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?
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.
To solve this problem, we need to use the concept of stoichiometry and the molar masses of the compounds involved.
1. Calculate the moles of H2O removed:
Step 1: Calculate the molar mass of MgSO4 and H2O.
- Molar mass of MgSO4 = 24.31 + 32.07 + (4 * 16.00) = 120.37 g/mol
- Molar mass of H2O = 2 * 1.01 + 16.00 = 18.02 g/mol
Step 2: Calculate the moles of H2O using the given weight of the hydrated MgSO4.
- Moles of H2O = Mass of H2O / Molar mass of H2O
= 3.211 g / 18.02 g/mol
Step 3: Calculate the moles of anhydrous MgSO4 remaining:
- Moles of anhydrous MgSO4 = Mass of anhydrous MgSO4 / Molar mass of MgSO4
= 1.570 g / 120.37 g/mol
2. Determine the formula for the hydrated MgSO4:
The formula for the hydrated MgSO4 can be determined by comparing the moles of H2O removed and the moles of anhydrous MgSO4 remaining. The ratio of moles between the two species will give us the subscripts in the formula.
In this case, the ratio is 1:7, which means for every 1 mole of anhydrous MgSO4, there are 7 moles of H2O. Therefore, the formula for the hydrated MgSO4 is MgSO4 * 7H2O.
Note: The "*" symbol in the formula represents the stoichiometric coefficient, which indicates the number of moles of water molecules.
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.