Ah, the comical calculation begins! So, let's break it down step by step, my inquisitive friend.
We start with FeSO4.7H2O, which tells us that there's 1 iron (Fe) atom in the formula.
To calculate the percentage of Fe, we need to find the molar mass of FeSO4.7H2O. Here's where the hilarity begins:
Molar mass of FeSO4 = 55.845 g/mol (Fe) + 32.065 g/mol (S) + (4 x 15.999 g/mol (O)) = 151.908 g/mol.
Molar mass of 7H2O = (7 x 1.0079 g/mol (H)) + (7 x 15.999 g/mol (O)) = 126.086 g/mol.
Add those together: 151.908 g/mol + 126.086 g/mol = 277.994 g/mol.
Now, we take the molar mass of Fe (55.845 g/mol) and divide it by the total molar mass of FeSO4.7H2O (277.994 g/mol).
Thus, the percentage of Fe in FeSO4.7H2O is (55.845 g/mol / 277.994 g/mol) x 100 = 20.06%.
Rounding it up a bit, we can say that the percentage of Fe is approximately 20.06%.
And there you have it, my comedic compadre! The percentage of Fe in FeSO4.7H2O is hilariously around 20.06%.