proportionally correct, so the reaction can still occur. Here, stoichiometry tells us that if we double the amount of acetone, the volumes of other reactants remain unchanged, but the amount of water must be reduced to keep the total solution volume the same.

Laboratory Techniques in Chemistry

Successful execution of experiments like the iodination of acetone relies heavily on the precise use of laboratory techniques. In chemistry labs, common techniques include measuring volumes, handling chemical reagents, mixing solutions, and controlling reaction conditions.
For instance, pipetting is used to measure out precise volumes of solutions, like the 10 mL measures of acetone, HCl, and I2 in our exercise. To prevent contamination, appropriate pipetting techniques must be followed. Mixing is another key technique, ensuring the reactants come into contact uniformly for the reaction to proceed efficiently.

Accuracy and Safety

Accuracy in measurement and calculation is crucial to maintain the desired concentration and reaction conditions. Safe handling of materials, proper waste disposal, and cleanliness are also essential parts of laboratory work. For example, when doubling the molarity of acetone in our mixture while keeping the volume and concentrations of other components constant, precision in measuring out the new volumes of acetone and water is crucial to achieving correct molarity.