Question

A 1.42 g sample of a pure compound, with formula M_2SO_4, was dissolved in water and treated with an excess of aqueous barium chloride, resulting in the precipitation of all the sulfate ion as barium sulfate. The precipitate was collected, dried, and found to weigh 2.3.3 g. Determine the atomic mass of M, and identify M.

Answer 1

BaSO4 precipitate formed

means all sulfate combines with Ba

233.43 g BaSO4 contain 96.06 g SO4^-2

2.33 g BaSO4 contain 2.33 x 96.06 / 233.43 = 0.96 g SO4^-2

means our M2SO4 contain 0.96 g SO4^-2

mass of M2 = 1.42 - 0.96 = 0.46 g

0.96 g SO4^-2  combines with 0.46 g M2

96.06 g SO4^-2  combines with 96.06 x 0.46 / 0.96 = 46.03 g M2

so mass of M = 46.03 / 2 =23 g

Na atomic weight = 23 g /mol

so M = Na