calculate the mass of iron in 2.5000g of feric nitrate nonahydrate Fe(NO3)3.9H2O
What percent Fe do you have in the nonahydrate? That will be (atomic mass Fe/molar mass nonahydrate)*100 = ?
(?/100) x 2.5000 g = mass Fe
To calculate the mass of iron in feric nitrate nonahydrate (Fe(NO3)3.9H2O), you need to determine the molar mass of Fe(NO3)3 and then use stoichiometry to find the mass of iron.
First, let's find the molar mass of Fe(NO3)3:
- The atomic mass of Fe (iron) is 55.845 g/mol.
- The atomic mass of N (nitrogen) is 14.007 g/mol.
- The atomic mass of O (oxygen) is 16.00 g/mol.
- Since there are three nitrate ions (NO3-) in the compound, we need to multiply the molar mass of nitrogen by 3.
- Finally, multiply the atomic mass of oxygen by 9 since there are nine water molecules (H2O) in the compound.
Calculating the molar mass of Fe(NO3)3:
Molar mass of Fe = 55.845 g/mol
Molar mass of N = 14.007 g/mol (multiply by 3 for 3 nitrogen atoms) = 42.021 g/mol
Molar mass of O = 16.00 g/mol (multiply by 9 for 9 oxygen atoms) = 144.00 g/mol
Total molar mass of Fe(NO3)3 = 55.845 g/mol + 42.021 g/mol + 144.00 g/mol = 241.866 g/mol
Now, using stoichiometry, we can calculate the mass of iron (Fe) in 2.5000 g of Fe(NO3)3.9H2O.
- Convert the given mass of Fe(NO3)3.9H2O to moles by dividing by the molar mass:
Moles of Fe(NO3)3.9H2O = 2.5000 g / 241.866 g/mol
- Since there is one iron atom in each Fe(NO3)3 molecule, the molar ratio is 1:1. Therefore, the moles of iron (Fe) will be the same as the moles of Fe(NO3)3.9H2O.
- Finally, calculate the mass of iron by multiplying the moles of Fe by the molar mass of Fe:
Mass of Fe = Moles of Fe × Molar mass of Fe = Moles of Fe × 55.845 g/mol
By following these steps, you can calculate the mass of iron in 2.5000 g of Fe(NO3)3.9H2O.