Problem 23 - Chemistry and Chemical Reactivity

Question

Analysis of Mixtures A mixture of \(\mathrm{CuSO}_{4}\) and \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) has a mass of 1.245 g. After heating to drive off all the water, the mass is only \(0.832 \mathrm{g} .\) What is the mass percent of \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) in the mixture? (See page \(95 .\) )

Step-by-Step Solution

Answer

The mass percent of \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) in the mixture is 91.8\%.

Step 1: Understand the Mixture Components

We have two components in the original mixture: \(\mathrm{CuSO}_{4}\) and \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\). The total mass of the mixture is given as 1.245 g. Heating the hydrated compound removes all water, resulting in a mass of 0.832 g.

Step 2: Write the Mass Equation

The mass of the anhydrous form \(\mathrm{CuSO}_{4}\) after heating is 0.832 g. Let the initial masses of \(\mathrm{CuSO}_{4}\) and \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) be \(m\) and \(n\) respectively. We know that: \[m + n = 1.245 \, \mathrm{g}\]

Step 3: Mass of Water Lost

The mass of water lost by heating is the difference between the original mass and the anhydrous mass: \[1.245 \, \mathrm{g} - 0.832 \, \mathrm{g} = 0.413 \, \mathrm{g}\]

Step 4: Calculate Moles of Water Lost

Since \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) loses 5 moles of water per mole, calculate the moles of water lost:\[\text{Molar mass of } \mathrm{H}_{2}\mathrm{O} = 18.015 \, \text{g/mol}\]\[\text{Moles of } \mathrm{H}_{2}\mathrm{O} = \frac{0.413}{18.015} \approx 0.0229 \, \text{mol}\]

Step 5: Determine Moles of \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\)

Since 5 moles of water come from 1 mole of \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\), the moles of \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) is:\[\frac{0.0229}{5} \approx 0.00458 \, \text{mol}\]

Step 6: Calculate Mass of \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\)

Calculate the mass for the moles of \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\):\[\text{Molar mass of } \mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O} = 249.685 \, \text{g/mol}\]\[n = 0.00458 \times 249.685 \approx 1.143 \, \mathrm{g}\]

Step 7: Mass Percent Calculation

Calculate the mass percent of \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) in the mixture:\[\text{Mass percent} = \left( \frac{1.143}{1.245} \times 100 \right) \approx 91.8\%\]

Key Concepts

Mixture Analysis

When analyzing mixtures, it is essential to identify the components and their respective quantities. In this case, the mixture contains \(\mathrm{CuSO}_{4}\) and its hydrated form, \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\). By understanding the combined mass of these substances before and after specific changes, like heating, we perform mixture analysis to determine proportions.

To do this effectively in the given problem, we first note that the total mass of the mixture before heating is 1.245 g and that heating leaves us with only 0.832 g. This initial step is crucial, as it allows for understanding the mass lost and enables further calculations.

Knowing the individual components in a mixture analysis can lead to identifying their specific contribution to the mixture's overall properties, such as the mass percentage of each component.

Hydrated Compounds

Hydrated compounds have water molecules weakly bonded within their crystalline structure. \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) is an example of a hydrated compound, as it contains five molecules of water per molecule of copper sulfate. This association affects the compound's properties, such as mass and color.

To convert a hydrated compound into its anhydrous form, heating can remove these water molecules. The resulting mass change helps identify how much of the compound was originally hydrated.

Hydrated compounds are essential in various applications, including chemical reactions, determining compound purity, and understanding moisture content in materials. Thus, appreciating how hydration affects a compound's mass is critical in solving related problems.

Molar Mass

The molar mass of a compound is the mass of one mole of its molecules, usually expressed in grams per mole (g/mol). For instance, the molar mass of \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) is calculated by summing the molar masses of each atom in its formula. Copper sulfate without water, \(\mathrm{CuSO}_{4}\), has a different molar mass.

Understanding molar mass is vital in chemistry because it enables the conversion of mass to moles, a fundamental step for stoichiometric calculations. This conversion is crucial in determining how much of a substance is present and in calculations involving reactions or changes in state, such as heating.

To find the mass of a specific number of moles, multiply the number of moles by the molar mass. This step is showcased in the problem when calculating the mass of \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) based on the determined moles.

Mass Loss through Heating

Mass loss through heating refers to the reduction in a substance's mass after being exposed to heat. This process often occurs when water or other volatile substances within a compound are driven off, as seen with \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\).

By heating, the water associated with the hydrated compound is released, resulting in the reduced mass measured. The difference in mass before and after heating highlights the quantity of water lost, allowing us to deduce the portion of the compound that was initially hydrated.

Understanding mass loss through heating is useful in evaluating the degree of hydration of a compound and can be applied in determining the purity of substances. The mass difference serves as a direct indicator of the volatile components that were present and their contribution to the initial mass.