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?