# 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?

• I guess that was because you took the concentration of iron in spinach from the absorbance graph by extrapolating substantially. You did an absorbance graph for 0.194 - 1.113 whereas your absorbency for spinach is only 0.015. So there may be some potential for errors there. – t.c Oct 14 '14 at 17:54

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