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

Answer:

1.345*10⁻⁴ mol/L

15.023 mg/L

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:

$0.025gFerrocene*\frac{1molFerrocene}{185.845g} *\frac{1mol Fe}{1molFerrocene}=1.345*10^{-4} molFe$

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

$\frac{1.345*10^{-4}molFe}{0.5L}= 2.69*10^{-4} mol/L$

• We can convert that value into mg/L:

$2.69*10^{-4} \frac{molFe}{L} *\frac{55.845g}{1molFe}*\frac{1000mg}{1g}=15.023 mg/L$