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} &amp; 
  \text{Absorbance} &amp; 
  V(\ce{HCl}) &amp;
  V(\ce{FeCl3}) &amp;
  V(\ce{KSCN}) \\
  \hline
  0.00005 &amp; 0.194 &amp; \pu{9 mL} &amp; \pu{1 mL} &amp; \pu{10 mL} \\
  0.0001  &amp; 0.424 &amp; \pu{8 mL} &amp; \pu{2 mL} &amp; \pu{10 mL} \\
  0.00015 &amp; 0.674 &amp; \pu{7 mL} &amp; \pu{3 mL} &amp; \pu{10 mL} \\
  0.0002  &amp; 0.89  &amp; \pu{6 mL} &amp; \pu{4 mL} &amp; \pu{10 mL} \\
  0.00025 &amp; 1.113 &amp; \pu{5 mL} &amp; \pu{5 mL} &amp; \pu{10 mL} 
\end{array}

With this data I've constructed a graph and a line of best fit.
\begin{align}
y &amp;= 4608x - 0.0322 &amp;
R^2 &amp;= 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 &amp;= 4608x - 0.0322\\
x &amp;= 0.0000102431
\end{align}

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

\begin{align}
0.0000102431 \times 55.845 &amp;= 0.00057202591\\
\pu{0.00057202591 g} &amp;= \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.)



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:
$$ m_\text{Fe,100g} = 7.4 \cdot 10^{-4}~\mathrm{mg\, g^{-1}} \cdot 100~\mathrm{g} = 0.0740~\mathrm{mg} $$