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)?

Answer 1

The percent by mass of water in iron (II) sulfate heptahydrate, FeSO4·7H2O, is 51.00%.

Step-by-step explanation:

To calculate the percent by mass of water in iron (II) sulfate heptahydrate, FeSO4·7H2O, we need to determine the molar mass of the compound and the molar mass of water. The molar mass of iron (II) sulfate heptahydrate is calculated as follows:

1. The molar mass of Fe is 55.85 g/mol.
2. The molar mass of S is 32.07 g/mol.
3. The molar mass of O is 16.00 g/mol.
4. The molar mass of H is 1.01 g/mol.
5. The molar mass of FeSO4 is equal to (55.85 g/mol + 32.07 g/mol + (4 * 16.00 g/mol)) = 151.91 g/mol.
6. The molar mass of FeSO4·7H2O, including the waters of crystallization, is equal to (151.91 g/mol + (7 * 18.02 g/mol)) = 247.14 g/mol.

Next, we determine the molar mass of the waters of crystallization:

1. The molar mass of water (H2O) is 18.02 g/mol.
2. The molar mass of the waters of crystallization (7H2O) is equal to (7 * 18.02 g/mol) = 126.14 g/mol.

Finally, we calculate the percent by mass of water in iron (II) sulfate heptahydrate:

(126.14 g/mol / 247.14 g/mol) x 100% = 51.00%

Answer 2

45.360% First, lookup the atomic weights of all involved elements: Atomic mass Iron = 55.845 Atomic mass Sulfur = 32.065 Atomic mass Hydrogen =1.00794 Atomic mass oxygen = 15.999 Now calculate the molar mass of FeSO4•7H2O 55.845 + 32.065 + 11 * 15.999 + 14 * 1.00794 = 278.01016 g/mol And the mass of 7H2O 14 * 1.00794 + 7 * 15.999 = 126.10416 Now calculate what percentage of 278.01016 is 126.10416 126.10416 / 278.01016 = 0.453595509 = 45.3595509% = 45.360%