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 KMnO₄ was 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 KMnO₄ solution 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

First of all you convert everything to moles.

In the titration you used 0.000224 mol (0.0100 M * 0.02235 L) KMnO4. 2 mol KMnO4 oxidize 5 mol of SnCl2 (5 SnCl2 + 2 KMnO4 + 16 HCl --> 2 MnCl2 + 5 SnCl4 + 8 H2O + 2 KCl) so 0.000224 mol oxidize 0.000559 mol of SnCl2.

You had 0.001351 (0.03 M * 0.04503 L) SnCl2 so when you added the SnCl2 solution you had this excess.

If you extract this from the total moles of SnCl2 you had oxidized the rest (0.001351 mol - 0.000792 mol).

For this oxidation you need 2/5 mol (see stoichiometry above) KMnO4 (0.02 M) (= 0.000792 * 2 / 5) 0.000317 mol. So after the first addition of KMnO4 (not titration) you had an excess of 0.000317 mol. In total you added 0.009006 mol (0.02 M * 0.4503 L) so the reduced amount by SO2 (2KMnO4 + 5SO2 + 2H2O → 2MnSO4 + 2KHSO4 + H2SO4) was 0.008689 mol (0.009006 mol - 0.000317 mol). 5 mol of SO2 reduce 2 mol of KMnO4 so 0.008689 mol KMnO4 were reduced by 0.021723 mol of SO2.
Each mole of SO2 contains one mol of S (Mr = 32.065) so in the begining you had 32.065 * 0.021723 mol = 0.696543 g of S
The sample you used had mass of 1.078 g, so there was 100 * 0.696543 g (S) / 1.078 g (compound) = 64.6 %