Answer:
Amount of elemental iron in sample A is 0.0405g.
Explanation:
Molar mass of C = 12 g/mol
Molar mass of H = 1 g/mol
Molar mass of Fe = 56 g/mol
Molar mass of O = 16 g/mol
So, molar mass of ferrous gluconate = [tex](12\times 12)g+(24\times 1)g+(1\times 56)g+(14\times 16)g=448g[/tex]
So number of mole of ferrous gluconate in 324 mg = [tex]\frac{0.324}{448}moles[/tex]
(number of moles = [tex]\frac{mass}{molar mass}[/tex])
As 1 mol of ferrous gluconate contains 1 mol of Fe therefore [tex]\frac{0.324}{448}moles[/tex] of ferrous gluconate contain [tex]\frac{0.324}{448}moles[/tex] of Fe.
So amount of elemental iron in sample A = [tex]\frac{0.324}{448}\times 56g=0.0405g[/tex]
