# What is the amount of nitrate ions in 20 g of Fe(NO3)3?

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?

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.

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