If 4.00 grams of the hydrate in #1 are heated how many grams of anhydrous salt would remain after all the water was liberated?
MgCo subscript 3 - 3H2O = molar mass of 108.31g/mol
I came up with the answer .668g MgCO3 anyhydrous salt would remain....
To find out how many grams of anhydrous salt would remain after all the water was liberated, we need to determine the percentage of water in the hydrate.
First, let's assume the hydrate has the formula MX·nH₂O, where MX represents the anhydrous salt and n represents the number of water molecules attached to each unit of the salt.
Next, we'll need the molar mass of the hydrate. We can determine this by adding the atomic masses of all the atoms present in one molecule of the hydrate. The atomic masses of each element can be found on the periodic table.
Once we have the molar mass of the hydrate, we need to determine the molar mass of the water component. We know that the molar mass of water (H₂O) is 18.015 g/mol.
To find the mass of the water in the hydrate, we need to subtract the molar mass of the anhydrous salt from the molar mass of the hydrate.
Now, we can calculate the percentage of water in the hydrate by dividing the mass of the water by the mass of the hydrate and multiplying by 100.
Finally, if 4.00 grams of the hydrate are heated, we can multiply this value by the percentage of water to find the mass of water that will be liberated. Subtracting this amount from the initial mass of the hydrate will give us the mass of the anhydrous salt remaining.
So to determine the answer, we first require the formula of the hydrate and the molar mass of the hydrate.