Question

A sample of \(0.6760 \mathrm{~g}\) of an unknown compound containing barium ions \(\left(\mathrm{Ba}^{2+}\right)\) is dissolved in water and treated with an excess of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\). If the mass of the \(\mathrm{BaSO}_{4}\) precipitate formed is \(0.4105 \mathrm{~g}\), what is the percent by mass of Ba in the original unknown compound?

Short answer

The percent by mass of Ba is approximately 35.73%.

Step-by-step solution

Determine moles of BaSO4

First, calculate the moles of \( \mathrm{BaSO}_{4} \) that formed. The molar mass of \( \mathrm{BaSO}_{4} \) is the sum of barium (137.33 g/mol), sulfur (32.07 g/mol), and four oxygens (4 * 16.00 g/mol), totaling to 233.39 g/mol. Use the formula: \( moles = \frac{\text{mass}}{\text{molar mass}} \). Thus, \( moles\ of\ \mathrm{BaSO}_{4} = \frac{0.4105\ g}{233.39\ g/mol} \approx 0.001759\ mol \).

Relate moles of Ba to moles of BaSO4

In \( \mathrm{BaSO}_{4} \), one mole of \( \mathrm{BaSO}_{4} \) contains one mole of \( \mathrm{Ba}^{2+} \) ions. Since we have \( 0.001759 \) moles of \( \mathrm{BaSO}_{4} \), we also have \( 0.001759 \) moles of \( \mathrm{Ba}^{2+} \).

Calculate mass of Ba in compound

The molar mass of barium is 137.33 g/mol. Calculate the mass of barium using the formula: \( \text{mass} = \text{moles} \times \text{molar mass} \). Thus, \( \text{mass of Ba} = 0.001759\ mol \times 137.33\ g/mol \approx 0.2415\ g \).

Calculate percent by mass of Ba

The percent by mass of barium in the compound is given by: \( \text{percent by mass} = \left( \frac{\text{mass of Ba}}{\text{mass of compound}} \right) \times 100 \). Hence, \( \text{percent by mass of Ba} = \left( \frac{0.2415\ g}{0.6760\ g} \right) \times 100 \approx 35.73\% \).

Key concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that involves relationships between reactants and products in a chemical reaction. It helps us understand how much of one substance is needed to react with another and allows the calculation of quantities in reactions. In this exercise, stoichiometry is used to relate moles of barium sulfate, \(\mathrm{BaSO}_{4}\), to moles of barium ions, \(\mathrm{Ba}^{2+}\), in the original compound.

Here, we calculate moles using the relation:

• Moles = \( \frac{\text{mass}}{\text{molar mass}} \)

Once the moles of \(\mathrm{BaSO}_{4}\) are known, stoichiometry allows us to deduce the moles of barium ions. This direct relationship is based on the formula \(\mathrm{BaSO}_{4}\), where one mole of barium sulfate contains one mole of barium ions. Thus, stoichiometry provides a straightforward way to measure and convert substances involved in chemical reactions.

Precipitation Reaction

A precipitation reaction occurs when two soluble substances in a solution react to form one or more insoluble products, called precipitates. In this problem, the precipitation reaction involves combining an unknown solution containing \(\mathrm{Ba}^{2+}\) ions with a solution of sodium sulfate, \(\mathrm{Na}_{2}\mathrm{SO}_{4}\).

During the reaction, insoluble barium sulfate, \(\mathrm{BaSO}_{4}\), forms as a precipitate:

• \(\mathrm{Ba}^{2+} (aq) + \mathrm{SO}_{4}^{2-} (aq) \rightarrow \mathrm{BaSO}_{4} (s)\)

The reaction demonstrates double displacement, where anions and cations exchange partners, and is characterized by the formation of a solid from aqueous reactants. Precipitation reactions are used analytically to detect the presence of specific ions in a solution.

Molar Mass Calculation

Molar mass is a measure of the mass of one mole of a given substance, typically expressed in grams per mole (g/mol). Calculating molar mass allows chemists to relate masses to moles, a critical step in stoichiometry calculations. In this exercise, the molar mass of barium sulfate, \(\mathrm{BaSO}_{4}\), is required.

Breaking down the molar mass calculation:

• Barium (\(\mathrm{Ba}\)): 137.33 g/mol
• Sulfur (\(\mathrm{S}\)): 32.07 g/mol
• Oxygen (\(\mathrm{O}\)): 4 * 16.00 g/mol = 64.00 g/mol

The sum of these values gives the molar mass of \(\mathrm{BaSO}_{4}\) as 233.39 g/mol. Understanding how to calculate molar mass is essential for converting between mass and moles, which forms the foundation of much of stoichiometric calculations.