a vial containing a sample of finely powdered iron ore is found to weigh 9.8117

Question

a vial containing a sample of finely powdered iron ore is found to weigh 9.8117 g. After most of the ore sample was poured into a beaker, the nearly empty vial weighed 6 9.6075 g. The ore is digested in HCl and diluted to 100.0 mL. A 25.00 mL aliquot of this sample solution is diluted to 1000.0 mL using volumetric glassware. Then, a 5.000 mL aliquot of this diluted sample solution is transferred to a 100.0 mL volumetric flask, buffered to pH 3.5, reduced with H2Q, complexed with phen, and diluted to the mark. The absorbance of this unknown test solution is read, and a calibration curve is used to determine its concentration to be 0.87472 mg/L. What is the wt% of iron in the ore?

[Iron ore: 34.269% Fe]

Please show the steps to get to the answer listed above.

Explanation / Answer

Concentration of Fe in the diluted phen complex = 0.87472 mg/L;

The dilution factor involved in this step is (5.00 mL)/(100.00 mL) = 1/20 and hence the concentration of Fe in the 5.00 mL aliquot is (0.87472 mg/L)*20 = 17.4944 mg/L.

25.00 mL sample was diluted to 1000.00 mL so that the dilution factor involved = (25.00 mL)/(1000.00 mL) = 1/40.

Therefore, concentration of Fe in the 25.00 mL aliquot = (17.4944 mg/L)*40 = 699.776 mg/L.

We have 100 mL solution so that the mass of Fe in the sample = (699.776 mg/L)*(100 mL)*(1 L/1000 mL) = 69.9776 mg = (69.9776 mg)*(1 g/1000 mg) = 0.0699776 g.

Mass of the sample taken = (9.8117 – 9.6075) g = 0.2042 g.

Percent of Fe in the sample = (0.069976 g)/(0.2042 g)*100 = 34.269% (ans).

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