Question

If 20.0g of MgSO4⋅7H2O is thoroughly heated, what mass of anhydrous magnesium sulfate will remain?

Answer 1

We know that :
MgSO4.7H2O + heat = MgSO4 + 7 H2O 
Also,
molar mass of MgSO4*7H2O = 246.47 g/mole
molar mass of MgSO4(anhydrous) = 120.415 g/mole

Therefore,
Remaining mass of anhydrous magnesium sulfate = (20 grams of MgSO4⋅7H2O) / (246.4746 grams of MgSO4⋅7H2O/mole) x (1 mole of MgSO4 / 1 mole of  MgSO4⋅7H2O) x (120.3676 grams of  MgSO4/mole )
                                                                                 = 7.7 grams of MgSO4

Answer 2

Answer: 9.8g

Explanation:

The calculation is based on the fact that all the water in the molecule will be removed.

i will, then, calculate the mass of water removed and then subtract it from the original 20.0 g of sample.

You can do that following these steps.

1) Calculate the number of moles of the hydrated magnsium sulfate, MgSO₄⋅7H₂O

number of moles = mass in grams / molar mass

The molar mass of MgSO₄⋅7H₂O is calcualted from the atomic mass of each atom times the number of atoms in the formula:

molar mass = 24.305 g/mol + 32.065 g/mol + 4×15.999g/mol + 7×2×1.008g/mol + 7×15.999 g/mol = 246.471 g/mol

⇒ moles of MgSO₄⋅7H₂O = 20.0g / 246.471 g/mol = 0.0811 moles

2) Calculate the number of moles of water, using the ratio from the chemical formula:

7 moles H₂O / 1 mol MgSO₄⋅7H₂O = x / 0.0811 mols MgSO₄⋅7H₂O

⇒ x = 0.0811 × 7 moles H₂O = 0.568 moles H₂O

3) Convert moles of H₂O to grams:

number of moles = mass in grams × molar mass = 0.568 moles × 18.015 g/mol = 10.2 g

4) Hence the mass remaining is 20.0g - 10.2g = 9.8g