Problem 83 - General Chemistry

Question

A chemist added an excess of sodium sulfate to a solution of a soluble barium compound to precipitate all of the barium ion as barium sulfate, \(\mathrm{BaSO}_{4}\). How many grams of barium ion are in a \(458-\mathrm{mg}\) sample of the barium compound if a solution of the sample gave \(513 \mathrm{mg} \mathrm{BaSO}_{4}\) precipitate? What is the mass percentage of barium in the compound?

Step-by-Step Solution

Answer

302.1 mg of barium ions; mass percentage of barium is 65.95%.

Step 1: Understand the reaction

The barium ions (\(\text{Ba}^{2+}\)) react with sulfate ions (\(\text{SO}_4^{2-}\)) to form barium sulfate (\(\text{BaSO}_4\)). This can be expressed by the balanced chemical equation: \( \text{Ba}^{2+} + \text{SO}_4^{2-} \rightarrow \text{BaSO}_4 \).

Step 2: Calculate moles of BaSO4 precipitate

Convert the mass of BaSO\(_4\) from mg to grams: \(513\, \text{mg} = 0.513\, \text{g}\). The molar mass of \(\text{BaSO}_4\) is approximately \(233.39\, \text{g/mol}\). Calculate the moles of \(\text{BaSO}_4\) using the formula: \[ \text{moles of BaSO}_4 = \frac{0.513\, \text{g}}{233.39\, \text{g/mol}} \approx 0.0022\, \text{mol} \].

Step 3: Determine moles of Ba ions

Since \(\text{BaSO}_4\) forms from \(\text{Ba}^{2+}\) in a 1:1 molar ratio, the moles of \(\text{Ba}^{2+}\) are the same as that of \(\text{BaSO}_4\). Thus, there are \(0.0022\, \text{mol}\) of \(\text{Ba}^{2+}\).

Step 4: Calculate mass of barium ions

The molar mass of barium (Ba) is approximately \(137.33\, \text{g/mol}\). Calculate the mass of barium ions using: \[ \text{mass of Ba}^{2+} = 0.0022\, \text{mol} \times 137.33\, \text{g/mol} \approx 0.3021\, \text{g} = 302.1\, \text{mg} \].

Step 5: Find mass percentage of barium

The mass percentage of barium is calculated using: \[ \text{Mass \, percentage of Ba} = \left(\frac{302.1\, \text{mg}}{458\, \text{mg}}\right) \times 100 \approx 65.95\% \].

Key Concepts

Precipitation Reaction

A precipitation reaction occurs when two soluble substances in aqueous solution react to form an insoluble solid, called a precipitate. In the case of barium sulfate (\(\text{BaSO}_4\)), the barium ions (\(\text{Ba}^{2+}\)) combine with sulfate ions (\(\text{SO}_4^{2-}\)) in solution to form insoluble barium sulfate. The reaction can be described by the balanced equation: \[\text{Ba}^{2+} + \text{SO}_4^{2-} \rightarrow \text{BaSO}_4\]
This type of reaction is important in separating different components of a solution since the precipitate can be filtered out. This property is used analytically to determine the concentration of ions in a solution by measuring the amount of precipitate formed.

• Start with a soluble barium compound and sodium sulfate.
• The sulfates react with barium ions to form a solid precipitate.
• This solid can be separated from the solution and measured.

Understanding this concept helps in obtaining quantitative information about ion concentrations through precipitation.

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions. It is crucial in determining how much of a substance is needed or formed using a balanced chemical equation. In our example with barium sulfate, stoichiometry allows us to predict mass and moles of products formed or consumed.

Using the equation \(\text{Ba}^{2+} + \text{SO}_4^{2-} \rightarrow \text{BaSO}_4\), we can say:

• One mole of barium ions reacts with one mole of sulfate ions to form one mole of barium sulfate.
• This 1:1 ratio is essential for calculating the required amounts of each reactant.
• The balanced equation ensures that matter is conserved during the reaction.

A systematic approach to stoichiometry will help in solving problems that involve converting masses of substances to moles and vice versa, ensuring all measurements relate precisely to the reactants and products involved.

Mole Concept

The mole concept provides a bridge between the mass of a substance and the number of particles it contains. It is a fundamental concept that allows chemists to work with the macro scale quantities they can measure and the micro scale quantities they can't. In any chemical reaction, the mole concept allows us to translate between:

• Mass of a compound and its number of moles using its molar mass.
• Moles of a reactant to moles of a product using a balanced chemical equation.

In our precipitation example, determining the number of moles of barium sulfate from its mass (\(0.513 \, \text{g}\)) required the use of its molar mass (\(233.39 \, \text{g/mol}\)). This conversion revealed the moles of barium ions derived from the sample, which is a vital step in tracking the progress of the reaction and calculating quantities consumed or produced.

Chemical Equations

Chemical equations are symbolic representations of chemical reactions. They describe the transformation of reactants into products, using symbols and formula units to depict the amounts involved. A well-balanced chemical equation reflects the conservation of mass and charge, ensuring that both sides of the equation are equal in the total number of each type of atom.
For the precipitation of \(\text{BaSO}_4\), the simplified equation is:\[\text{Ba}^{2+} + \text{SO}_4^{2-} \rightarrow \text{BaSO}_4\]This equation highlights the 1:1 relationship between barium ions and sulfate ions. Understanding chemical equations involves:

• Interpreting the symbols and determining the reactants and products.
• Balancing the equation to ensure the mass and charge are conserved.
• Using the balanced equation to dissect stoichiometric relationships, aiding in quantitative analyses of chemical quantities.

By mastering these equations, one can predict the outcomes of chemical reactions and calculate the necessary or resulting amounts of substances involved.