posted 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?
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
I will be happy to critique your thinking or work.
How do I change the weight of the water removed into moles?
moles water=masswater/molmass of water
where molmas is 18g/mole (you need to figure it more accurately, 2 H and one O)
Would the weight of the water removed be 3.211g - 1.570 g?
yes, of course.