Data Sheet #3 identification code of unknown 5.3534899 14.3336 20.37a0 1.02181.0
ID: 1086584 • Letter: D
Question
Data Sheet #3 identification code of unknown 5.3534899 14.3336 20.37a0 1.02181.0801 mass of crucible plus sample,g mass of empty crucible,g mass of hydrate sample, g mass of crucible andsample after fist heatig ad4 83 14.83140 3810 after second heating and cooling, g after third heating and cooling g mass of water lost, g mass percent of water in samples formula of anhydrous salt molar mass of anhydrous salt, g mol" mass of anhydrous salt, g number of moles of anhydrous salt remaining in the oSi94Y O.Sli9 20.14 /0 crucible, mol number of moles of water lost, mol average number of moles of water per mole of hydrate, mol formula of hydrateExplanation / Answer
Formula of anhydrous salt=MgSO4
molar mass of anhydrous salt (g/mol)=120.366 g/mol
mass of anhydrous salt=(mass of crucible +sample after final heating)-(mass of crucible)
for trial 1,mass of anhydrous salt=14.8340g-14.3236g=0.5104g
for trial 2,mass of anhydrous salt=21.3810g-20.8720g=0.509g
No of moles of anhydrous salt remaining in the crucible (after evaporating water)=mass of anhydrous salt/molar mass of anhydrous salt
for trial 1, No of moles of anhydrous salt =(0.5104g/120.366 g/mol)=0.00424 mol
for trial 2.No of moles of anhydrous salt=(0.509g/120.366 g/mol)=0.00423 mol
no of moles of water lost=mass of water lost/molar mass of water
trial 1) no of mol of water lost=0.5194g/18.015g/mol=0.0288 mol
trial 2)no of mol of water lost=0.5119g/18.015g/mol=0.0284 mol
mol of hydrate=(same as mol of anhydrous salt as 1 mol anhydrous salt is present per mol hydrate)
No of mol of water per mol hydrate=mol of water /mol of hydrate
trial 1) No of mol of water per mol hydrate=0.0288 mol/0.00424 mol=6.79
trial 2)No of mol of water per mol hydrate=0.0284 mol/0.00423mol=6.71
Average No of mol of water per mol hydrate=(6.79+6.71)/2=6.75=7 (approx)
formula of hydrate=MgSO4.7H2O
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