an iron supplement is used to treat anaemia and 50mg of iron 2 is required per tablet.if the iron compound used in the tablet is FeSO4.7H2O,what mass of this compound would be required per tablet to provide the desired amount of iron 2?
50 mg Fe x (molar mass FeSO4.7H2O/atomic mass Fe) = ?mg FeSO4.7H2O is what the final equation is.
Intermediate equation is
50 mg Fe x (1 mol Fe/atomic mass Fe) x (1 mol FeSO4.7H2O/1 mol Fe) x (molar mass FeSO4.7H2O/1mol FeSO4.7H2O) = ? mg FeSO4.7H2O
To find the mass of the iron compound required per tablet, we need to determine the molar mass of FeSO4·7H2O and use the stoichiometry to calculate the mass of FeSO4·7H2O that provides 50mg of iron(II).
Here's how we can do it step by step:
Step 1: Find the molar mass of FeSO4·7H2O.
To calculate the molar mass of FeSO4·7H2O, we need to sum up the atomic masses of all the elements in the compound.
Atomic masses:
- Fe: 55.845 g/mol
- S: 32.06 g/mol
- O: 16.00 g/mol (there are four oxygen atoms in FeSO4·7H2O)
- H: 1.00784 g/mol (there are 14 hydrogen atoms in FeSO4·7H2O)
Molar mass of FeSO4·7H2O:
= (55.845 g/mol) + (32.06 g/mol) + 4(16.00 g/mol) + 7(1.00784 g/mol)
= 278.04176 g/mol
Step 2: Calculate the mass of FeSO4·7H2O required per tablet.
Since the molar mass is given as grams per mole, we can use it to convert 50mg of iron(II) to the corresponding mass of FeSO4·7H2O.
Number of moles of iron(II) = Mass of iron(II) / Molar mass of Fe(II)
= 50 mg / (55.845 g/mol)
= 0.893 mol (approx.)
From the balanced chemical equation, we know that 1 mole of FeSO4·7H2O yields 1 mole of Fe(II). Therefore, the molar ratio of FeSO4·7H2O to Fe(II) is 1:1.
Hence, the mass of FeSO4·7H2O required per tablet would also be approximately 50mg.
Thus, approximately 50mg of the iron compound FeSO4·7H2O is required per tablet to provide the desired amount of 50mg of iron(II).