The formula of the unknown iron compound that you analyzed by titrating with KMn

Question

The formula of the unknown iron compound that you analyzed by titrating with KMnO_4 was Fe(NH_4)_2(SO_4)_2*6H_2O. The three class average was 14.5 % Fe with a 92% precision. a) Calculate the class' % error: b) Comment on the quality of the class' results: c) Two lab teams obtained results of 8.2 and 9.5% Fe. Explain whether their results have been caused by the following errors: i) buret tip was not filled initially: ii) buret was wet when initially filled with KMnO_4: iii) the magnetic spinner was initially turned on too fast spattering some of the Iron compound to spatter on to the sides of the flask and thus not be titrated: d) In the titration MnO_4^-1 reacts in aqueous addle solution to produce Mn^+2. If you titrated a solution containing Sn^+2 to the same faint pink end point, write and balance the equation:

Explanation / Answer

a) Calculate the molar mass of Fe(NH4)2(SO4)2 *6H2O by summing up the atomic masses of the elements multiplied by the respective co-efficient as below:

Molar mass = [55.85 + 2*(14.0067 + 4*1.008) + 2*(32.065 + 4*15.9994) + 6*(2*1.008 + 15.9994)] g/mol = 392.145 g/mol.

Molar mass of Fe = 55.85 g/mol.

Therefore, percentage Fe in the sample = (55.85 g/mol)/(392.145 g/mol)*100 = 14.242%

Percent error in calculation = absolute value – experimental value/(absolute value)*100 = 14.242 – 14.5/(14.242)*100 = 1.811% (ans).

b) For experimental determination of molecular formula of a compound, an error margin of 3-5% is tolerable. The experimentally obtained results showed less than 2% error and hence the quality of the class data was within limits and pretty good.

c i) Consider a 50.00 mL buret was used by the two groups. For class A burets, the entire volume of liquid from the zero minicus to the tip of the buret reads as 50.00 mL. Since the two groups didn’t fill the burets till the tip, the groups will register/read a higher volume of KMnO4 dispensed and hence the calculated moles of Fe2+ oxidized. Thus, the error will lead to a higher value of percentage Fe. Hence, not filling the buret to the tip cannot cause the percentage Fe to be low.

ii) The buret was initially wet and hence a slightly lower volume of KMnO4 was required to fill the buret completely. Since KMnO4 is coloured and water is colourless, the whole volume of liquid in the buret will be coloured. In other words, the KMnO4 will be diluted and hence the actual strength will be lower than the calculated strength. A higher volume of KMnO4 will be required, but in the end, calculations based on number of moles will yield the same result for percentage Fe. Hence, the given error cannot cause percentage Fe to be low.

iii) This will definitely affect the percentage Fe and the calculated results will be low. The reason is some amount of Fe is lost and hence the results will be lower.

d) Write down the half reactions:

MnO4- + 8 H+ + 5 e- -------> Mn2+ + 4 H2O (reduction half)

Sn2+ ---------> Sn4+ + 2 e- (oxidation half)

Multiply the oxidation half by 5 and the reduction half by 2 and add

2 MnO4- + 5 Sn2+ + 16 H+ --------> 2 Mn2+ + 5 Sn4+ + 8 H2O (balanced).

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