Problem 89 - Chemistry The Molecular Science

Question

You wish to make a \(0.200-\mathrm{M}\) solution of \(\mathrm{CuSO}_{4}(\mathrm{aq})\). Calculate the mass of \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) required to make \(0.500 \mathrm{~L}\) of solution.

Step-by-Step Solution

Answer

24.972 grams of \(\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}\) are needed.

Step 1: Understand the Problem

We need to find the mass of copper(II) sulfate pentahydrate, \( \mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O} \), required to prepare a \(0.200-\mathrm{M}\) solution in \(0.500 \mathrm{~L}\) of solution.

Step 2: Calculate Moles of \(\mathrm{CuSO}_4\) Required

The concentration formula is \(M = \frac{\text{moles of solute}}{\text{liters of solution}}\). To find the moles needed: \[\text{Moles of } \mathrm{CuSO}_4 = 0.200 \text{ M} \times 0.500 \text{ L} = 0.100 \text{ moles}\]

Step 3: Find Molar Mass of \(\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}\)

Calculate the molar mass using atomic masses: - \( \mathrm{Cu} = 63.55 \text{ g/mol} \)- \( \mathrm{S} = 32.07 \text{ g/mol} \)- \( \mathrm{O} = 16.00 \text{ g/mol (4 for sulfate, 5 for water)}\)- \( \mathrm{H} = 1.01 \text{ g/mol (5 for water)}\)\(\mathrm{CuSO}_4 = 63.55 + 32.07 + (4 \times 16.00) = 159.62 \text{ g/mol}\)\(5\mathrm{H}_2\mathrm{O} = 5 \times (2\times1.01 + 16.00) = 90.10 \text{ g/mol}\)Total molar mass = 159.62 + 90.10 = 249.72 \text{ g/mol}.

Step 4: Calculate Mass Required

Use the moles calculated in Step 2 and molar mass from Step 3:\[\text{Mass} = 0.100 \text{ moles} \times 249.72 \text{ g/mol} = 24.972 \text{ g}\]Thus, 24.972 grams of \(\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}\) are required.

Key Concepts

Molarity

When we talk about molarity, we are referring to the concentration of a solution. Molarity, often represented by the symbol \( M \), is defined as the number of moles of solute per liter of solution. This concept helps us understand how much solute is present in a given volume of solvent, making it a fundamental measurement in chemistry.
To compute molarity, use the formula:

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

For example, in the original exercise, we wanted a 0.200 M solution with a volume of 0.500 L. To find the amount of solute needed, you rearrange the formula to solve for moles:

• \( \text{moles of } \mathrm{CuSO}_4 = 0.200 \times 0.500 = 0.100 \text{ moles} \)

By understanding molarity, you will have a key tool for preparing solutions with precise concentrations.

Copper(II) Sulfate

Copper(II) sulfate is a versatile chemical compound often used in laboratories. It can be found in its pentahydrate form, \( \mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O} \), which includes water molecules as part of its structure. This compound appears as blue crystals and is soluble in water, making it useful for various chemical reactions and experiments.
Some common applications of copper(II) sulfate include:

• Preparing solutions for electroplating processes
• Acting as a reagent in chemical analysis
• Serving as an electrolyte in copper refining

Understanding its hydrated form is crucial, as the presence of water affects its weight and how it will be measured for experiments.

Molar Mass Calculation

The molar mass of a compound is the total mass of one mole of its molecules. To calculate it, you add up the atomic masses of all atoms in the formula. In the original exercise, we determined the molar mass of \( \mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O} \), a compound that includes copper, sulfur, oxygen, and water molecules.
Here's a breakdown of the steps for calculating molar mass:

• Copper (Cu): 63.55 g/mol
• Sulfur (S): 32.07 g/mol
• Oxygen (O): 16.00 g/mol
• Water (\( 5 \mathrm{H}_2 \mathrm{O} \)): 90.10 g/mol (calculated as \( 5 \times (2 \times 1.01 + 16.00) \))

After calculating these, sum them all:

• \( \mathrm{CuSO}_4 = 63.55 + 32.07 + (4 \times 16.00) = 159.62 \text{ g/mol} \)
• Combine with water: \( 159.62 + 90.10 = 249.72 \text{ g/mol} \)

Using the total molar mass, you can then determine the mass required for a specific number of moles, as shown in the exercise.