Virogen
Jan18-12, 06:33 PM
Any ideas what the inherent flaw in this experiment is?
1. The problem statement, all variables and given/known data
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?
1. The problem statement, all variables and given/known data
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?