Asked by Marty

A hydrated MgSO4 salt, weighing 3.211g, is heated in a crucible until reaching a constant weight. The weight of the anhydrous MgSO4 is 1.570 grams.

1. Calculate the moles of H20 removed and the moles of anhydrous MgSO4 remaining in the crucible.

2. What is the formula for the hydrated MgSO4?
Answered by bobpursley
You can calculate the weight of the removed water. Change that to moles.

Next, change the grams of anhydrous sulfate salt to moles.

Compare the ratio of those two: that is the prefix x on the hydraded salt

MgSO4.xH2O

I will be happy to critique your thinking or work.
Answered by Marty
How do I change the weight of the water removed into moles?
Answered by bobpursley
moles water=masswater/molmass of water

where molmas is 18g/mole (you need to figure it more accurately, 2 H and one O)
Answered by Marty
Would the weight of the water removed be 3.211g - 1.570 g?
Answered by bobpursley
yes, of course.
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