posted by Ben
5.34g of salt of formula M2SO4 (M is a metal)were dissolved in water. The sulfate ion was precipitated by adding excess barium chloride solution when 4.66g of barium sulfate (BaSO4) were obtained.
a)How many moles of sulfate ion were precipitated as barium sulfate?
b)How many moles of M2SO4 were in the solution?
c)What is the formula mass of M2SO4?
d)What is the RAM of M? Identify M.
a. mols = grams/formula mass BaSO4 = about 4.66/233 = estimated 0.02
b. mols M2SO4 = mols BaSO4 since
M2SO4 ==> BaSO4
c. mols = grams/formula mass so
formula mass = grams/mols = about 5.34/0.02 = about 267.
d. If M2SO4 = 267 then
2M must be 267-formula mass SO4 = about 267-96 = about 171 so M = about 171/2 = about 85.5 Look that up on the periodic table.