The ethyl mercaptan, C2H5SH, concentration in a mixture wasdetermined by shaking a 1.534-g sample with 60.0 mL of 0.01293 M I2in a tightly stoppered flask:
2C2H5SH + I2 →2C2H5SSC2H5 +2I- + 2H+
The excess I2 was back-titrated with 10.72 mL of 0.01425 MNa2S2O3:
2S2O32- + I2 →S4O62- + 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)
