Problem 116 - General Chemistry: Principles and Modern Applications

Question

Under appropriate conditions, copper sulfate, potassium chromate, and water react to form a product containing \(\mathrm{Cu}^{2+},\) \(\mathrm{CrO}_{4}{^2}{^-},\) and \(\mathrm{OH}^{-}\) ions. Analysis of the compound yields \(48.7 \% \mathrm{Cu}^{2+}, 35.6 \% \mathrm{CrO}_{4}{^2}{-},\) and \(15.7 \% \mathrm{OH}^{-}\). (a) Determine the empirical formula of the compound. (b) Write a plausible equation for the reaction.

Step-by-Step Solution

Answer

The empirical formula of the compound in the given conditions is \(\mathrm{Cu}_{4}( \mathrm{CrO}_{4}) \mathrm{OH}_{5}\). The plausible reaction equation is \(\mathrm{CuSO}_{4} + 2 \mathrm{K}_{2} \mathrm{CrO}_{4} + 5 \mathrm{H}_{2} \mathrm{O} \rightarrow 4\mathrm{Cu}_{4}( \mathrm{CrO}_{4} )\mathrm{OH}_{5} + \mathrm{K}_{2} \mathrm{SO}_{4}\).

Step 1: Determine molar ratios

To determine the empirical formula of the compound, assume a 100 g sample of the compound. This allows us to treat the percentage composition as grams. This means that we have 48.7 g of \(\mathrm{Cu}^{2+}\), 35.6 g of \(\mathrm{CrO}_{4}{^2}{^-}\), and 15.7 g of \(\mathrm{OH}^{-}\). Then, convert this mass into moles by dividing by the molar mass. For \(\mathrm{Cu}^{2+}\), this gives a value of \(48.7 / 63.55 \approx 0.766\) mol. For \(\mathrm{CrO}_{4}{^2}{^-}\), this gives a value of \(35.6 / 194.19 \approx 0.183\) mol. For \(\mathrm{OH}^{-}\), this gives a value of \(15.7 / 17.01 \approx 0.923\) mol.

Step 2: Calculate the empirical formula

The empirical formula is given by the ratios of these moles. To get the smallest whole numbers, divide all the mole values by the smallest of the three. In this case, the smallest value is that of \(\mathrm{CrO}_{4}{^2}{^-}\), which gives ratios of 4 for \(\mathrm{Cu}^{2+}\), 1 for \(\mathrm{CrO}_{4}{^2}{^-}\), and 5 for \(\mathrm{OH}^{-}\). Thus, the empirical formula of the compound is \(\mathrm{Cu}_{4}( \mathrm{CrO}_{4}) \mathrm{OH}_{5}\).

Step 3: Write the reaction equation

Now, using the reactants copper sulfate, potassium chromate, and water, and under appropriate conditions, we may suggest a plausible equation for the reaction. With the empirical formula, \(\mathrm{CuSO}_{4} + 2 \mathrm{K}_{2} \mathrm{CrO}_{4} + 5 \mathrm{H}_{2} \mathrm{O} \rightarrow 4\mathrm{Cu}_{4}( \mathrm{CrO}_{4} )\mathrm{OH}_{5} + \mathrm{K}_{2} \mathrm{SO}_{4}\) is plausible, having in mind that ions in water solution are present.

Key Concepts

Molar Ratios

When trying to find the empirical formula of a compound, molar ratios are essential. They help us understand the simplest form of a chemical formula by showcasing the smallest whole-number relationships between the involved elements. To determine the molar ratio, you'll first need to convert the percentage composition into a weight-based approach. Imagine you have 100 grams of the compound. This lets you directly translate the percentages into grams.

For instance:

• Copper: 48.7 grams.
• Chromate: 35.6 grams.
• Hydroxide: 15.7 grams.

Next, you convert the mass of each component into moles by dividing it by its molar mass. The molar mass is the weight of one mole of a substance. For example, the molar mass of copper (\{\mathrm{Cu^{2+}}\}) is \(63.55 \text{ g/mol}\), and you do similarly for chromate and hydroxide ions. Once you have the molar amounts:

• 0.766 mol of copper.
• 0.183 mol of chromate.
• 0.923 mol of hydroxide.

You find the smallest value and divide all the molar amounts by it. For these examples, dividing each by \(0.183\) simplifies to ratios which give you the empirical formula of \(\mathrm{Cu}_{4}( \mathrm{CrO}_{4}) \mathrm{OH}_{5}\). This step ensures the compound's representation in its simplest ratio.

Chemical Reaction Equation

A chemical reaction equation is a symbolic representation of a chemical reaction. It shows the starting materials, or reactants, and the results, known as products. These help understand how elements interact and transform into different substances. For your reaction of copper sulfate with potassium chromate and water to form the compound \(\mathrm{Cu}_{4}( \mathrm{CrO}_{4}) \mathrm{OH}_{5}\), the reaction equation has been calculated as:

• Reactants: \(\mathrm{CuSO}_{4}\), \(2 \mathrm{K}_{2} \mathrm{CrO}_{4}\), \(5 \mathrm{H}_{2} \mathrm{O}\).
• Products: \(4\mathrm{Cu}_{4}( \mathrm{CrO}_{4} )\mathrm{OH}_{5}\), \(\mathrm{K}_{2} \mathrm{SO}_{4}\).

Each part of the chemical equation is balanced to make sure that the number of each type of atom is the same on both sides of the equation. This respect for the Law of Conservation of Mass underscores the idea that matter is neither created nor destroyed in chemical reactions, merely transformed. Understanding these transformations aids in predicting the outcome of chemical interactions, vital for comprehending more complex chemistry topics.

Percentage Composition

Percentage composition tells us how much of each element is present in a compound, often given in percentage terms. It's a crucial starting point in determining empirical formulas because it allows the conversion of elemental percentages into mass, and then into moles for further analysis.

To find the percentage composition, the mass of each component is divided by the total mass of the compound, then multiplied by 100 to convert into a percentage. For instance, in the compound analyzed:

• Copper makes up 48.7%.
• Chromate contributes 35.6%.
• Hydroxide comprises 15.7%.

These percentages reflect the different amounts of each ion in 100 g of the compound. When attempting to find the empirical formula, you use this data to calculate the number of moles of each constituent, guiding the determination of the simplest atomic ratio between them. This is why percentage composition is instrumental in the exploration of a compound's inherent structure.