Question

A 1.803-g sample of gypsum, a hydrated salt of calciumsulfate, CaSO₄, is heated in a crucible until a constantmass is reached. The mass of the anhydrous CaSO₄ salt is1.426-g. Calculate the percent by mass of water in the hydratedcalcium sulfate salt. Also, calculate the moles of water removedand the moles of anhydrous CaSO₄ remaining in thecrucible. And, finally, What is the formula of the hydrated calciumsulfate; that is, what is the whole-number mole ratio of calciumsulfate to water?

Please show all calculations. I know its a long problem, butdefinately worth a 7+ rating if you can help me out.Thanks!!!!!!!!!!!!

Answer 1

Hello,
          I am acurrent AP Chemistry student and I like practicing Chemproblems. This problem will not help you, but also me. Ok, sohere we go.


Mass of Hydrated CaSO_{4 =1.803g   (Given)
Mass of Water = Mass of Hydrated} CaSO_{4 - Mass of Anhydrous} CaS<sub>4

Mass of Water = 1.803g -1.426g = 0.377g

% Mass of water in Hydrated Calcium Sulfate =</sub> x 100
So,

% Mass of water =   x 100 = 20.9 %

Moles of H₂O = Mass of H₂O / Molar Mass ofH₂O

Moles of H₂O =

Molar Mass of anhydrous CaSO<sub>4 =136.15 g / mol

Moles of</sub> anhydrous CaSO_{4 =} = 0.0105 mol CaSO4

Moles Hydrated Salt = 1.803g / 136.15g = 0.01324 g
Whole-number mole ratio of calcium sulfate to water = = 1.58
Now multiple the ratio by 5 to get 7.9 and then round the ratio to8.


So the formula would be like this : 5 CaSO_{4 ^.}   8H₂O

Hope this helped!
Thx.