107: 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 g of barium sulfate was obtained, what was the mass percentage of barium in the sample?
Solution: Here, we are going to calculate the mass percentage of barium in the given sample. Step1: The reaction between barium ion and sulfuric acid takes place in the following manner: Ba (aq) + H SO (aq) --------> BaSO (aq) + 2H (aq) + 2 4 4 From the above equation it is clear that 1 mol of barium sulfate is produced from 1 mol of barium ion. Step2: Given, mass of barium sulfate = 0.2815 g Molar mass of barium sulfate = 233.4 g/mol Therefore, number of moles of barium sulfate = given mass / molar mass = 0.2815 / 233.4 = 0.0012 mol Step3: So, number of moles of barium ion used = number of moles of barium sulfate formed = 0.0012 mol. 2+ 2+ Now, 1 mol Ba ion = molar mass of Ba ion. Therefore, 0.0012 mol of Ba ion = 0.0012 x molar mass of Ba ion. 2+ 2+ = 0.0012 x 137.3 g [molar mass of Ba = 137.3 g/mol] = 0.165 g Step4: Mass percentage of Ba ion = (Mass of Ba ion / mass of sample) X 100 % = (0.165 / 3.455) X 100 % = 0.048 X 100 % = 4.8 % Thus, percentage of barium in the sample is 4.8 %.