Chapter 4 Stoichiometry: Quantitative Information about Chemical Reactions - Study Questions - Page 179c: 22

Theoretical yield: $9.37\ g$ Percent yield: $92.27\%$

Number of moles of $CH_3SH$: $10.0\ g\div 48.11\ g/mol=0.208\ mol$ Theoretical yield: $0.208\ mol \times 1_{CH_3COSCH_3}/2_{CH_3SH}\times 90.14\ g/mol=9.37\ g$ Percent yield: $8.65/9.37\times100\%=92.37\%$