Question

21 Twenty dietary iron tablets with a total mass of 20.131 g were ground and mixed thoroughly. Then 2.998 g of the powder were dissolved in HNO3 and heated to convert all the iron (FeSO4.7H20) to Fe2O3.XH20. Addition of NH3 precipitated Fe,O3.XH20, which was ignited to give 0.264 g of Fe2O3 (M.wt. = 159.69 g/mol). What is the average mass of FeSO4.7H20 (M.wt. = 278.01 g/mol) in each tablet? * (2 points)

Answer 1

Conversion of FeSO4, 7H2O to Fe2O3

2FeSO4·7H2O → Fe2O3 + SO2 + SO3 + 14H2O

This means one mole of Fe2O3 is formed from 2 moles of FeSO4.7H20

Moles of Fe2O3 formed = mass of Fe2O3/molar mass

Mass of Fe2O3 = 0.264 g

Molar mass = 159.69 g/mol

Fe2O3 moles = 0.264g/(159.69 g/mol)

= (0.264/159.69)mol

= 0.00165 mol

As per balanced reaction equation, 1 mol of Fe2O3 will form from 2 moles of FeSO4.7H2O

Therefore,

Moles of FeSO4.7H2O = 2 × 0.00165 = 0.003306 mol

Molar mass of FeSO4.7H2O = 278.01g/mol

Mass of FeSO4.7H2O = moles × molar mass

= 0.003306 mol × (278.01 g/mol)

= 0.9192 g

Mass of FeSO4.7H2O = 0.9192 g per weight of sample (per 2.998g).

Mass of tablet sample taken for analysis = 2.998g

Total mass of 20 tablets = 20.131g

Amount of FeSO4.7H2O in 20 tablet,

= 20.131g × (0.9192g/2.998g)

= 6.172g

Amount of FeSO4.7H2O per tablet = 6.17 2/20

= 0.3086 g

Amount of FeSO4.7H2O per each tablet = 0.309 g