Question

If you digested 25 mg of pure ferrocene in nitric acid and then diluted the resulting solution up to 500 mL with 1% aqueous hydrochloric acid, what would be the concentration of iron in solution (provide an answer in both mg/L and mol/L)?

Answer 1

1.345*10⁻⁴ mol/L

15.023 mg/L

Step-by-step explanation:

The chemical formula of ferrocene is Fe(C₅H₅)₂, thus its molecular weight is:

55.845 g/mol + 10*12g/mol + 10 *1g/mol = 185.845 g/mol

• The moles of Fe contained in 25 mg (or 0.025 g) of ferrocene are:

`0.025gFerrocene*(1molFerrocene)/(185.845g) *(1mol Fe)/(1molFerrocene)=1.345*10^(-4) molFe`

• The final volume is 500 mL, or 0.5 L. So the iron concentration in mol/L is:

`(1.345*10^(-4)molFe)/(0.5L)= 2.69*10^(-4) mol/L`

• We can convert that value into mg/L:

`2.69*10^(-4) (molFe)/(L) *(55.845g)/(1molFe)*(1000mg)/(1g)=15.023 mg/L`