Question

A \(6.977 \mathrm{~g}\) sample of a mixture was analysed for barium ion by adding a small excess of sulphuric acid to an aqueous solution of the sample. The resultant reaction produced a precipitate of barium sulphate, which was collected by filtration, washed, dried and weighed. If \(0.4123 \mathrm{~g}\) of barium sulphate was obtained, what was the mass percentage of barium in the sample? a. \(6.952 \%\) b. \(3.476 \%\) c. \(4.376 \%\) d. \(1.738 \%\)

Short answer

The mass percentage of barium in the sample is 3.476% (option b).

Step-by-step solution

Write the Reaction Equation

The reaction between barium ions and sulphuric acid forms barium sulphate as per the equation: \( \text{Ba}^{2+} + \text{SO}_4^{2-} \rightarrow \text{BaSO}_4 \).

Calculate Molar Masses

Determine the molar mass of barium sulphate \( \text{BaSO}_4 \). The atomic masses are approximately: Ba = 137.33 g/mol, S = 32.07 g/mol, and O = 16.00 g/mol. Therefore, the molar mass of barium sulphate is \( 137.33 + 32.07 + 4 \times 16.00 = 233.39 \text{ g/mol} \).

Calculate Moles of BaSO4

Use the mass of barium sulphate obtained to find the moles. \( \text{Moles of BaSO}_4 = \frac{0.4123 \text{ g}}{233.39 \text{ g/mol}} \approx 0.001767 \text{ mol} \).

Relate Moles of BaSO4 to Moles of Ba

Since the stoichiometry between Ba and BaSO4 is 1:1, the moles of barium are also \(0.001767 \text{ mol} \).

Calculate Mass of Barium

Calculate the mass of barium using its moles and atomic mass. \( \text{Mass of Ba} = 0.001767 \text{ mol} \times 137.33 \text{ g/mol} = 0.2427 \text{ g} \).

Calculate Mass Percentage of Barium in the Sample

Use the formula \( \text{Mass percentage} = \left(\frac{\text{mass of Ba}}{\text{total sample mass}}\right) \times 100 \). So, \( \left(\frac{0.2427 \text{ g}}{6.977 \text{ g}}\right) \times 100 \approx 3.476\% \).

Select Correct Answer

The mass percentage of barium in the sample is closest to option b, \(3.476\%\).

Key concepts

Stoichiometry

Stoichiometry is the aspect of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It involves using the coefficients from balanced equations to determine the proportions of substances that participate in the reaction.
To understand stoichiometry, one must first write the balanced chemical equation for the reaction. For instance, in the given exercise, the reaction is between barium ions and sulfate ions: - Equation: \( \text{Ba}^{2+} + \text{SO}_4^{2-} \rightarrow \text{BaSO}_4 \)This reaction is balanced as one barium ion combines with one sulfate ion to form one barium sulfate molecule, reflecting a 1:1 stoichiometric ratio. Knowing the reaction's stoichiometry allows us to relate the amounts of each substance, making it possible to calculate the mass percentage of barium in the sample through subsequent steps.

Molar Mass

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It is vital for converting mass into moles, a central aspect of stoichiometry.
Calculating the molar mass involves summing the atomic masses of all atoms present in a compound. For barium sulfate (\( \text{BaSO}_4 \)), we calculate as follows:

• Barium (Ba): 137.33 g/mol
• Sulfur (S): 32.07 g/mol
• Oxygen (O): 16.00 g/mol (four oxygen atoms contribute 4 x 16.00 g/mol)

Thus, the molar mass of \( \text{BaSO}_4 \) is \( 137.33 + 32.07 + 4 \times 16.00 = 233.39 \text{ g/mol} \). With this information, you can convert the precipitate's mass to moles, a necessary step to find the mass percentage of barium.

Barium Ion Analysis

Analyzing barium ions requires understanding the stoichiometric ratio in their conversion to barium sulfate. The analysis starts by finding the number of moles of \( \text{BaSO}_4 \) formed using the measured mass: - Moles of \( \text{BaSO}_4 \) = \( \frac{0.4123 \text{ g}}{233.39 \text{ g/mol}} \approx 0.001767 \text{ mol} \)Since the stoichiometric ratio between barium ions and barium sulfate is 1:1, the moles of barium ions originally present in the mixture is equal to the moles of \( \text{BaSO}_4 \): - Moles of \( \text{Ba}^{2+} \) = 0.001767 mol
This calculation is crucial, as it directly leads to determining the mass of barium in the original sample and, consequently, the mass percentage of barium.

Precipitation Reaction

A precipitation reaction occurs when two soluble salts react in solution to form one or more insoluble products, known as precipitates. In our example, when barium ions (\( \text{Ba}^{2+} \)) meet sulfate ions (\( \text{SO}_4^{2-} \)), barium sulfate (\( \text{BaSO}_4 \)) precipitates out of the solution, due to its poor solubility in water.
This reaction helps isolate the barium ions from the aqueous solution in a measurable form. Collecting, washing, drying, and weighing the resulting precipitate allows us to trace back and quantify the presence of barium in the original sample. Understanding precipitation reactions is critical in qualitative and quantitative analysis in chemistry as they often provide a straightforward method for separating and analyzing ions within a mixture.