How to determine the iron content in spinach with a spectrometer?

I used a spectrometer to find the absorbency of 5 solutions with different iron concentrations.


\begin{array}{rrrrr}
 \ce{Fe (III)} / \pu{mol L-1} & 
 \text{Absorbance} & 
 V(\ce{HCl}) &
 V(\ce{FeCl3}) &
 V(\ce{KSCN}) \\
 \hline
 0.00005 & 0.194 & \pu{9 mL} & \pu{1 mL} & \pu{10 mL} \\
 0.0001 & 0.424 & \pu{8 mL} & \pu{2 mL} & \pu{10 mL} \\
 0.00015 & 0.674 & \pu{7 mL} & \pu{3 mL} & \pu{10 mL} \\
 0.0002 & 0.89 & \pu{6 mL} & \pu{4 mL} & \pu{10 mL} \\
 0.00025 & 1.113 & \pu{5 mL} & \pu{5 mL} & \pu{10 mL} 
\end{array}


With this data I've constructed a graph and a line of best fit.
\begin{align}
y &= 4608x - 0.0322 &
R^2 &= 0.9994
\end{align}


I took $\pu{5.10g}$ of spinach leafs and burned them, then added $\pu{10.00 mL}$ of $\ce{HCl}$ and $\pu{10 mL}$ of $\ce{KSCN}$ then filtered the contents (so ashes weren't in the beaker).


The absorbency of this spinach extract was $0.015$.


My question is: How do I calculate how much iron is in this sample? I need to determine the mass of iron per $\pu{100 g}$ of spinach.


Using the line of best fit question, solve for $V(\ce{Fe(III)}) = x$ concentration:
\begin{align}
0.015 &= 4608x - 0.0322\\
x &= 0.0000102431
\end{align}


Convert $\pu{mol/L}$ to $\pu{g/L}$ to $\pu{mg/L}$:


\begin{align}
0.0000102431 \times 55.845 &= 0.00057202591\\
\pu{0.00057202591 g} &= \pu{0.572 mg}
\end{align}


But this is clearly wrong because it's not taking account the amount of spinach I started with ($\pu{5.10 g}$). Anytime I try to include the amount, either by multiplying the density my answer is way off.


Is it reasonable that I'm getting a number above $\pu{10 mg}$? The iron amount reported by the USDA says that there is $\pu{2.7 mg}$ of iron per $\pu{100 g}$ of spinach.


Can anyone guide me in the correct direction?




Here's how I approached the problem.


Plot the Data
=============


Below is the plot I quickly hacked together, and a fit of the form $g(x) = a\cdot x$. (Note that I do want the line to go through (0,0) because the absorption at zero concentration should be zero. And it's a pity you didn't check that with a measurement... One never knows whether the calibration was successful.)


![Plot of the data given by OP with a linear regression through (0,0)](https://i.stack.imgur.com/j8A4v.png)


The parameter found was $a=4.43236~\mathrm{L\, mmol^{-1}} = 4.43236~\mathrm{mM^{-1}}$.


Calculate the Concentration of Fe Ions Present
==============================================


You have measured an absorbance of 0.015 with a corresponding concentration of $c\_\text{meas} = 3.38\cdot 10^{-3} ~ \mathrm{mmol\, L^{-1}}$.


The mass of iron in the solution is given by
$$ m\_\ce{Fe} = c\_\text{meas} \cdot M\_\ce{Fe} \cdot V = 0.00378~\mathrm{mg} $$
with the volume $V=0.02~\mathrm{L}$ and molar mass $M\_\text{Fe} = 55.845~\mathrm{g\, mol^{-1}}$.


Calculate Mass Percentage
=========================


The mass of iron in one gram of spinach is easily calculated:
$$ w\_\text{Fe,spinach} = \frac{0.00378~\mathrm{mg}}{5.10~\mathrm{g}} = 7.4 \cdot 10^{-4} ~\mathrm{mg\, g^{-1}} $$


(I could just calculate away the units, but I was too lazy to do it here.)


The mass in $100~\mathrm{g}$ of spinach follows accordingly:
$$ m\_\text{Fe,100g} = 7.4 \cdot 10^{-4}~\mathrm{mg\, g^{-1}} \cdot 100~\mathrm{g} = 0.0740~\mathrm{mg} $$


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