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

Fe (III) concentration (in mol/L)   Absorbance  Ingredients
0.00005                             0.194      9mL HCl + 1mL FeCl3 + 10mL KSCN
0.0001                              0.424      8mL HCl + 2mL FeCl3 + 10mL KSCN
0.00015                             0.674      7mL HCl + 3mL FeCl3 + 10mL KSCN
0.0002                              0.89       6mL HCL + 4mL FeCl3 + 10mL KSCN
0.00025                             1.113      5mL HCL + 5mL FeCl3 + 10mL KSCN

With this data I've constructed a graph and a line of best fit.

y = 4608x - 0.0322
R^2 = 0.9994

I took 5.10g of spinach leafs and burned them, then added 10.00mL of HCl and 10mL of 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 100g of spinach.

My attempts:

Using the line of best fit question, solve for Fe(III) concentration:

0.015 = 4608x - 0.0322
x = 0.0000102431

Convert mol/L to g/L to mg/L

0.0000102431 * 55.845 = 

0.00057202591 0.00057202591g = 0.572mg

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

Can anyone guide me in the correct direction?

  • $\begingroup$ Determine the mass of iron per 100g of spinach, so 0.910048mg * 100/5.10. $\endgroup$ – t.c Oct 14 '14 at 17:47
  • $\begingroup$ Molar mass of iron is 56g/mol by the way. $\endgroup$ – t.c Oct 14 '14 at 17:48
  • $\begingroup$ Is it reasonable that I'm getting a number above 10mg? The iron amount reported by the USDA says that there is 2.7mg of iron per 100g of spinach. $\endgroup$ – Dave Chen Oct 14 '14 at 17:52
  • 2
    $\begingroup$ 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. $\endgroup$ – t.c Oct 14 '14 at 17:54
  • $\begingroup$ But if I have a small [], shouldn't I get a very small amount of iron? $\endgroup$ – Dave Chen Oct 14 '14 at 17:56

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)

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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