What is the amount of nitrate ions in $20.0\:\mathrm{g}$ of $\ce{Fe(NO3)3}$?

The chemical formula for the nitrate ion is $\ce{NO3}$, I think.

The molar mass of $\ce{Fe(NO3)3}$ is $242~\mathrm{g~mol^{-1}}$ of which $186\ \mathrm{g\ mol^{-1}}$ belong to $\ce{NO3}$. That means that around $76.85\:\%$ of the substance is nitrate ion.

$20.0\ \mathrm g$ of the substance are equivalent to $20.0\ \mathrm g/242\ \mathrm{g\ mol^{-1}} = 0.0826~\mathrm{mol}$ of $\ce{Fe(NO3)3}$.

Since roughly $76.85\:\%$ are nitrate ions, there are about $0.06~\mathrm{mol}$ of nitrate ions in those $20\ \mathrm g$ of substance.

One mole is $6.02\times 10^{23}$, so if I multiply $(0.06\ \mathrm{mol})(6.02\times 10^{23}\ \mathrm{mol^{-1}}) = 3.612\times10^{22}$.

My Answer: There are $3.612\times10^{22}$ nitrate ions in $20.0\:\mathrm{g}$ of $\ce{Fe(NO3)3}$.

However, that is wrong. The options in the website are:

  • $1.49\times 10^{23}$
  • $4.98\times 10^{21}$
  • $60.0$
  • $8.25\times10^{21}$

And my answer is not even close to any of them.

What did I do wrong?


1 Answer 1


The molar mass of iron (III) nitrate is $m(\ce{Fe(NO3)3}\approx 242~\mathrm{g\, mol^{-1}}$. Pay close attention to the unit.

You correctly calculated the amount of substance of iron (III) nitrate to be $n(\ce{Fe(NO3)3}= 0.0826~\mathrm{mol}$

Now you should ask yourself the question: How many nitrate ions are in one formula unit of iron (III) nitrate?

There are three $\ce{NO3^-}$ per every $\ce{Fe(NO3)3}$.

What does that mean for the number of moles of nitrate ions?

It means that $n(\ce{Fe(NO3)3} = \frac13\cdot n(\ce{NO3^-})$, so you have to multiply the number of moles by three. $n(\ce{NO3^-}) = 0.248~\mathrm{mol}$

Now you know the number of moles of nitrate ions and you simply have to multiply with Avogadro's constant.

$N(\ce{NO3^-}) = n(\ce{NO3^-}) \cdot L = 1.49\cdot10^{23}$


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