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 SO₄^2-. 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___%.

Answer 1

The solution is as follows

5 moles of SO2 reacts with 2 moles of KMnO4 to form 2 moles of SO4^2-

5 moles of SnCl2 reacts with 2 moles of KMnO4 to reduce MnO4- to Mn2+

Original moles of KMnO4 = 0.02 x 0.3112 = 0.0062 moles

Final moles of KMnO4 required to react with excess of SnCl2 = 0.01 x 0.02095 = 0.00021 moles

moles of excess SnCl2 reacted = 0.00053 moles

Initial moles of SnCl2 added = 0.03 x 0.03112 = 0.000934 moles

moles of unreacted KMnO4 = 0.000934 - 0.00053 = 0.000404 moles

therefore, moles of KMnO4 reacted with SO2 = 0.0062 - 0.000404 = 0.0058 moles

moles of SO2 = 0.0145 moles

grams of SO2 = 64.07 x 0.0145 = 0.929 g

grams of sample = 2.712 g

Thus, the percentage of sulfur in the original sample of the organic compound that had been burned is 34.26 %.