A 103.7 mg ore sample containing iron is digested with acid, filtered and dilute

Question

A 103.7 mg ore sample containing iron is digested with acid, filtered and diluted to 200.0 mL in a volumetric flask. A 10.00 mL aliquot is transferred to a 100.00 mL volumetric flask containing hydroxylamine hydrochloride, sodium acetate and diluted to the mark. The resulting orange-red solution had an absorbance of 0.426 at 508 nm in a 1 cm cell. If the molar absorptivity of the iron-phenanthroline complex at this wavelength is 1.89 x 10^3 L/mol. cm, determine the percentage of iron in the ore sample.

Explanation / Answer

103.7 mg (ore)---(Digested)-------->200 mL (Std Flask)----->10mL-------->100mL (added reagents)

A = 0.426

= 1.89 x 103 L mol-1 cm-1

l = 1 cm

A = cl

c = A/cl = 0.426/1.89 x 103 L mol-1 cm-1 x 1 cm

c = 2.25x10-4 mol/L

Amount in 100 mL = 2.25x10-4 mol/L x 0.1L = 2.25x10-5 mol

10 mL of solution contains 2.25x10-5 mol

200 mL contains 2.25x10-5 mol x 20 = 4.5x10-4 mol

So amount of iron present is 4.5x10-4 mol x 55.84g/mol = 251.28x10-4 g = 25.128 mg

103.7 mg of ore containes 25.128 mg of Fe, so the percentage of Fe in the ore is

25.125x100/103.7 = 24.22%

so the percentage of Fe in the ore is 24.22%

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