Problem 88 - Chemistry The Molecular Science

Question

You need to make a \(0.300-\mathrm{M}\) solution of \(\mathrm{NiSO}_{4}(\mathrm{aq}) . \mathrm{Cal}-\) culate the mass of \(\mathrm{NiSO}_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) you should put into a 0.500-L volumetric flask.

Step-by-Step Solution

Answer

39.43 g of \( \text{NiSO}_4 \cdot 6 \text{H}_2\text{O} \) is needed.

Step 1: Understand the Goal

We are tasked with making a 0.300 M solution of \( \text{NiSO}_4 \) using \( \text{NiSO}_4 \cdot 6 \text{H}_2\text{O} \) in a 0.500 L flask. We need to find out how much of the hydrated salt is required.

Step 2: Calculate Moles Required

Calculate the number of moles of \( \text{NiSO}_4 \) required to make the solution. Using the formula for molarity \( M = \frac{\text{moles of solute}}{\text{liters of solution}} \), we have:\[ \text{Moles of } \text{NiSO}_4 = 0.300 \text{ M} \times 0.500 \text{ L} = 0.150 \text{ moles} \]

Step 3: Calculate Molar Mass of Hydrated Compound

The molar mass of \( \text{NiSO}_4 \cdot 6 \text{H}_2\text{O} \) is calculated by adding the molar masses of its components:- \( \text{NiSO}_4 \): \( 58.69 + 32.07 + (4 \times 16.00) = 154.76 \text{ g/mol} \)- \( 6 \text{H}_2\text{O} \): \( 6 \times (2 \times 1.01 + 16.00) = 108.12 \text{ g/mol} \)- Total molar mass: \( 154.76 + 108.12 = 262.88 \text{ g/mol} \)

Step 4: Calculate Mass of Hydrated NiSO4 Required

Use the molar mass to find the mass of \( \text{NiSO}_4 \cdot 6 \text{H}_2\text{O} \) needed. Multiply the moles calculated by the molar mass:\[ \text{Mass} = 0.150 \text{ moles} \times 262.88 \text{ g/mol} = 39.43 \text{ g} \]

Key Concepts

Molarity Calculation

Molarity is an important concept in chemistry, especially when preparing solutions. It allows us to know the concentration of a substance in a given volume of liquid. Molarity (\( M \)) is defined as the number of moles of solute per liter of solution. This can be written as \( M = \frac{\text{moles of solute}}{\text{liters of solution}} \).

For example, when tasked with preparing a 0.300 M solution of \( \mathrm{NiSO}_{4} \), we need to dissolve this solute in a specific volume. In this case, it's 0.500 liters. By using the formula, we can calculate the number of moles required: \( 0.300 \times 0.500 = 0.150 \text{ moles} \).

• Ensure you understand the volume your solution will have because it's directly related to how many moles you need.
• Accurate calculations are key, as any variation might alter the solution's properties.

Hydrated Compounds

Hydrated compounds, like \( \mathrm{NiSO}_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O} \), are prevalent in chemical formulas where water molecules are associated with the compound. These waters of hydration are part of the crystal structure. Understanding how to calculate the molar mass of such compounds is crucial for solution preparation.

For example, to calculate the molar mass of \( \mathrm{NiSO}_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O} \), you need to sum the molar masses of all atoms present in both the anhydrous salt and the water molecules:

• \( \mathrm{NiSO}_{4} \): Adding up masses of Nickel (\( 58.69 \text{ g/mol}\)), Sulfur (\( 32.07 \text{ g/mol}\)), and Oxygen (\( 4 \times 16.00 \text{ g/mol}\)).
• Water component: \( 6 \times (2 \times 1.01 + 16.00) = 108.12 \text{ g/mol} \)
• Total for hydrated salt: \( 154.76 + 108.12 = 262.88 \text{ g/mol} \)

Incorporate the hydrated water in all calculations to avoid underestimating the mass needed for solutions.

Stoichiometry

Stoichiometry involves calculating the quantities of reactants and products in chemical reactions. It's also applied when preparing solutions, allowing the precise conversion between moles, masses, and solution concentrations.

When determining how much \( \mathrm{NiSO}_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O} \) is needed, we use the moles of \( \mathrm{NiSO}_{4} \) (0.150 moles from previous calculations) and multiply it by the molar mass of the hydrated compound (\( 262.88 \text{ g/mol} \)). This gives us the exact mass of hydrated compound to use: \( 0.150 \times 262.88 = 39.43 \text{ g} \).

• Understanding stoichiometry ensures that correct masses are used, preventing errors.
• Always ensure that your measurements reflect real experimental conditions, including hydrated states.
• Faulty stoichiometry could lead to miscalculations, affecting experimental outcomes.

Applying stoichiometric principles is fundamental in chemistry to ensure precision and accuracy in preparing solutions.