Question

You are given a solid that is a mixture of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) and \(\mathrm{K}_{2} \mathrm{SO}_{4}\) A \(0.205-\mathrm{g}\) sample of the mixture is dissolved in water. An excess of an aqueous solution of \(\mathrm{BaCl}_{2}\) is added. The \(\mathrm{BaSO}_{4}\) that is formed is filtered, dried, and weighed. Its mass is 0.298 g. What mass of \(\mathrm{SO}_{4}^{2-}\) ion is in the sample? What is the mass percent of \(\mathrm{SO}_{4}^{2-}\) ion in the sample? What are the percent compositions by mass of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) and \(\mathrm{K}_{2} \mathrm{SO}_{4}\) in the sample?

Short answer

The mass of SO₄²⁻ ion in the sample is 0.123g, the mass percent of SO₄²⁻ ion in the sample is 60.0%, and the percent composition by mass of the mixture of Na₂SO₄ and K₂SO₄ in the sample is 40.0%.

Step-by-step solution

Use stoichiometry to find the mass of SO4^2- ion in the sample

First, we have to find the molar mass of BaSO4. Using the periodic table, we find that the molar mass of BaSO4 is \(137.33 + (32.07 + 4(16.00)) = 233.43\) g/mol. Next, we have to find the number of moles of BaSO4 formed, which is given by \(\frac{\text{mass of BaSO4}}{\text{molar mass of BaSO4}}\). So, we have: Number of moles of BaSO4 = \(\frac{0.298 \,g}{233.43 \,g/mol} = 1.28 \times 10^{-3}\) mol. Since the stoichiometric ratio of BaSO4 to SO4^2- is 1:1, this means that the same number of moles of SO4^2- are present in the sample. Now, we find the mass of SO4^2- by using the molar mass of SO4^2-, which is given by \(32.07 + 4(16.00) = 96.07\) g/mol: Mass of SO4^2- = (number of moles of SO4^2-) × (molar mass of SO4^2-) = \(1.28 \times 10^{-3} mol \times 96.07 g/mol = 0.123\) g.

Calculate the mass percent of SO4^2- ion in the sample

To find the mass percent of SO4^2- ion in the sample, we can use the formula: Mass percent of SO4^2- = \(\frac{\text{mass of SO4}^{2-}}{\text{total mass of sample}} \times 100\%\). Mass percent of SO4^2- = \(\frac{0.123\,g}{0.205\,g} \times 100\% = 60.0\%\)

Determine the percent compositions by mass of Na2SO4 and K2SO4 in the sample

We'll start by determining the mass of the Na2SO4 and K2SO4 part in the sample. Since the sample is comprised of only Na2SO4 and K2SO4, the mass of these parts is the difference between the total mass of the sample and the mass of SO4^2- ion: Mass of Na2SO4+K2SO4 = Total mass - Mass of SO4^2- = \(0.205\,g - 0.123\,g = 0.082\,g\) Now, we know that the mass percent of SO4^2- ion in the sample is 60.0%. Therefore, the mass percent of Na2SO4+K2SO4 is the remaining percentage: Mass percent of Na2SO4+K2SO4 = \(100\% - 60.0\% = 40.0\%\) Finally, we don't have enough information to further break down the percentage composition for Na2SO4 or K2SO4 individually. So, we can only report that the percent composition by mass of the mixture of Na2SO4 and K2SO4 is 40.0%. In summary: - The mass of SO4^2- ion in the sample is 0.123g - The mass percent of SO4^2- ion in the sample is 60.0% - The percent composition by mass of the mixture of Na2SO4 and K2SO4 in the sample is 40.0%

Key concepts

Mass Percent Calculation

The mass percent calculation is an essential part of understanding the composition of chemical mixtures. It provides a way to express how much of a specific component is present in a given mixture.
In the context of stoichiometry problems, such as this example with a mixture containing \( \mathrm{Na}_2 \mathrm{SO}_4 \) and \( \mathrm{K}_2 \mathrm{SO}_4 \), knowing the mass percent helps us understand the relative amount of sulfate ions present. Let's look at how this was calculated.

To find the mass percent of sulfate ions \((\mathrm{SO}_4^{2-})\) in the sample:

• First, calculate the mass of sulfate ions in the sample using the stoichiometry data provided. This involves multiplying the number of moles of sulfate by its molar mass.
• Next, use the formula:
  \[ \text{Mass percent of SO}_4^{2-} = \left( \frac{\text{mass of SO}_4^{2-}}{\text{total mass of sample}} \right) \times 100\% \]
• The result shows that sulfate ions make up 60.0% of the total mass of the sample, indicating that these ions are a significant component of the mixture.

Understanding mass percent gives insight into the concentration and allows comparisons between different substances in a mixture.

Molar Mass

Molar mass serves as a bridge between the microscopic scale of molecules and ions and the macroscopic scale of grams and moles. It is a fundamental tool in stoichiometry.
When solving stoichiometry problems, molar mass allows us to convert between mass and number of moles, essential for finding relationships in a chemical reaction.

In our example, molar mass is used in several steps:

• The molar mass of \( \mathrm{BaSO}_4 \) is calculated by summing the atomic masses of barium, sulfur, and oxygen: \( 137.33 + (32.07 + 4 \times 16.00) = 233.43 \ \text{g/mol} \).
• This value is necessary for determining the number of moles of \( \mathrm{BaSO}_4 \) produced, as we divide the mass of \( \mathrm{BaSO}_4 \) by its molar mass.
• The stoichiometry of the reaction indicates an equal number of \( \mathrm{SO}_4^{2-} \) moles, so converting back with the molar mass of \( \mathrm{SO}_4^{2-} \) gives us its mass in the sample.

Molar mass is paramount in stoichiometry to precisely measure and predict quantities in chemical reactions.

Precipitation Reaction

Precipitation reactions are a type of chemical reaction where two soluble salts in solution react to form an insoluble salt, known as the precipitate.
Such reactions are governed by solubility rules and are often used for purification or separation processes in chemistry.

In the given problem, the precipitation reaction occurs between the dissolved sulfate ions \((\mathrm{SO}_4^{2-})\) and \( \mathrm{BaCl}_2 \), resulting in the formation of \( \mathrm{BaSO}_4 \):
\[ \mathrm{Ba}^{2+} + \mathrm{SO}_4^{2-} \rightarrow \mathrm{BaSO}_4 \ (\text{solid}) \]
Here are some key points about this reaction:

• In this reaction, \( \mathrm{BaSO}_4 \) precipitates out as a solid because it is not soluble in water. This property allows it to be filtered, dried, and weighed, providing a practical method to measure the amount of sulfate ions present.
• The reaction follows a 1:1 molar ratio, meaning each mole of sulfate reacts with one mole of barium to produce one mole of barium sulfate.
• Understanding the nature of the reaction aids in the proper calculation of the mass of \( \mathrm{SO}_4^{2-} \) ions and how much is being separated during the process.

Precipitation reactions are crucial in analytical chemistry for isolating and quantifying specific ions in a mixture.