An organic compound contains carbon, hydrogen, and sulfur. A sample of it with a

Question

An organic compound contains carbon, hydrogen, and sulfur. A sample of it with a mass of 2.712 g was burned in oxygen to give gaseous CO2, H2O, and SO2. These gases were passed through 311.2 mL of an acidified 0.0200 M KMnO4 solution, which caused the SO2 to be oxidized to SO42-. Only part of the available KMnO4was reduced to Mn2+. Next, 31.12 mL of 0.0300 M SnCl2 was added to 31.12 mL portion of this solution, which still contained unreduced KMnO4. There was more than enough added SnCl2 to cause all of the remaining MnO4- in the 31.12 mL portion to be reduced to Mn2+. The excess Sn2+ that still remained after the reaction was then titrated with 0.0100 M KMnO4, requiring 0.02095 L of the KMnO4solution to reach the end point. Based upon all this data, the percentage of sulfur in the original sample of the organic compound that had been burned is___%.

Explanation / Answer

The reaction with SnCl2 is

5 SnCl2 + 2 KMnO4 + 16 HCl --> 2 MnCl2 + 5 SnCl4 + 8 H2O + 2 KCl

moles of KMnO4= 0.01*0.02059= 0.000206 moles

since 2 moles of this oxidizes 5 moles of SNCl2, SnCl2= 0.000206*5=0.00103 moles

SnCl2 in 0.03M and 31.12cml =(31.12/1000)*0.03=0.000934

Balance= 0.00103-0.000934=9.6X10-5

So KMnO4 required= 9.6X10-5*2/5=0.00003856 moles

Total KMnO4 added = (31.12/1000)*0.02=0.000622

amount reduced by SO2 by the reaction

2KMnO4 + 5SO2 + 2H2O 2MnSO4 + 2KHSO4 + H2SO4

2 moles of KMnO4 reduces 5 moles of SO2

0.000622*5/2= 0.001556 moles of SO2

Each mole of SO2 contains 32 gms of sulfur

hence,0.001556 moles contain =0.001556*32 =0.0497gms of sulfur

the sample is 2.712

Hence, % of sulfur =100*0.0497/2.712=2.3%

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