Question

A mixture of MnSO4 and MnSO4 x 4H2O 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 MnSO4x4H2O are presented in the mixture? what is the mass prevent of MnSO4 x 4H2O in the mixture?

Answer 1

Mass of mixture=2.005 g

Mass remaining after driving off all the water=1.780 g

That means that the mass of MnSO₄ remaining after driving off water=1.780 g

So mass of water=2.005 g - 1.780 g=0.225 g

Molar mass of water=2xMolar mass of H+Molar mass of O

=2x1 g/mol+16 g/mol=2 g/mol+16 g/mol=18 g/mol

Number of moles of water driven off=Mass of water/molar mass

=0.225 g/18 g/mol=0.0125 mol

We can see from the formula MnSO₄x4H₂O that for every 4 mol H₂O there is 1 mol MnSO₄ associated with it

So for 1 mol H₂O there will be (1/4) mol MnSO₄ associated with it

For 0.0125 mol H₂OO there will be (1/4)x0.0125 mol=0.003125 mol MnSO₄ associated with it

Molar mass of MnSO₄=Molar mass of Mn+Molar mass of S+4xMolar mass of O=54.9 g/mol+32 g/mol+4x16 g/mol=54.9 g/mol+32 g/mol+64 g/mol=150.9 g/mol

So mass of MnSO₄.4H₂O present in the mixture=number of moles of MnSO₄ associated with waterx Molar mass of MnSO₄+Mass of water driven off

=0.003125 mol x 150.9 g/mol+0.225 g

=0.472 g+0.225 g=0.697 g

Mass percent of MnSO₄.4H₂O in mixture

=(Mass of MnSO₄.4H₂O/Mass of mixture)x100

=(0.697 g/2.005 g)x100=34.76 %