Vincent Yee // 01 November 2006 // Chemistry 111
Experiment 15CD : Synthesis and Analysis of a Compound of Iron
Introduction
This experiment was a continuation of the experiment 15, parts A and B in which an unknown compound of iron was analyzed in order to determine the empirical formula of the chemical reaction tested. Part C focused on determining the percent oxalate in the unknown solution. In order to determine the oxalate, a titration of the oxalate with a standard solutino of potassium permaganate was prepared.
Part C used three trials for experimental accuracy in which a standard of permaganate was titrated into a solution of sodium oxalate. The amount of permaganate titrated into the sodium oxalate in order to turn it just barely purple was used to compute the molarity of the permaganate. From the molarity of the permaganate, the mass of initial unknown iron compound was used to compute the percent Oxalate in the unknown.
Part D focused on the percent water in the composition. The mass of the vial and the unknown iron sample were weighed and then heated for 90 minutes. The difference of the masses is determined to be the mass of water that evaporated. From the mass of the water, the percent water of the unknown was computed.
From the percent oxalate and the percent water, combined with the percent iron, the percent potassium could be determined by difference. Then from the four percentages, an empiracal formula was determined by difference analysis. Mole ratios were used to determine the empirical formula.
Using the obtained formula, the balanced chemical equation was determined.
Data
Parte CExp. / Na2C2O4(g) / Flask(g) / BuretInitial(mL) / Buretfinal(mL) / Difference
1 / 0.204 / 120.72 / 0.5 / 31.9 / 31.4
2 / 0.217 / 122.461 / 1.1 / 34.2 / 33.1
3 / 0.208 / 102.328 / 1.2 / 32.3 / 31.1
Exp. / Unknown Iron(g) / BuretInitial(mL) / Buretfinal(mL) / Difference
1 / 0.185 / 1.8 / 24.9 / 23.1
2 / 0.2 / 10.4 / 35.3 / 24.9
3 / 0.208 / 0.2 / 25.6 / 25.4
Parte D
Exp / InitialVial(g) / InitialVial + Unk(g) / FinalVial + Heat(g) / Mass of sample
JV1 / 13.082 / 14.7 / 14.329 / 1.618
JV2 / 13.116 / 14.68 / 14.515 / 1.564
JV3 / 13.359 / 15.02 / 14.838 / 1.661
Calculations
In part C, the amount of permaganate titrated into the sodium oxalate was determined by taking the difference in the final and initial reading. From the mass of the sodium oxalate sample, the molarity of the permaganate was determined.
Exp10.204 Na2C2O4(g) x 1 mole Na2C2O4 x 1 mole H2C2O4 x 1 mole MnO4 x
134 g Na2C2O4 1 moleNa2C2O4 1 mole H2C2O4
1 = Molarity of permaganate
Volume of KMnO4 Delivered
Exp. / Na2C2O4(g) / Na2C2O4(moles) / H2C2O4(moles) / MnO4(moles) / MolarityKMnO41 / 0.204 / 0.001522388 / 0.001522388 / 0.001522388 / 0.048483696
2 / 0.217 / 0.001619403 / 0.001619403 / 0.001619403 / 0.048924561
3 / 0.208 / 0.001552239 / 0.001552239 / 0.001552239 / 0.049911216
The average molarity was calculated as 0.049106491.
From the average molarity, the unknown iron sample was analyzed:
Exp 1 0.0231 L KMnO4 x Molarity KMnO4 x 1 mole H2C2O4 x 1 mole C2O4 x
1 mole MnO4- 1 mole H2C2O4
88.0g C2O4 x 1 x 100 = percent oxalate.
1 mole C2O4 mass of sample (.185g)
Exp. / Unknown Iron(g) / BuretInitial(mL) / Buretfinal(mL) / Difference1 / 0.185 / 1.8 / 24.9 / 23.1
2 / 0.2 / 10.4 / 35.3 / 24.9
3 / 0.208 / 0.2 / 25.6 / 25.4
KMnO4(L) / KMnO4(moles) / H2C2O4(moles) / C2O4(moles) / C2O4(g) / percent / %Oxalate
0.0231 / 0.00113436 / 0.00113436 / 0.00113436 / 0.099823675 / 0.539587433 / 53.95874329
0.0249 / 0.001222752 / 0.001222752 / 0.001222752 / 0.107602143 / 0.538010716 / 53.80107164
0.0254 / 0.001247305 / 0.001247305 / 0.001247305 / 0.109762829 / 0.527705908 / 52.77059081
The average percent oxalate from the three trials was 53.5%
Moving on from the percent oxalate analysis, the experiment determined the percent water. From the difference of the final mass of the vial from the initial mass of the vial obtained the mass of the sample. The mass of water was determined by the difference of the initial vial mass before heat and the mass of the vial after the 90 minutes of heat.
Part DExpmnt / InitialVial(g) / InitialVial + Unk(g) / FinalVial + Heat(g) / Mass of sample / Difference
JV1 / 13.082 / 14.7 / 14.329 / 1.618 / 0.371
JV2 / 13.116 / 14.68 / 14.515 / 1.564 / 0.165
JV3 / 13.359 / 15.02 / 14.838 / 1.661 / 0.182
Then the percent water was calculated from the difference of the vial before and after heat divided by the mass of the sample. Multiplying by 100 gave the percent in proper form. The Average percent of water was determined by taking the average of the three trials. The deviation was determined by the differnce of the average percent value and the calculated percent water. The absolute value function was then applied to the deviatin to present a positive value. The Average deviation was an average of the three trials’ deviation. The Relative Average deviation was obtained from the product of the average deviation divided by the average percent water, which was then multiplied by 100 to be presented in traditional form.
water(x/100) / %water / Avg % water / Deviation / Avg Deviation / Rel Avg Dev / Rel Avg Dev %0.22929542 / 22.9295 / 14.81222 / 8.11731950 / 5.411546334 / 0.36534329 / 36.534328
0.10549872 / 10.5498 / 4.26235102
0.10957254 / 10.9572 / 3.85496847
Since the percent water has now been calculated to be 14.8 percent, the percent potassium could be calculated. The following formula was used:
%K = 100% - %Fe - %H2O - %Oxalate.
% Iron / %Oxalate / Avg % water / %potassium9.495 / 53.5101352 / 14.81222314 / 22.18264166
From part F, assuming the unknown had 100g, the following element fell into the following amounts in grams:
Iron / Oxalate / Water / Potassium9.495 / 53.5101352 / 14.81222314 / 22.18264166
The mole ratios resulted from the amount of grams divided by the molecular weight of the specific element:
Moles Fe = 9.495 g = 0.170mol Moles C2O4 = 53.5101g = 0.608mol
55.845 g/mol 88.018 g
Moles Water = 14.81222 g = 0.822mol Moles K = 22.1826 g = 0.567mol
18.0148 g 39.098 g
Using empirical data, the empirical formula was determined: K3Fe(C2O4)3 * 4 . H2O.
Using the iron element as a reference, the potassium and the oxalate had roughly three times the amount and water had rouhgly four times the amount, relative to the iron. Using that logic, the empirical formula was devised.
Using the empirical formula, a balanced equation can be found:
FeCl3(aq) + 3(K2C2O4)(aq) K3Fe(C2O4)3(s) * 4H2O(l) + 3KCl(aq)
From Experiment 15A the theoretical yield can be determined by the volume of the Fe(III)Cl used (15.6mL) and it’s molarity (1.50M).
Moles Fe(III)Cl = (.0156L) * (1.50M) = 0.0234 moles
The number of moles of K2C2O4 is determined by the mass of the sample : 17.610 g The molecular weight of K2C2O4 is 166.214 and the mole ratio is: 0.105. Since the Fe(III)Cl had a lower molar amount, the iron(III) cloride was the limiting reactant. Since one mole of iron(III) cloride was used to create one mole of the iron product, 509.2522 grams of product was theoretically produced. The percent yield of product is equal to:
[(Actual yield) / (Theoretical Yield)] *100
Actual Yield / Th. Yield / Act / Theo. / Percent yield509.2522 / 491.2374 / 1.036672289 / 103.6672289
Using the given formula in class: K3Fe(C2O4)3 * 3H2O, molar ratios were determined and used as theoretical values. Multiplying the number of moles by the molecular weight produced a number of grams. Adding all the calculated amounts in grams produced the molar ratios.
Molar amounts / Mol. Weight(g) / Molar amount(g) / Fraction of 1 / Molar Percent.3 mole K * / 39.098 / 117.294 / 0.238772536 / 23.87725365
1 mole Fe * / 55.845 / 55.845 / 0.113682305 / 11.36823051
3 mole C2O4 * / 88.018 / 264.054 / 0.537528291 / 53.75282908
3 mole H2O / 18.0148 / 54.0444 / 0.110016868 / 11.00168676
Total / 491.2374 / 100
Discussion
The theoretical values produced values that were relative to the actual produced values. Possibly due to experimental or calculated error, the number of moles “actually” created was one more than the theoretical value. Having four moles of water produced, rather than the theoretical three did skew the molar percentages. The deviation from the expected values was withing tolerable ranges, however. The produced values did not indicate a large mechanical error in the formula.
In theory, the actual yield should have been less than the theoretical yield, however, due to some mechanical or experimental error, the percent yield was more than 100%, at 103%. If the actual yield had been less than the theoretical yield, the phenomena could be explained by the fact that the experiment was not completely lossless. There will be moments when 100% of a sample is not transferred completely for weighing and therefor some sample or test product would be lost; creating less than an 100% yield.
Answers to Questions
The solution in part C must be titrated to the palest color of purple possible because the target volume would create a very pale purple. Too much of a titrate would start to skew the results with too high of a volume reading titrated. Too much titrate would create too high of an oxalate percentage because the result would signify that more liquid was required to turn the solution the pale purple color. If oxidation numbers were assigned to equation 2 in the lab manual (page 104):
+7 -2 +1 +4 -2 +1 +2 +4 -2 +1 -2
MnO4- (aq) + H2C2O4 (aq) + 2H+ (aq) Mn2+ (aq) + 2CO2 (g) + 4H2O (l)
This reaction is as redox reaction because the magnesium loses the electrons of the oxygen and becomes a positive ion, while the hydrogen ions combine with oxygen and gain electrons, thus resulting in the electron-transfer reaction, typical of redox reactions. The reducing agent is Maganese because it donates electrons and Oxygen is the oxidizing agent because it accepts electrons.
If the glass vial was not completely dry and had a trace of water in it, then the results would be too high of a reading for percent water composition. The sample would have the weight of the added water and therefor more water would evaporate, creating a higher percent of the sample.
If the 400 mL beaker were dirty with some unknown substance, then when the vials are heated, the “dirt” could evaporate and settle in the unknown vials and distort the measurement readings and cause an impure sample. The tare feature should not be used in this part for measurement because when the samples are finished being heating for the apporpriate time, there is not a way to separate the samples into another container without risk of losing sample left behind in the vial. Therefore the samples must be weighed in their vials without taring the scale.
Conclusion
The unknown iron sample had a composition of 9.50% iron, 53.5% Oxalate, 14.8% Water, and 22.2% Potassium.