It is often possible to change a hydrate into an anhydrous compound by beating i
ID: 1048486 • Letter: I
Question
It is often possible to change a hydrate into an anhydrous compound by beating it to drive off the water (dehydration). A 39.19 gram sample of a hydrate of CaSO_4 was heated thoroughly in a porcelain crucible, until its weight remained constant After heating, 30.99 grams of the anhydrous compound remained What is the formula of the hydrate? Use a period instead of a dot in the formula of the hydrate __________ A student is running an experiment in which 70.7 grams of MnSO_4 is needed, but the only jar of reagent in the lab is labelled management (II) sulfate monohydrate. How many grams of the hydrate must the student weigh out in order to get the desired amount of the anhydrous compound? ___ g MnSO_4.H_2O One of the hydrates of Ca(NO_3)_2 is calcium nitrate tetrahydrate A 52.9 gram sample of Ca(NO_2)_2.4H_2O was heated thoroughly in a porcelain crucible, until its weight remained constant. After heating, how many grams of the anhydrous compound remained? _______Explanation / Answer
1) Let the formula of hydrate is CaSO4.xH2O, where x represents the number of water molecules attached
Molar mass of anhydrous CaSO4 = 40 + 32 + 4 * 16 = 136 gm/mol
Molar mass of hydrous CaSO4.xH2O = 136 + 18x
Since number of moles will be same we can write
39.19/(136+18x) = 30.99/136
136 * (39.19-30.99) = 30.99 * 18 * x
x = 1.9992
Hence number of water molecules attached is equal to 2
Formula of Hydrate = CaSO4.2H2O
2)
Molar mass of anhydrous MgSO4 = 24 + 32 + 4 * 16 = 120 gm/mol
Molar mass of monohydrous MgSO4.H2O = 120 + 18 = 138 gm/mol
Mass of hydrated MgSO4 required = Mass of anhydrous sample * Molar mass of monohydrous/Molar mass of anhydrous sample
=> 70.7 * 138/120
=> 81.305 grams
3)
Molar mass of Ca(NO3)2.4H2O = 40 + (14+48)*2 + 4 * 18 = 40 + 124 + 72 = 236 gm/mol
Molar mass of Ca(NO3)2 = 40 + (14+48)*2 = 164 gm/mol
Grams of anhydrous compound left = 52.9 * 164/236
=> 36.761 grams
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