How many grams of the hydrate must the student weigh out in order to get the desired amount of the anhydrous compound?

Question

A student is running an experiment in which 51.0 grams of MnSO₄ is needed, but the only jar of the reagent in the lab is labeled manganese(II) sulfate pentahydrate.

Answer 1

molar mass of H₂O = 18 g/mol

molar mass of MnSO₄ = (55 + 32 + (16 * 4)) g/mol = 151 g/mol

molar mass of MnSO₄.5H₂O = (55 + 32 + (16 * 4) + (5 * 18)) g/mol = 241 g/mol

mass of MnSO₄.5H₂O = mass of MnSO₄ * (molar mass of MnSO₄.5H₂O/molar mass of MnSO₄)

51g MnSO₄ * (241/151) = 81.397 g

Answer 2

Hi Julissa A.,

Even as a student, when you prepare for a lab you should be aware that there may be issues with your supplied materials. So for anhydrous materials, you would want to be aware if they were prone to pick up water or CO₂ (or both!) from the air; and for hydrates, you might want to be aware that they might either lose or acquire more water. So you would check the literature and find that MnSO₄ forms anhydrous, monohydrate, tetrahydrate, pentahydrate, perhaps hexahydrate, and heptahydrate salts. Each of these is stable within a particular ambient humidity range at a particular temperature. Let the user beware, especially of color, texture, etc. variations within a source jar! The tip-off in this question was the phrase, "is labeled Mn (II) sulfate pentahydrate".

Just sayin! (There's nothing wrong with Saqibul C.'s calculations.)

-- Cheers, --Mr. d.