Question

A 3.455-g sample of a mixture was analyzed for barium ion by adding a small excess of sulfuric acid to an aqueous solution of the sample. The resultant reaction produced a precipitate of barium sulfate, which was collected by filtration, washed, dried, and weighed. If \(0.2815 \mathrm{~g}\) of barium sulfate was obtained, what was the mass percentage of barium in the sample?

Short answer

The mass percentage of barium in the sample is 4.79%.

Step-by-step solution

Find the mole ratio of Ba in BaSO4

The balanced chemical equation for the formation of barium sulfate (BaSO4) is as follows: \[ Ba^{2+} + SO_4^{2-} \rightarrow BaSO_4 \] From this equation, we can see that 1 mole of barium ion (Ba) reacts with 1 mole of sulfate ions (SO4) to produce 1 mole of barium sulfate (BaSO4). Hence, the mole ratio of Ba in BaSO4 is 1:1.

Calculate the molar mass of BaSO4

To calculate the molar mass of BaSO4, we can add the molar masses of its individual elements (Ba, S, and O). Molar mass of Ba = 137.33 g/mol Molar mass of S = 32.07 g/mol Molar mass of O = 16.00 g/mol Therefore, molar mass of BaSO4 = 137.33 g/mol (Ba) + 32.07 g/mol (S) + 4(16.00 g/mol) (O) = 233.43 g/mol

Calculate the moles of BaSO4

Given that the mass of BaSO4 obtained is 0.2815 g, we can calculate the moles of BaSO4 using the formula: moles of BaSO4 = (mass of BaSO4) / (molar mass of BaSO4) moles of BaSO4 = (0.2815 g) / (233.43 g/mol) = \(1.206 \times 10^{-3}\) moles

Calculate the mass of Ba in BaSO4

Since the mole ratio of Ba in BaSO4 is 1:1, the moles of Ba in BaSO4 are equal to the moles of BaSO4, which is \(1.206 \times 10^{-3}\) moles. Now, we can find the mass of Ba using the formula: mass of Ba = moles of Ba × molar mass of Ba mass of Ba = \((1.206 \times 10^{-3} \text{ moles}) \times (137.33 \text{ g/mol})\) = 0.1657 g

Calculate the mass percentage of Ba in the sample

Now that we have the mass of Ba in the sample, we can calculate the mass percentage using the formula: mass percentage of Ba = \((\frac{mass\:of\:Ba}{mass\:of\:sample}) \times 100\%\) mass percentage of Ba = \((\frac{0.1657 \mathrm{~g}}{3.455 \mathrm{~g}}) \times 100\%\) = 4.79% So the mass percentage of barium in the sample is 4.79%.