2 Part Question: An analytical chemist is given 1.546 g of a solid coordination

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2 Part Question:

An analytical chemist is given 1.546 g of a solid coordination complex that definitely contains a Cox(C2O4)y complex (with the Co in the Co2+ oxidation state) and Li counter-ions, and possibly also water. His mission is to determine its correct empirical formula.

Part I) He takes the Cox(C2O4)y solid and chips off a 0.5103 g solid sample. He manages to dissolve this sample into a solution of about 150 mL of 2 M sulfuric acid. He then titrates a 0.101 M solution of K2Cr2O7 into this Cox(C2O4)y analyte solution. The analyte solution remains colorless until the endpoint, at which point it turns yellow. The volume of 0.101 M K2Cr2O7 solution titrated in this experiment is 11.00 mL. What is the weight % of oxalate in the original sample and what is the mmol of oxalate/100.0 g of complex?

Remember to first balance the following equation:
_ K2Cr2O7 (aq) + _ H2C2O4 (aq) + _ H+(aq) _ Cr3+(aq) + _ CO2 (g) + _ H2O(l) + _ K+(aq)

a) 7.67 wt% & 87.1 mmol b) 47.9 wt% & 544 mmol c) 19.2 wt% & 218 mmol d) 57.5 wt% & 654 mmol e) 6.39 wt% & 72.6 mmol

Part II) After thinking it over, the analytical chemist decides that a precipitation would be more accurate (and easier) than spectrophotometry. Hence, he dissolves a 0.5168 g sample of Cox(C2O4)y in 100.0 mL of aqueous solution and makes sure the Co2+ is freed from the Cox(C2O4)y with the addition of acid. He then adds an excess of ammonium phosphate, (NH4)3PO4, resulting in the precipitation of a red solid powder. Once completely dried, the mass of the red solid powder of Co3(PO4)2 is measured to be 0.1356 g. Determine the amount of Co2+ in the original sample.

From the determination of the amount of Co2+ in the original Cox(C2O4)y complex and your answer for C2O42- in the previous question, which one of the following answers appears to be the full empirical formula?

a) Li6[Co2(C2O4)6]·13H2O b) Li6[Co2(C2O4)5]·18H2O c) Li3[Co(C2O4)3]·6H2O d) Li16[Co(C2O4)9]·24H2O e) Li4[Co(C2O4)3]·6H2O

Explanation / Answer

We need the balanced equation between Cr2O7 2- and C2O4 2-. First balance each half-reaction.

Cr2O7 2- ==> 2Cr3+ . . .add 7 H2O to right to balance O.
Cr2O7 2- ==> 2Cr3+ + 7H2O . . .add 14H+ to the left to balance H
Cr2O7 2- + 14H+ ==> 2Cr3+ + 7H2O . . .add 6e- to the left to balance the charge
Cr2O7 2- + 14H+ + 6e- ==> 2Cr3+ + 7H2O


C2O4 2- ==> 2CO2 . . .add 2e- to the right to balance the charge.
C2O4 2- ==> 2CO2 + 2e- . . .multiply this entire equation by 3 to give us 6e- on the right. Then add it to the first equation.

. .3C2O4 2- ==> 6CO2 + 6e-
+ Cr2O7 2- + 14H+ + 6e- ==> 2Cr3+ + 7H2O
======================================...
. .3C2O4 2- + Cr2O7 2- + 14H+ ==> 6CO2 + 2Cr3+ + 7H2O

Now we know that it takes 3 moles of C2O4 2- to react with 1 mole of Cr2O7 2-.

moles Cr2O7 2- = M Cr2O7 2- x L Cr2O7 2- = (0.101)(0.01100) = 0.00111 moles Cr2O7 2-

0.00111 moles Cr2O7 2- x (3 moles C2O4 2- / 1 mole Cr2O7 2-) = 0.00333 moles C2O4 2-

0.00333 moles C2O4 2- x (88.0 g C2O4 2- / 1 mole C2O4 2-) = 0.293 g C2O4 2-

% C2O4 2- = (g C2O4 2- / g sample) x 100 = (0.293 / 0.5103 g) x 100 = 57.4%


(0.00333 moles C2O4 2- / 0.5168 g complex) x (1000 mmoles / 1 mole) = 6.44 mmoles C2O4 2- / g complex = 644 mmoles C2O4 2- / 100 g complex

CHOICE D is correct.
======================================...


3Co2+(aq) + 2PO4 3-(aq)==> Co3(PO4)2(s)

0.1356 g Co3(PO4)2 x (1 mole Co3(PO4)2 / 366.7 g Co3(PO4)2) = 0.0003698 moles Co3(PO4)2

0.0003698 moles Co3(PO4)2 x (3 moles Co2+ / 1 mole Co3(PO4)2) = 0.001109 moles Co2+

0.001109 moles Co2+ x (58.93 g Co2+ / 1 mole Co2+) = 0.06538 g Co2+

(0.06538 g Co2+ / 0.5168 g sample) x 100 = 12.65% Co2+

Then the complex contains 12.65% Co2+ and (from Part 1) 57.4% C2O4 2-.

In 100 g of complex there are 12.65 g Co2+ and 57.4 g of C2O4 2-. Convert those to moles.

12.65 g Co2+ x (1 mole Co2+ / 58.93 g Co2+) = 0.215 moles Co2+
57.4 g C2O4 2- x (1 mole C2O4 2- / 88.02 g C2O4 2-) = 0.652 moles C2O4 2-

That gives us a C2O4 2- / Co2+ mole ratio of 3:1. The empirical formula of the complex ion is
Co(C2O4)3 4-. If the counterion is lithium (+1 charge), there must be 4 of them to balance the -4 charge on the complex ion. So our empirical formula is now Li4Co(C2O4)3.

I'm going with choice E

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