Question

The ethyl mercaptan, C2H5SH, concentration in a mixture was determined by shaking a 1.534-g sample with 60.0 mL of 0.01293 M I2 in a tightly stoppered flask:

2C₂H₅SH + I₂ → 2C₂H₅SSC₂H₅ + 2I^- + 2H⁺

The excess I2 was back-titrated with 10.72 mL of 0.01425 M Na₂S₂O₃:

2S₂O₃^2- + I₂ → S₄O₆^2- + 2I^-

Calculate the percentage of C₂H₅SH (FM = 62.13 g/mol) in the mixture.

(a) 5.67% (b) 3.32% (c) 1.24% (d) 4.33%

(please show work)

Answer 1

From the equation of the reaction between Thiosulfate and Iodine shows that 1 mol of Iodine is reduced by 2 moles of Thiosulfate.

Number of moles of Thiosulfate used for titration is 0.01072 L * 0.01425 M = 0.00015276 mol.

So, excess amount of Iodine present in the solution is 0.00015276 mol/2 = 0.00007638 mol

Amount of Iodine initially added to the solution is 0.060 L * 0.01293 M = 0.0007758 mol

So, Iodine reacting with ethyl mercaptan is (0.0007758 - 0.00007638) mol = 0.00069942 mol

From the equation of the reaction between Thiosulfate and Mercaptan shows that 1 mol of Iodine is reduced by 2 moles of Mercaptan.

Hence, number of moles of Mercaptan present in the sample is 0.00069942 mol/2 = 0.00034971 mol

Molar mass of Mercaptan is 62.13 g/mol

So, mass of Mercaptan present in the sample is 62.13 g/mol * 0.00034971 mol = 0.022 g

The percentage of C₂H₅SH in the mixture is

         (0.022 / 1.534) * 100 % ~1.24%