Question

What is the percent by mass of water in iron (II) sulfate heptahydrate, FeSO4*7H2O (or what percent of the molar mass of FeSO4*7H2O is due to the waters of crystallization?

Please include work so I can understand thanks.

Answer 1

Ok, so this question is asking you to find out what percent mass of water makes up the compound itself.

Step 1: The first step is to find the molecular mass of your compound.

We know that in the compound there is 1 Iron, 1 Sulfur, 11 Oxygen, and 14 Hydrogen, each of these elements have a corresponding molecular weight that can be found on your periodic table. Once each atomic weight has been isolated (off the periodic table) you must multiply that number by how ever many atoms of that species are present in your compound.

So for example. There are 11 Oxygens in your compound. Oxygen has an atomic weight of 15.9994g/mol according to the periodic table, now we must multiply the 15.9994g/mol by 11. This gives us 175.9934 g/mol. Continue on finding the atomic masses of the rest of the compounds constituents, multiply them by their frequencies. (HINT: Hydrogen x 14, Iron x 1, Sulfur x 1 )

Once all of the masses have been calculated you must add them all together! This gives you the molecular mass of your compound. You should get around 278.0 g/mol.

Step 2: WATER!! So by the same means as above we want to find out the molecular mass of one water molecule. It is 18.0g/mol, BUT there are 7 in our compound, so this number must be multiplied by 7. 126.0g/mol.

Step 3: Divide your mass of water by the mass of the whole compound.

126.0g/mol / 278.0g/mol = 0.453 (the units can be dropped because they have now cancelled out), Last thing to do is multiply 0.453 x 100% which gives us a final percentage of 45% water in your compound:)

Hopefully this helps!