mixture of MnSO4 and MnSO4*4H20 has a mass of 2.005 g. After heating to drive off all the water the mass is 1.780 g. How many grams of MnSO4*4H20 are present in the mixture?

Question

A mixture of MnSO4 and MnSO4*4H20 has a mass of 2.005 g. After heating to drive off all the water the mass is 1.780 g. How many grams of MnSO4*4H20 are present in the mixture? What is the mass percent of MnSO4*4H20 in the mixture?

I have no idea how to start this, and Im not sure what Im really calculating. Anything would be very helpful. Thank you!

Answer 1

You are supposed to be calculating how many grams of MnSO₄•4H₂O were originally present.

mass of H₂O = 2.005 g - 1.780 g = 0.225 g H₂O

moles H₂O = 0.2250 g x 1 mole H₂O/18.0 g = 0.0125 moles H₂O

moles MnSO₄•4H₂O = 0.0125 mol H₂O x 1 mol MnSO₄•4H₂O/4 mol H₂O = 0.003125 moles MnSO₄•4H₂O

mass MnSO₄•4H₂O = 0.003125 moles MnSO₄•4H₂O x 223.06 g/mol = 0.6971 g

mass % MnSO₄•4H²O = mass MnSO₄•4H₂O / total mass (x100%) = 0.6971 g / 2.005 g (x100%) = 34.77%