Question

An ore of barium contains \(\mathrm{BaCO}_{3}\). A 1.542 g sample of the ore was treated with \(\mathrm{HCl}\) to dissolve the \(\mathrm{BaCO}_{3}\). The resulting solution was filtered to remove insoluble material and then treated with \(\mathrm{H}_{2} \mathrm{SO}_{4}\) to precipitate \(\mathrm{BaSO}_{4}\). The precipi- (a) How tate was filtered, dried, and found to weigh \(1.159 \mathrm{~g}\). many moles of barium were in the ore sample? (b) How many grams of barium were in the ore sample? (c) What is the percentage by mass of barium in the ore? (d) If the molarity and volume of the sulfuric acid needed for the precipitation was noted, could you get the same answers?

Short answer

The number of moles of barium in the ore sample is equal to the moles of \(\mathrm{BaSO}_{4}\) precipitated. The mass of barium can be obtained using the moles of barium and its atomic mass. The percentage by mass of barium in the ore is calculated by comparing the mass of barium to the mass of the ore. With the molarity and volume of \(\mathrm{H}_{2} \mathrm{SO}_{4}\), it is possible to cross-check these answers.

Step-by-step solution

Calculate the molar mass of \(\mathrm{BaSO}_{4}\)

First, calculate the molar mass of \(\mathrm{BaSO}_{4}\) by adding the atomic masses of barium (Ba), sulfur (S), and oxygen (O). The atomic masses (in g/mol) are approximately: Ba = 137.33, S = 32.07, O = 16.00. Thus, the molar mass of \(\mathrm{BaSO}_{4}\) is \(137.33 + 32.07 + 4(16.00) = 233.39 \, \text{g/mol}\).

Calculate the moles of \(\mathrm{BaSO}_{4}\)

To find the number of moles of the \(\mathrm{BaSO}_{4}\) precipitate, divide the mass of the precipitate (1.159 g) by the molar mass of \(\mathrm{BaSO}_{4}\) as calculated in Step 1 (\(233.39 \text{g/mol}\)). \[\text{Moles of } \mathrm{BaSO}_{4} = \frac{1.159 \text{ g}}{233.39 \text{ g/mol}}\]

Determine moles of barium

Since one mole of \(\mathrm{BaSO}_{4}\) contains one mole of barium, the moles of barium are equal to the moles of \(\mathrm{BaSO}_{4}\) precipitate. Use the result from Step 2.

Calculate the mass of barium in the ore sample

The mass of barium is calculated by multiplying the moles of barium (from Step 3) by the atomic mass of barium (137.33 g/mol). \[\text{Mass of Ba} = \text{moles of Ba} \times 137.33 \text{ g/mol}\]

Calculate the percentage by mass of barium in the ore

Find the percent by mass of barium in the ore by dividing the mass of barium (from Step 4) by the mass of the ore sample (1.542 g) and multiplying by 100%. \[\text{Percentage by mass of Ba} = \left(\frac{\text{Mass of Ba}}{1.542 \text{ g}}\right) \times 100\%\]

Discuss molarity and volume of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) usage

If the molarity and volume of the sulfuric acid used were known, you can calculate the moles of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) that reacted. Since \(\mathrm{BaSO}_{4}\) has a 1:1 molar ratio with \(\mathrm{BaCO}_{3}\), and there is also a 1:1 molar ratio with \(\mathrm{H}_{2} \mathrm{SO}_{4}\), one could calculate the amount of barium in the sample. However, this requires assuming that all \(\mathrm{H}_{2} \mathrm{SO}_{4}\) reacted with \(\mathrm{BaCO}_{3}\) and that the sample is pure \(\mathrm{BaCO}_{3}\).

Key concepts

Barium Sulfate Precipitation

Barium sulfate precipitation is a process central to gravimetric analysis in chemistry, particularly when determining the quantity of barium in a sample. It involves adding sulfuric acid to a solution containing barium ions, leading to the formation of a white, insoluble solid—barium sulfate ((BaSO_4)). This reaction is crucial as (BaSO_4)'s lack of solubility enables its separation from the mixture by filtration. During the reaction, for every mole of barium ions, one mole of (BaSO_4) forms. This 1:1 stoichiometric relationship between barium ions and the barium sulfate precipitate is essential for the subsequent calculations of molar mass and percentage by mass, giving a clear path to quantifying the barium content in the original sample.

Understanding this mechanism is crucial as any impurities or incomplete reactions can affect the accuracy of the gravimetric analysis. Careful preparation and handling of materials are pivotal to ensure that the only precipitate formed is (BaSO_4), and all barium present is accounted for in this solid form.

Molar Mass Calculation

Calculating the molar mass is a fundamental process in chemistry, serving as a bridge between the microscopic world of atoms and molecules and the macroscopic world we can measure. The molar mass of a compound, like barium sulfate ((BaSO_4)), is determined by summing the masses of the individual elements that make up the compound, based on the number of each type of atom present. In our example, the atomic mass of barium, sulfur, and oxygen must be accurately added, considering the exact number of atoms of each in the chemical formula of (BaSO_4).

To get the total molar mass, we use the atomic mass unit (amu), where one amu is defined as one twelfth of the mass of a carbon-12 atom. The precision in determining these atomic masses directly affects the accuracy of our molar mass calculation, which in turn impacts all subsequent quantitative results in our gravimetric analysis.

Percentage by Mass

The percentage by mass (or mass percent) is a measure that depicts the amount of a certain substance within a mixed sample as a proportion of the total sample mass. It's an invaluable concept in chemistry for representing concentration, especially in gravimetric analysis. To find the percentage by mass of barium in the ore, you divide the mass of the barium determined in the ore sample by the total mass of the sample and then multiply by 100 to get a percentage. This calculation provides a clear understanding of the purity of the ore in terms of barium content.

It's crucial to ensure that the measurement of the ore mass and the mass of the isolated barium is accurate. Any error here will directly skew the final percentage by mass value, leading to possible inaccuracies in conclusions drawn from the data.

Stoichiometry

Stoichiometry is the part of chemistry that refers to the quantitative relationships of substances as they participate in chemical reactions. In the context of barium sulfate precipitation from a barium carbonate ore, stoichiometry allows us to relate the measurable mass of the precipitate to the moles of reactants and products. Since the reaction between barium carbonate and sulfuric acid proceeds in a 1:1 mole ratio, stoichiometry provides the basis for all the mass and molar calculations, leading up to the determination of the percentage by mass of barium in the ore sample.

Precise stoichiometric calculations are vital in predicting the amount of product formed from a given quantity of reactant and vice versa. Gravimetric analysis strongly depends on the stoichiometric ratios outlined in the balanced chemical equations for ensuring accurate and meaningful results.