How to solve this chemical question?

Question

Mass of MgCO3 hydrate = 0.9987 g.

Constant mass of anhydrous MgCO3 = 0.4831 g

How many moles of water were in the MgCO3 hydrate?

Answer 1

Hey Ooyeon,

So, we know the total mass of the hydrate compound, as well as the anhydrous mass of MgCO3, which is the mass of MgCO3 in the sample without any water. Therefore, to find the amount of water present in the sample, we subtract the mass of anhydrous MgCO3 from the mass of the hydrate MgCO3.

0.9987g - 0.4831g= 0.5156 g water present in the sample

After finding the amount of water we have, all we have to do is convert to moles of water using the molar mass of water.

0.5156g water *1 mole water/ 18.00g water= 0.0286 moles water in the MgCO3 hydrate.

Hope this helps!