Asnwer the following question in bold: If you weigh out exactly 0.150 g of the u

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Asnwer the following question in bold:

If you weigh out exactly 0.150 g of the unknown KwFex(C2O4)y(H2O)z to be dissolved in the 100 mL volumetric flask, and the concentration of Fe(III) in the 50 mL volumetric flask is calculated to be 1.27 x 10-4 M, what is the mass (in grams) of iron that you would have in a 100.0 g sample of your unknown

Part III. Measurement of the Amount of Iron in the KwFeCO4(H20), Coordination Compound Separate the Iron(III) from the Coordination Compound. Accurately measure about 0.15 g of the coordination compound synthesized earlier in these series of experiments. Transfer the sample to a 100 mL volumetric flask. Add 5 mL of 6 M H2SO4 (Caution: avoid skin contact) and 20 mL of 0.5 M CaCl. Swirl the solution. A mixed precipitate of CaC204 and CaSO4 forms. Dilute the mixture to the mark with distilled water Invert the volumetric flask several times, and allow the mixture to stand so that the precipitate will settle. Do not disturb Prepare the Stock Unknown Iron(II) Solution. Add about 15 mL of distilled water to a 50 mL volumetric flask and add a small scoop of ascorbic acid. Without disturbing the precipitate in the 100 mL volumetric flask from the previous step use a 10 mL graduated pipet and carefully withdraw a little more than 2 mL of the solution. Dispense exactly 2.00 mL of the solution into 6 CHM 31 the 50 mL volumetric flask. Then add, with a graduated cylinder, about 20 mL of 0.0020 M bipyridyl solution and dilute to the mark with the buffer solution (pH 4.7) Determine the Amount of Iron(II) in the Unknown Test Solution. Prepare a set of five unknown test solutions in the same manner as for the standard solution. Determine the Absorbance (A) of these five unknown test solutions. At least three of the unknown test solutions should have A values that lie with the range of the A values for the set of standard solutions used to generate the Beer's Law plot. If the data for your unknowns does not meet this criteria prepare additional unknown test solutions: vary the initial volumes of the stock unknown solution and of water as needed Concentration of the tris-Bipyridyliron(lI) Complex in Unknown Test Solutions. Read the Beer's law graph to determine the molar concentration of tris-bipyridyliron II) complex in each of your unknown test solutions (remember, at least three solutions must have absorbance values that lie within the range of the Beer's law plot of the standard solutions) Calculate the Iron(III) in the Original Unknown KFe(C20 (H2O) Solution. Use the molar concentrations of the tris-bipyridyliron(II) complex for at least three of the test solutions to back- calculate for the molar concentration of iron(II) in the original solution that contained the coordination compound. In the calculation consider the two dilutions to prepare the original solution: a 2.00 mL to 50.0 mL dilution to prepare the stock unknown iron(II) solution and further dilution of the stock solution for each of the unknown test solutions to a volume of 10.00 mL Calculate the Amount of Iron in the KFe(C,O(HO) Coordination Compound. Calculate the number of moles of iron in the measured mass of the coordination compound used to make the stock unknown solution. Calculate both the mass and number of moles of iron that would be present in 100 g of the compound.

Explanation / Answer

Atomic mass of iron = 55.845 g/mol.

The molarity of iron (as Fe(III)) in the sample is 1.27*10-4 M. The volume of the sample = 50 mL.

Therefore, mole(s) of iron in the 50 mL sample = (volume of sample in L)*(molarity of iron) = (50 mL)*(1 L/1000 mL)*(1.27*10-4 M) = 6.35*10-6 mole.

Since the co-ordination compound was originally dissolved in 100 mL volumetric flask, hence, the number of mole(s) of iron in the sample = (6.35*10-6 mole)*(100 mL/50 mL) = 1.27*10-5 mole.

0.150 g of the compound contains 1.27*10-5 mole iron; therefore, 100.0 g of the sample will contain (1.27*10-5 mole)*(100.0 g/0.150 g) = 8.467*10-3 mole iron

Mass of iron in 100.0 g sample = (8.467*10-3 mole)*(55.845 g/mol) = 0.4728 g ? 0.473 g (ans).

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