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Ethyl Mercaptan (C2H5SH) is a colorless or yellowish liquid or a gas with a pungent, garlic or skunk-like odor. The concentration of this compound was determined by shaking a 3.85 g sample with 58.00 mL of 0.01204 M lodine (12) in a tightly stoppered flask. The reaction is shown below. 2 C2H5SH + 12 → C2H5SSC2H5 + 2 1¯ + 2H* The excess 12 was back titrated with 19.86 mL of 0.01437 M Na2S2O3 according to the reaction: S406²-(aq) ²(aq) ➜ 21 ¹2(aq) + 2 S₂032- Calculate the percentage C₂H5SH (MW = 62.14 g/mol). (aq) +

Ethyl Mercaptan (C2H5SH) is a colorless or yellowish liquid or a gas with a pungent, garlic or skunk-like odor. The concentration of this compound was determined by shaking a 3.85 g sample with 58.00 mL of 0.01204 M lodine (12) in a tightly stoppered flask. The reaction is shown below. 2 C2H5SH + 12 → C2H5SSC2H5 + 2 1¯ + 2H* The excess 12 was back titrated with 19.86 mL of 0.01437 M Na2S2O3 according to the reaction: S406²-(aq) ²(aq) ➜ 21 ¹2(aq) + 2 S₂032- Calculate the percentage C₂H5SH (MW = 62.14 g/mol). (aq) +

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Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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Ethyl Mercaptan (C₂H5SH) is a colorless or yellowish liquid or a gas with a pungent, garlic or skunk-like odor. The concentration of this compound was determined by shaking a 3.85 g sample with 58.00 mL of 0.01204 M lodine (12) in a tightly stoppered flask. The reaction is shown below. 2 C2H5SH + 12 → C2H5SSC2H5 +2 I¯ + 2H+ The excess 12 was back titrated with 19.86 mL of 0.01437 M Na2S2O3 according to the reaction: 12(aq) + 2 S₂032 (aq) → 21" (aq) +S406²-(aq) Calculate the percentage C₂H5SH (MW = 62.14 g/mol).
Transcribed Image Text:Ethyl Mercaptan (C₂H5SH) is a colorless or yellowish liquid or a gas with a pungent, garlic or skunk-like odor. The concentration of this compound was determined by shaking a 3.85 g sample with 58.00 mL of 0.01204 M lodine (12) in a tightly stoppered flask. The reaction is shown below. 2 C2H5SH + 12 → C2H5SSC2H5 +2 I¯ + 2H+ The excess 12 was back titrated with 19.86 mL of 0.01437 M Na2S2O3 according to the reaction: 12(aq) + 2 S₂032 (aq) → 21" (aq) +S406²-(aq) Calculate the percentage C₂H5SH (MW = 62.14 g/mol).
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