Problem 11 - Chemistry for Allied Health

Question

Calculate how many grams of each solute would be required in order to make the given solution. a. \(3.40 \mathrm{~L}\) of a \(0.780 \mathrm{M}\) solution of iron(III) chloride, \(\mathrm{FeCl}_{3}\) b. \(60.0 \mathrm{~mL}\) of a \(4.10 \mathrm{M}\) solution of calcium acetate, \(\mathrm{Ca}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}\)

Step-by-Step Solution

Answer

For FeCl3: 429.79 grams; For Ca(CH3COO)2: 38.91 grams.

Step 1: Understand Molarity Concept

Molarity ( ext{M}) is defined as the number of moles of solute per liter of solution. The formula for molarity is given by \( ext{M} = \frac{\text{moles of solute}}{\text{liters of solution}} \).

Step 2: Calculate Moles of Solute for FeCl3

For the solution of iron(III) chloride, given volume (\(3.40 \text{ L}\)) and molarity (\(0.780 \text{ M}\)), calculate moles: \( ext{moles of } \text{FeCl}_3 = 0.780 \text{ M} \times 3.40 \text{ L} = 2.652 \text{ moles} \).

Step 3: Convert Moles to Grams for FeCl3

Find the molar mass of \(\text{FeCl}_3\) (iron(III) chloride). Atomic masses: \(\text{Fe} = 55.85\), \(\text{Cl} = 35.45\), so \( \text{molar mass of } \text{FeCl}_3 = 55.85 + 3(35.45) = 162.2 \text{ g/mol} \). Then calculate grams: \( \text{grams of } \text{FeCl}_3 = 2.652 \text{ moles} \times 162.2 \text{ g/mol} = 429.7944 \text{ grams} \).

Step 4: Calculate Moles of Solute for Ca(CH3COO)2

For the solution of calcium acetate, convert 60.0 mL to liters: \(60.0 \text{ mL} = 0.060 \text{ L}\). Given molarity (\(4.10 \text{ M}\)), calculate moles: \( ext{moles of } \text{Ca(CH}_3\text{COO)}_2 = 4.10 \text{ M} \times 0.060 \text{ L} = 0.246 \text{ moles} \).

Step 5: Convert Moles to Grams for Ca(CH3COO)2

Find the molar mass of \(\text{Ca(CH}_3\text{COO)}_2\) (calcium acetate). Atomic masses: \(\text{Ca} = 40.08\), \(\text{C} = 12.01\), \(\text{H} = 1.01\), \(\text{O} = 16\). Molar mass calculation: \(40.08 + 4(12.01) + 6(1.01) + 4(16) = 158.17 \text{ g/mol}\). Then, calculate grams: \( \text{grams of } \text{Ca(CH}_3\text{COO)}_2 = 0.246 \text{ moles} \times 158.17 \text{ g/mol} = 38.90782 \text{ grams} \).

Key Concepts

Solute Mass Calculation

To determine the mass of a solute needed for a specific chemical solution, it's essential first to compute the number of moles of solute that will be present. The general procedure involves two key steps.

• **Calculate the number of moles:** Use the equation for molarity: \[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}} \]This allows you to find out how many moles of the solute are needed based on the solution's volume and concentration.
• **Convert moles to mass:** Once you have the number of moles, convert it to grams by multiplying by the solute's molar mass (formula weight).

Applying these steps ensures that you know precisely how much solute to weigh out for preparing your solution.

Molar Mass

Molar mass is a crucial concept as it bridges the gap between the quantity of a substance in moles and its mass in grams. Each element has a specific atomic mass, typically found on the periodic table, and the molar mass of a compound is the sum of the atomic masses of all atoms in its formula.
For example, to calculate the molar mass of iron(III) chloride (\text{FeCl}_3):

• Iron (Fe): 1 atom \(\times 55.85\text{ g/mol}\)
• Chlorine (Cl): 3 atoms \(\times 35.45\text{ g/mol}\)

Add them up:\[ 55.85 + 3(35.45) = 162.2 \text{ g/mol} \] The same principle applies to any compound. Knowing how to accurately calculate molar mass is essential in converting moles to grams, which is imperative for solution preparation.

Chemical Solution Preparation

Once the solute mass is calculated, the next step is to prepare the chemical solution accurately. Chemical solution preparation involves several key actions that ensure safety and precision:

• **Weighing the Solute:** Use an analytical balance for precise measurement. Ensure that all equipment is clean and dry.
• **Dissolving the Solute:** Add the weighed solute to a volumetric flask or beaker. Generally, part of the solvent (often water) is added initially to help dissolve the solute.
• **Diluting to the Desired Volume:** Once the solute is completely dissolved, add more solvent to reach the total desired volume, as specified in your experiment.

Finally, mix thoroughly to ensure uniform distribution throughout the solution. This preparation process is crucial for maintaining the accuracy and effectiveness of the intended chemical reaction or analysis.