Question

400 milligram iron capsule contain 100 milligram ferocious fumerate [CnCOO)${}_2$ Fe, mol mass = 170] Calculate % of iuon in capsule?

Answer 1

Molecular Mass of $(CHCOO{)}_{2}Fe=170$

As given;

$Fe$ present in $100mg$ of $(CHCOO{)}_{2}Fe$:

$=[\frac{MolarmassofFe}{TotalMolecularmass}\times 100]mg$

$=[\frac{56}{170}\times 100]mg$

$=\frac{560}{17}mg=32.9mg$

$Fe$ present in $400mg$ of $(CHCOO{)}_{2}Fe=\frac{32.9mg}{400mg}$

% of $Fe$ in capsule= $\frac{32.9}{400}\times \frac{100}{10}=\frac{82.25}{10}$%

$=8.225$%