Problem 67 - Chemistry: The Central Science in SI Units, Global Edition

Question

(a) How many grams of ethanol, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\), should you dissolve in water to make \(1.00 \mathrm{~L}\) of vodka (which is an aqueous solution that is \(6.86 \mathrm{M}\) ethanol)? (b) Using the density of ethanol \((0.789 \mathrm{~g} / \mathrm{mL})\), calculate the volume of ethanol you need to make \(1.00 \mathrm{~L}\) of vodka.

Step-by-Step Solution

Answer

(a) To make 1.00 L of vodka with 6.86 M ethanol, you need to dissolve 315.56 g of ethanol in water. (b) To make 1.00 L of vodka, you need 400 mL of ethanol (approximately).

Step 1: Calculate the moles of ethanol required for 1.00 L of vodka.

To find out how many grams of ethanol are needed, first, we have to find the number of moles of ethanol required to make 1.00 L of vodka. The molarity (M) is the number of moles of solute (in this case, ethanol) per liter of solution. Therefore, for 6.86 M of ethanol, we have: Moles of ethanol = molarity × volume of the solution Moles of ethanol = \(6.86 \, \text{moles/L} \times 1.00 \, \text{L}\)

Step 2: Calculate the mass of ethanol required for 1.00 L of vodka.

Now that we have the moles of ethanol required, we can find the mass by using the molecular weight of ethanol. The molecular weight of \(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OH}\) is approximately 46 g/mol. So, to find the mass of ethanol needed, we multiply the moles of ethanol by its molecular weight: Mass of ethanol = moles of ethanol × molecular weight of ethanol Mass of ethanol = moles of ethanol × 46 g/mol

Step 3: Calculate the volume of ethanol required for 1.00 L of vodka using the density.

Now that we have the mass of ethanol, we can calculate its volume using the given density (0.789 g/mL). To find the volume, we need to divide the mass by the density: Volume of ethanol = mass of ethanol ÷ density Volume of ethanol = mass of ethanol ÷ 0.789 g/mL Now we can calculate the mass and volume of ethanol required to make 1.00 L of vodka using the given molarity and density of ethanol.

Step 4: Answers

(a) Mass of ethanol required: Moles of ethanol = \(6.86 \, \text{moles/L} \times 1.00 \, \text{L} = 6.86 \, \text{moles}\) Mass of ethanol = \(6.86 \, \text{moles} \times 46 \, \text{g/mol} = 315.56 \, \text{g}\) So, to make 1.00 L of vodka with 6.86 M ethanol, you need to dissolve 315.56 g of ethanol in water. (b) Volume of ethanol required: Volume of ethanol = \(315.56 \, \text{g} ÷ 0.789 \, \text{g/mL} = 400 \, \text{mL}\) (approximately) Therefore, to make 1.00 L of vodka, you need 400 mL of ethanol.

Key Concepts

Molarity

Molarity is a way of expressing the concentration of a solute in a solution. It is defined as the number of moles of a solute per liter of solution. The formula for molarity
can be represented as:
\[M = \frac{n}{V}\]where \(M\) is the molarity, \(n\) is the number of moles of solute, and \(V\) is the volume of the solution in liters.

To prepare 1.00 L of a 6.86 M ethanol solution:

• We need to calculate the moles of ethanol by multiplying the desired molarity (6.86 M) by the volume of the solution (1.00 L).
• This simple multiplication gives us the total moles of ethanol needed, which is 6.86 moles.

The concept of molarity is essential in solution preparation because it provides a clear indication of concentration, ensuring precise measurements are used in experiments and solutions.

Density Calculations

Density is a measure of how much mass is contained in a given volume. It is commonly expressed as grams per milliliter (g/mL) or kilograms per liter (kg/L). The formula to calculate density is:
\[\text{density} = \frac{\text{mass}}{\text{volume}}\]When you know the density and mass of a substance, you can find its volume using the rearranged formula:
\[\text{volume} = \frac{\text{mass}}{\text{density}}\]For ethanol:

• The given density is 0.789 g/mL.
• If we need to prepare 1.00 L of vodka, first calculate the mass of ethanol and divide it by its density to find the required volume.
• This allows us to determine that 315.56 g of ethanol results in 400 mL of volume due to the given density.

Density calculations are vital in chemistry because they link the mass of a liquid to its volume, facilitating accurate solution preparation.

Grams to Moles Conversion

Converting grams to moles is a fundamental skill in chemistry that involves using the molar mass of a compound. The molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). The formula used for this conversion is:
\[\text{moles} = \frac{\text{mass}}{\text{molar mass}}\]In the example of ethanol:

• The molecular formula is \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) with a molar mass of approximately 46 g/mol.
• By taking the mass of ethanol calculated as necessary for the solution, we divide it by 46 g/mol to find the moles.
• This step is critical in linking amounts in grams to the number of molecules or "beaker" quantities like moles.

This conversion is essential for stoichiometric calculations and making sure chemical reactions and solutions are calculated accurately.

Volume Measurement

Volume measurement is crucial in solution preparation and is often given in milliliters (mL) or liters (L). In preparing solutions, accurate volume measurement ensures that the correct concentrations are achieved.
When you have the mass of a liquid and its density, you can use those measurements to find the volume needed using:
\[\text{volume} = \frac{\text{mass}}{\text{density}}\]For calculating the volume of ethanol needed:

• First, determine how much mass of ethanol you have calculated for solution preparation.
• Use the density of 0.789 g/mL to find the precise volume of ethanol required.
• This approach prevents errors in concentration due to incorrect volume measurements.

Accurate volume measurement ensures that solutions meet desired specifications and are consistent in concentration, which is crucial for reproducible results in laboratory settings.