Calculate the percentage of water in NH4Fe(SO4)2 x 12 H2O using the theoretical method.
%H2O = (7*molar mass H2O/molar mass NH4Fe(SO4)2.12 H2O)*100 = ?
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To calculate the percentage of water in a compound using the theoretical method, we need to know the formula of the compound.
The compound NH4Fe(SO4)2 x 12 H2O consists of ammonium ions (NH4+), iron (Fe) ions, sulfate ions (SO4^2-), and water (H2O) molecules. The "x 12 H2O" in the formula means that there are 12 water molecules associated with each molecule of NH4Fe(SO4)2.
To find the percentage of water, we need to determine the mass of water and the total mass of the compound.
First, we need to calculate the molar mass of NH4Fe(SO4)2 x 12 H2O:
- The molar mass of each element can be found on the periodic table:
- Nitrogen (N) has a molar mass of 14.01 g/mol.
- Hydrogen (H) has a molar mass of 1.01 g/mol.
- Iron (Fe) has a molar mass of 55.85 g/mol.
- Sulfur (S) has a molar mass of 32.07 g/mol.
- Oxygen (O) has a molar mass of 16.00 g/mol.
- Calculate the molar mass of NH4Fe(SO4)2:
- (14.01 g/mol) x 2 (for NH4) + (55.85 g/mol) + (32.07 g/mol) x 2 (for SO4) = 132.13 g/mol.
Next, we calculate the total molar mass of NH4Fe(SO4)2 x 12 H2O:
- Multiply the molar mass of NH4Fe(SO4)2 by the number of moles of this compound:
- 132.13 g/mol x 1 = 132.13 g/mol.
- Calculate the molar mass of 12 H2O:
- (1.01 g/mol x 2) x 12 = 24.24 g/mol.
- Add the molar mass of NH4Fe(SO4)2 to the molar mass of 12 H2O:
- 132.13 g/mol + 24.24 g/mol = 156.37 g/mol.
To find the percentage of water in NH4Fe(SO4)2 x 12 H2O, we divide the molar mass of 12 H2O by the total molar mass of the compound and multiply by 100%:
- (24.24 g/mol / 156.37 g/mol) x 100% = 15.51%.
Therefore, NH4Fe(SO4)2 x 12 H2O contains approximately 15.51% water.