If it is known that the iron(ii) salt is FeSO4.7H20 calculate the %Fe from the formula
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%.
To calculate the percentage of iron (Fe) in the compound FeSO4·7H2O, you first need to determine the molar mass of Fe in the compound and the overall molar mass of the compound.
The molar mass of Fe is calculated from the periodic table. Iron has an atomic weight of approximately 55.85 g/mol. Since there is only one Fe atom in FeSO4, the molar mass of Fe is also 55.85 g/mol.
Next, you calculate the molar mass of the entire compound FeSO4·7H2O.
The molar mass of FeSO4 is found by adding up the atomic masses of its constituent elements:
Fe: 1 × 55.85 g/mol = 55.85 g/mol
S: 1 × 32.07 g/mol = 32.07 g/mol
O: 4 × 16.00 g/mol = 64.00 g/mol
The molar mass of water (H2O) is calculated by adding the atomic masses of hydrogen (H) and oxygen (O) atoms:
H: 2 × 1.01 g/mol = 2.02 g/mol
O: 1 × 16.00 g/mol = 16.00 g/mol
Since there are seven water molecules in FeSO4·7H2O, the molar mass of water is multiplied by 7:
7 × (2.02 g/mol + 16.00 g/mol) = 126.14 g/mol
Now calculate the overall molar mass of the compound by adding the molar mass of FeSO4 and 7 times the molar mass of water:
55.85 g/mol + 126.14 g/mol = 182.99 g/mol
To determine the percentage of Fe in the compound, you need to divide the molar mass of Fe by the molar mass of the compound and multiply by 100:
(55.85 g/mol / 182.99 g/mol) × 100% ≈ 30.51%
Therefore, the percentage of Fe in FeSO4·7H2O is approximately 30.51%.