Hydrated crystals lab, need help with questions?

anyone who can answer the questions using the data table will get best answer and points. im really lost in this chapter and unsure of what to do.
please and thank you
Mass of Evaporating dish = 46.64g
Mass of Evaporating dish + hydrated MgSO4 = 49.46
Mass of Hydrated MgSO4 = 2.82g
Mass of Evaporating dish + anhydrous MgSO4 = 48.47g
Mass of anhydrous MgSO4 = 1.83g
mass of H2O in the hydrate = 0.99g
Things i need to find:
Moles of anhydrous MgSO4
Moles of H2O in the hydrate

Questions i need to answer:
1.Compare the number of moles or anhydrous MgSO4, to the number of moles of water in the hydrate. Use the ratio of these two values to predict a forumula for hydrated MgSO4.
2.The method used i this experiment to find the percentage of water in the hydrated crystals is not suitable for all hydrates. Give at least two reasons why this may be so.
3.How would your experimental results be affected if you did not use a desiccator. What is a desiccator? how does it work?

hydrated MgSO4 = 2.82 g
anhydrous MgSO4 = 1.83
moles of anhydrous MgSO4 = (24+32+4*16)=120g/mol,
so 1.83 g is 1.83/120 = 0.01525 moles
0.99g of water is 0.99/18 =0.055 moles, so ratio is 0.01525:0.055 = 1:3.6 moles, so the formula is approx. MgSO4*4H2O (a bit odd, as it's normally 7 moles of water per mole of MgSO4)
2) not sure what method you used, but in some cases water forms strong complexes with metal ions (particulalry transition metal ions) and cannot be removed easily
3) desiccator is a device which maintains low water vapor (a chemical is used that strips water from the air, like a dehumidifier). Sometimes it can also be kept under vacuum
1.83/120.37 = .0152 mols anhydrous MgSO4

2.82-1.83= .99g H2O

.99/18g/mol= 0.055 mols H2O

.015:.055 3 to 11 ratio

#2 I have no clue what your experiment was so.....

#3 The use of a desiccator protected your anhydrous product from any humidity in the air. Product is stored in them and the air is vaccumed out.

Good Luck!

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