1. The data listed below was determined from the dehydration of a hydrate of cal
ID: 875454 • Letter: 1
Question
1. The data listed below was determined from the dehydration of a hydrate of calcium sulfate Mass of empty crucible = 17.274 g Mass of crucible and calcium sulfate hydrate = 19.084 g Mass of crucible and anhydrous calcium sulfate = 18.705 g a. Write the chemical formulas for water and calcium sulfate. From the formulas Calculate the molar masses of water and anhydrous calcium sulfate. b. Calculate the mass of water removed from the hydrate during the drying process from the difference between the mass of the crucible and hydrate and the mass of the crucible and the anhydrous salt. e. Use the mass of water (part 1b) and the molar mass of water to calculate the moles of water in the sample. d. Determine the mass of anhydrous calcium sulfate by subtracting the mass of the empty crucible from the mass of the crucible containing the anhydrous compound. e. Use the molar mass of the anhydrous compound and the mass of the anhydrous compound (part 1d) to determine the number of moles of anhydrous salt. f. Determine the ratio of the number of moles water in the hydrate to the number of moles of anhydrous salt in the hydrate by dividing the answer from part 1c by the answer from part le. This value is the number of moles of water in the hydrate for each mole of the anhydrous salt. Use this value to write the chemical formula for the hydrate.Explanation / Answer
item a.
water H2O----> H: 2x1=2gr/mol
O: 1x16=16gr/mol
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18gr/mol
Anhydrous calcium sulfate CaSO4 ----> Ca: 1x40=40g/mol
S: 1x32=32g/mol
O: 4x16= 64g/mol
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136gr/mol
item b.
mass of water removed = mass of crucible calcium sulfate hydrate - mass of crucible and anhydrous salt
mass of water removed = 19.084 - 18.705 =0.379gr
item c
mass of water removed = 0.379gr
molar mass of water= 18gr/mol
1 mol of water has 18 gr
x mol of water has 0.379gr
x=(0.379 gr x 1mol)/18gr
x=0.021mol
item d.
mass of anhydrous calcium sulfate = mass of crucible and anydrous calcium sulfate - mass of empty crucible
mass of anhydrous calcium sulfate = 18.705 - 17.274 = 1.431 gr
item e
mass of anhydrous=1.431gr
molar mass of anhydrous=136gr/mol
1mol of anhydrous has 136 gr
x mol of anhydrous has 1.431gr
x=(1.431gr x 1mol)/136gr = 0.010mol
item f
number of moles of water in the hydrate for each mole of the anhydrous salt = 0.021/0.010 = 2.1
CaSO4.2H2O is named calcium sulfate dihydrate or calcium sulfate-2-water
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