Colorimetric iron determination in fortified cereal
To determine the iron content of a fortified breakfast cereal, a student grinds 20.0 g of cereal, digests it in acid, and makes the solution up to 100.0 cm³. A 10.0 cm³ aliquot is treated with excess potassium thiocyanate and made up to 50.0 cm³; the red complex absorbs at 480 nm. Calibration standards, treated identically, give absorbances: 0.00 mg dm⁻³, 0.000; 2.00, 0.175; 4.00, 0.350; 6.00, 0.610; 8.00, 0.700; 10.00, 0.875. The cereal sample gives absorbance 0.440. The packet states 14 mg iron per 100 g. A second student notices that the 6.00 mg dm⁻³ cuvette had a fingerprint on it and wants to discard that reading. The blank shows a small absorbance of 0.020 when measured separately, but the student did not subtract it.
1. Construct a calibration line from the data. Decide, with justification, whether the 6.00 mg dm⁻³ point should be rejected, and use the blank-corrected sample absorbance to find the iron concentration in the measured solution.
Read the worked answer
Check the increments in absorbance per 2.00 mg dm⁻³: 0–2: 0.175 2–4: 0.175 4–6: 0.260 6–8: 0.090 8–10: 0.175
The first, second and last intervals are consistent with a slope of about 0.0875 per mg dm⁻³. The 6.00 point lies well above the trend: its expected absorbance is about 0.525, but it reads 0.610. A fingerprint on the cuvette would scatter light and increase the apparent absorbance, which is the right direction to explain the high reading. There is a physical reason to reject it, and the remaining five points are linear.
Using 2.00 and 10.00 mg dm⁻³: slope = (0.875 – 0.175)/(10.00 – 2.00) = 0.700/8.00 = 0.0875 per mg dm⁻³. Intercept: 0.175 – 0.0875 × 2.00 = 0.000.
Sample absorbance after blank correction: A = 0.440 – 0.020 = 0.420. Concentration = 0.420/0.0875 = 4.80 mg dm⁻³.
Without blank correction the value would be 0.440/0.0875 = 5.03 mg dm⁻³, about 5% higher. The blank must be subtracted.
2. Calculate the iron content in mg per 100 g of cereal and compare it with the label value of 14 mg per 100 g. Decide whether the result shows that digestion was complete, and suggest one check.
Read the worked answer
The measured solution volume is 50.0 cm³ = 0.0500 dm³. Mass of iron in the measured solution: m = cV = 4.80 × 0.0500 = 0.240 mg.
This solution came from a 10.0 cm³ aliquot of the 100.0 cm³ digest. The dilution factor from digest to measured solution is 100.0/10.0 = 10, so the digest contains: 0.240 × 10 = 2.40 mg Fe.
This came from 20.0 g of cereal, so per 100 g: 2.40 × (100/20.0) = 12.0 mg Fe per 100 g.
The label says 14 mg per 100 g. The experimental value is about 86% of the label value, which is plausible but on the low side. It does not prove that digestion was complete; incomplete extraction, loss during transfer, or matrix effects in the cereal digest could all lower the result. Nor is the result absurdly high, so gross contamination is unlikely. A good check is a spike-recovery test: add a known amount of iron standard to a separate portion of cereal before digestion and see whether the measured increase matches the added amount. Repeating the digestion with a more vigorous acid treatment, or using a certified reference material, would also test completeness.
3. Explain why the absorbance is measured at 480 nm rather than at 650 nm, and why excess thiocyanate is added. State one reason the blank absorbance must be subtracted.
Read the worked answer
The iron(III)–thiocyanate complex is red. A red solution absorbs blue-green light and transmits red light. Its absorption maximum is in the blue-green region near 480 nm, so measuring there gives a large absorbance change for a small change in iron concentration. At 650 nm, in the red part of the spectrum, the complex absorbs very little because red light is transmitted; the sensitivity would be poor and the calibration line shallow.
Thiocyanate is added in excess to push the complex-formation equilibrium towards the product: Fe³⁺ + SCN⁻ ⇌ [Fe(SCN)]²⁺ Excess ligand helps ensure that essentially all the iron is present as the coloured complex, so the absorbance is proportional to the total iron concentration rather than to only a fraction of it.
The blank contains the reagents and cuvette but no added iron. It may absorb slightly because of impurities, turbidity or the cuvette itself. Subtracting the blank removes this constant contribution; otherwise every reading, including the sample, would be too high. The blank value is also needed to check that the calibration line passes close to the origin.