An Erlenmeyer flask from the previous experiment containing the solution of the

Question

An Erlenmeyer flask from the previous experiment containing the solution of the iron-oxalate complex after the titration with potassium permanganate was retained. The mass of the iron-oxalate complex used in this titration was 0.1416 grams.

The solution was made up to 250 mL exactly, and then 20 mL of the solution was ultimately diluted to exactly 100 mL. Thiocyanate was added to produce a deep red color and the absorbance of the solution at 450 nm was measured.

According to the calibration curve, the absorbance corresponded to a concentration of Fe3+ equal to 2.306E-4 M.

What is the concentration of Fe3+ in the 250 mL flask?

Concentration of Fe3+ in the 250 mL flask = M

What is the number of moles, and therefore the mass, of Fe3+ in the flask?

Moles of Fe3+ in the 250 mL flask = mol

Mass of Fe3+ in the 250 mL flask = gram

Calculate the mass percent of iron in the complex.

Percent iron in the complex = %

A 0.4619 gram sample of the iron oxalate complex was heated in an oven at about 75oC for 60 minutes. Upon cooling and reweighing the sample, it was determined that the mass lost was0.0497 grams.

What is the percent by mass of water in the iron-oxalate complex?

Percent water in the complex = %

If the percent by mass of oxalate in the complex had previously been determined to be 54.44%, what is the percent by mass of potassium in the complex?

Percent potassium in the complex = %

PLEASE SHOW ALL WORK

Explanation / Answer

=(2.610e-4 mol Fe+3 / 1 L diluted sol'n) (100 mL diluted sol'n / 20 mL undiluted sol'n)
= 13.05e-4 mol Fe+3 / 1 L undiluted solution

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n = (13.05e-4 mol Fe+3 / 1 L undiluted solution) (0.25 L undiluted solution)
= 3.2625e-4 mol Fe+3

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m = (3.2625e-4 mol Fe+3) (MW(Fe) g/mol)

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The fraction of iron in the complex is (mass of iron) / (mass of complex)
= [m / (0.1603 g)] (100% / 1)

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If all that was lost was water, then
% water = (0.0614 g) / (0.5546 g) (100% / 1)

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Suppose you have 100 g of complex. Calculate mass of oxalate and then moles of oxalate

(100 g complex) (0.53 g oxalate / 1 g complex) = 53 g oxalate

x (1 mol oxalate / 88 g oxalate) = 0.60 mol oxalate

Next use stoichiometry to calculate moles of potassium, and then use MW to calculate mass:

(0.60 mol oxalate) (3 mol K+ / 3 mol ox) = 0.60 mol K+
x (39 g K / 1 mol K) = 23.4 g K

and finally the concentration (wt/wt) of potassium is (mass of potassium) / (mass of complex)

= (23.4 g K) / (100 g complex) (100% / 1)

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