DATA: 2 g Unknown number. 2/, O 1 Mass of crucible lid 2 Mass of crucible and hy

Question

DATA: 2 g Unknown number. 2/, O 1 Mass of crucible lid 2 Mass of crucible and hydrate lid 2L 56lBg 3 Mass of crucible and residue lid (1st heating) 4 Mass of crucible and residue lid (2nd heating) 2) 55 Sog 5 Mass of crucible and residue lid (3rd heating 02.4953 USE THE LAST HEATING PERFORMED FOR ALL CALCULATIONS SHOW YOUR WORK WORK ALONE CALCULATIONS: 1. If we use good technique (avoid spills, etc.) the following will be true: hen mass of hydrate mass of the residue mass of the water lost of water lost by From the data above, calculate the mass of the anhydrous compound the residue) and the mass the appropriate masses. Copy the descriptions for the lines you are using from the table. Mass of Mass of Mass of hydrate (starting material Mass of Mass of Mass of anhydrous compound (residue) Mass of Mass of Mass of water lost 2. Use the data and calculations above to determine the percent water (by mass) in the hydrated compound 3. If a different sample of your contains 0.815 g of Mgso, how many grams of H20 does it contain? Use the mass percentage information from #2.

Explanation / Answer

1)Mass of crucible and hydrate+lid=21.9607g

Mass of crucible +lid=21.0278g

mass of hydrate=0.9329g

mass of anhydrous compound(residue)`=Mass of crucible and residue+lid(2nd heating)-mass of crucible and lid=21.5580g-21.0278g=0.5302g

Mass of crucible and hydrate+lid=21.9607g

Mass of crucible and residue+lid(2nd heating)=21.5580g

mass of water lost=21.9607g-21.85580g=0.4027g

2)% water (by mass)=(mass of water lost/mass of hydrate)*100=0.4027/0.9329*100=43.1665%

3)mass of sample=0.815g

mass of water=43.1665% of 0.815g=43.1665/100 *0.815=0.3518g

4)mass of water#1=0.4027g

moles of water=0.4027g/molar mass of water=0.4027g/18g/mol=0.0223 moles

moles of MgSO4=mass of residue/molar mass of MgSO4=0.5302g/120.366g/mol=0.00440mol

moles of water=0.0223 moles/0.00440mol=5.0

formula=MgSO4.5H2O

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