Mass of chromium in ammonium dichromate What mass of chromium is in $\pu{100 mL}$ of $\pu{0.0400 mol/L}$ $\ce{(NH4)2Cr2O7}$?


The specific concept is how to find the mass of a certain ion/element in a substance, given the concentration. I am at a total loss to this.

At first I tried to find the molar mass of the chromium itself, which resulted in roughly $\pu{104 g/mol}$. I then tried to find the amount of moles in the concentration, which is 

$$\pu{0.1 L} \cdot \pu{0.04 mol/L} = \pu{0.004 mol}$$

Then I multiplied the molar mass of chromium from the amount of moles in the concentration, and I got $\pu{0.208 g}$.

This is not the answer. The answer was $\pu{0.800 g}$

I then tried a different approach. I tried to find the molar mass of my original substance, then divided the molar mass of chromium from the molar mass of ammonium chromate, which will give me the amount of chromium per ammonium chromate. Then I thought all I needed to do was to multiply it by the concentration. This just messed up everything and I was back to square one.

How do I approach this problem?
To find the mass of chromium, one needs to know the amount $n$ ($m = nM$). Each formula unit of ammonium dichromate $\ce{(NH4)2Cr2O7}$ contains $2$ $\ce{Cr}$, therefore

$$n(\ce{Cr}) = 2 \cdot n(\ce{(NH4)2Cr2O7})$$

The amount of ammonium dichromate can be found from its concentration (as you correctly did):

$$n(\ce{(NH4)2Cr2O7}) = C(\ce{(NH4)2Cr2O7}) \cdot V$$

Putting it all together, the final expression for the mass of chromium is

\begin{align}
m(\ce{Cr}) &amp;= 2 \cdot C(\ce{(NH4)2Cr2O7}) \cdot V \cdot M(\ce{Cr}) \\
&amp;= 2 \cdot \pu{0.04 mol L-1} \cdot \pu{0.1 L} \cdot \pu{51.996 g mol-1} \\
&amp;= \pu{0.416 g}
\end{align}

The answer deviates with the one provided by the textbook; however, if you were asked to find the mass of chromium(VI) oxide $\ce{CrO3}$ instead, it fits perfectly:

\begin{align}
m(\ce{CrO3}) &amp;= 2 \cdot C(\ce{(NH4)2Cr2O7}) \cdot V \cdot M(\ce{CrO3}) \\
&amp;= 2 \cdot \pu{0.04 mol L-1} \cdot \pu{0.1 L} \cdot \pu{99.99 g mol-1} \\
&amp;= \pu{0.80 g}
\end{align}

I deduced $\ce{CrO3}$ by backtracking to the molecular weight of the compound, which should be around $\pu{100 g mol-1}$, and among common chromium substances this is the only one that comes to mind.