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Virogen
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Any ideas what the inherent flaw in this experiment is?
In the lab (actual lab), we combined copper(II) sulfate pentahydrate with barium nitrate to form barium sulfate (precipitate). Every group got a 115% or higher yield than stoichiometry would predict.
Ba(NO3)2 + CuSO4 5H2O --> Cu(NO3)2 + BaSO4 + 5H2O
Mass of barium nitrate: 1 g
Mass of copper(II sulfate pentahydrate) 0.75 g
Moles of copper(II) sulfate pentahydrate: 0.75g/249.71g/mol = 0.00300 mol
Moles of barium nitrate: 1g/261.35 g/mol = 0.00383 mol
The limiting reagent is thus copper(II) sulfate pentahydrate, and because of the 1:1 mole ratio, 0.00300 mol of barium sulfate should be produced.
0.00300 mol x 233.39 g/mol = 0.700g
However, all my students recorded at least 0.8 g... I'm very confused. It must have something to do with the water/pentahydrate?? But I can't see that mathematically if I try to remove the water completely. Any ideas what the flaw in this experiment is?
Homework Statement
In the lab (actual lab), we combined copper(II) sulfate pentahydrate with barium nitrate to form barium sulfate (precipitate). Every group got a 115% or higher yield than stoichiometry would predict.
Ba(NO3)2 + CuSO4 5H2O --> Cu(NO3)2 + BaSO4 + 5H2O
Mass of barium nitrate: 1 g
Mass of copper(II sulfate pentahydrate) 0.75 g
Moles of copper(II) sulfate pentahydrate: 0.75g/249.71g/mol = 0.00300 mol
Moles of barium nitrate: 1g/261.35 g/mol = 0.00383 mol
The limiting reagent is thus copper(II) sulfate pentahydrate, and because of the 1:1 mole ratio, 0.00300 mol of barium sulfate should be produced.
0.00300 mol x 233.39 g/mol = 0.700g
However, all my students recorded at least 0.8 g... I'm very confused. It must have something to do with the water/pentahydrate?? But I can't see that mathematically if I try to remove the water completely. Any ideas what the flaw in this experiment is?