Problem 120 - General Chemistry

Question

A sample of \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) was heated to \(100^{\circ} \mathrm{C}\), where it lost water and gave another hydrate of copper(II) ion that contained \(29.76 \%\) Cu. An \(85.42-\mathrm{mg}\) sample of this new hydrate gave \(93.33 \mathrm{mg}\) of barium sulfate precipitate when treated with a barium nitrate solution. What is the formula of the new hydrate?

Step-by-Step Solution

Answer

The formula of the new hydrate is CuSO₄·3H₂O.

Step 1: Determine Moles of BaSO₄

First, find the molar mass of barium sulfate (BaSO₄) which is approximately 233.4 g/mol. Using the given mass of BaSO₄ (93.33 mg or 0.09333 g), calculate the moles: \[ \text{moles of BaSO}_4 = \frac{0.09333}{233.4} \approx 0.0004 \text{ moles} \]

Step 2: Relate Moles of BaSO₄ to Moles of CuSO₄

Each mole of copper(II) sulfate (CuSO₄) reacts to form one mole of BaSO₄. Therefore, the moles of CuSO₄ in the hydrate sample is also 0.0004 moles.

Step 3: Calculate Mass of Cu in the New Hydrate

Using the percentage of copper in the new hydrate (29.76%), calculate the mass of copper: \[ \text{mass of Cu} = 0.08542 \times 0.2976 = 0.02542 \text{g or 25.42 mg} \]

Step 4: Find Moles of Cu

Calculate the moles of copper using its atomic mass (63.55 g/mol): \[ \text{moles of Cu} = \frac{0.02542}{63.55} \approx 0.0004 \text{ moles} \]

Step 5: Calculate Moles of CuSO₄ in Hydrate

Since the moles of Cu is approximately equal to moles of CuSO₄ due to sample consistency, the moles of CuSO₄ in the sample is again 0.0004 moles.

Step 6: Determine Mass of CuSO₄ in the Sample

The molar mass of CuSO₄ is approximately 159.6 g/mol. Calculate the mass in 0.0004 moles: \[ \text{mass of CuSO}_4 = 0.0004 \times 159.6 \approx 0.0638 \text{g or 63.8 mg} \]

Step 7: Calculate Total Mass of Water in the New Hydrate

Find the mass of water by subtracting mass of CuSO₄ from total mass of hydrate: \[ \text{mass of water} = 85.42 - 63.8 = 21.62 \text{mg} \]

Step 8: Convert Mass of Water to Moles

The molar mass of water (H₂O) is 18.02 g/mol. Calculate moles of water: \[ \text{moles of water} = \frac{0.02162}{18.02} \approx 0.0012 \text{ moles} \]

Step 9: Determine Coefficient of Water in the Hydrate

From moles calculated: \[ x = \frac{0.0012}{0.0004} = 3 \] Therefore, the new hydrate formula is CuSO₄·3H₂O.

Key Concepts

Copper(II) Sulfate

Copper(II) sulfate, commonly represented as CuSO₄, is a chemical compound with significant importance in various industrial and laboratory applications. It typically exists as an ionic salt, forming complex hydrates, which are compounds with water molecules incorporated into their structure.

• CuSO₄ can appear in different hydrated forms, with the pentahydrate ( ext{CuSO}_{4} imes 5 ext{H}_{2} ext{O}) being the most commonly encountered variant. This is the familiar bright blue solid we see in chemistry labs.
• Upon heating, it can lose all or some of its water content, which results in different hydrate forms.

Understanding the hydration state of copper(II) sulfate is crucial as it affects physical properties like color, solubility, and mass.
When hydrates lose water, the process can be efficiently described by mass changes and percentage composition, forming the basis for calculating new chemical formulas of hydrates.

Mole Calculation

Mole calculations are essential in determining the composition and reactions of chemical substances. The mole is a basic unit in chemistry representing Avogadro's number (6.022 imes 10^{23}) of particles, allowing chemists to quantify substances.

• When calculating moles, the molar mass of a substance, the mass of one mole of the substance, is fundamental. For instance, the molar mass for barium sulfate (BaSO₄) is approximately 233.4 g/mol.
• In our practice problem, an important step is converting the mass of the barium sulfate precipitate to moles, using the formula:

\[ ext{moles of BaSO}_4 = \frac{ ext{mass (g)}}{ ext{molar mass (g/mol)}} \]
This step is crucial because it links the experimental data of mass with theoretical calculations, allowing us to relate it to the moles of other compounds involved, like copper(II) sulfate.

Barium Sulfate Precipitation

Precipitation reactions occur when two solutions react to form an insoluble solid called a precipitate. In this exercise, barium sulfate (BaSO₄) emerges as a precipitate through the interaction of compounds in solution.

• When copper(II) sulfate reacts with a barium compound like barium nitrate, the sulfate ions ( SO_{4}^{2-}) in solution meet the barium ions ( Ba^{2+}), creating insoluble barium sulfate.
• This process is not only essential for forming solid materials but is also a step used to ensure certain ions or elements' presence and quantities can be precisely measured.

By measuring the mass of precipitates formed, one can backtrack to determine the amount of the original reactants involved, like copper(II) sulfate in this case.
This is a common method used in gravimetric analysis, illustrating how chemical reactions can be analyzed quantitatively.

Chemical Formula Determination

Determining the chemical formula of a compound, particularly hydrates, requires a precise understanding of mole relationships and mass data. Hydrates have water molecules incorporated into their crystalline framework, making them unique.

• The first step involves calculating the moles of the involved components, such as BaSO₄ and CuSO₄, from the given mass data.
• Then, with the help of mole ratio calculations, you can discern the number of water molecules in the hydrate, as depicted in this sample: the mass of the copper in the sample correlates with mole calculations to verify the water content.
• Finally, comparing the molar quantities leads to the revelation of a coefficient, which signifies the number of water molecules attached to each mole of CuSO₄, in this case, three in CuSO₄·3H₂O.

By following this systematic process, the hydrate's formula is discerned, providing vital insights into its chemical identity.