Answer
a. $165.403\text{ grams}$
b. $3.023\text{ mol $C_2H_3Cl_3O_2$}$
c. $5.5*10^{22}\text{ atoms Cl}$
d. $5.5*10^{22}\text{ atoms Cl}$
e. $1.6\text{ grams $C_2H_3Cl_3O_2$}$
f. $1.37*10^{-19}\text{ grams $C_2H_3Cl_3O_2$}$
Work Step by Step
a. $\text{ Molar Mass} = 2*12.0107+3*1.00794+3*35.453+2*15.9994=165.403\text{ grams}$
b. $500.\text{ grams $C_2H_3Cl_3O_2$}*\frac{1\text{ mol $C_2H_3Cl_3O_2$}}{165.403\text{ grams $C_2H_3Cl_3O_2$}}=3.023\text{ mol $C_2H_3Cl_3O_2$}$
c. $2.0*10^{-2}\text{ mol $C_2H_3Cl_3O_2$}*\frac{165.403\text{ grams $C_2H_3Cl_3O_2$}}{1\text{ mol $C_2H_3Cl_3O_2$}}=3.3\text{ grams $C_2H_3Cl_3O_2$}$
d. $5.0\text{ grams $C_2H_3Cl_3O_2$}*\frac{1\text{ mol $C_2H_3Cl_3O_2$}}{165.403\text{ grams $C_2H_3Cl_3O_2$}}*\frac{3\text{ mol Cl}}{1\text{ mol $C_2H_3Cl_3O_2$}}*\frac{6.0221*10^{23}\text{ atoms Cl}}{1\text{ mol Cl}}=5.5*10^{22}\text{ atoms Cl}$
e. $1.0\text{ grams Cl}*\frac{1\text{ mol Cl}}{35.453\text{ grams Cl}}*\frac{1\text{ mol $C_2H_3Cl_3O_2$}}{3\text{ mol Cl}}*\frac{165.403\text{ grams $C_2H_3Cl_3O_2$}}{1\text{ mol $C_2H_3Cl_3O_2$}}=1.6\text{ grams $C_2H_3Cl_3O_2$}$
f. $500.\text{ molecules $C_2H_3Cl_3O_2$}*\frac{1\text{ mol $C_2H_3Cl_3O_2$}}{6.0221*10^{23}\text{ molecules $C_2H_3Cl_3O_2$}}*\frac{165.403\text{ grams $C_2H_3Cl_3O_2$}}{1\text{ mol $C_2H_3Cl_3O_2$}}=1.37*10^{-19}\text{ grams $C_2H_3Cl_3O_2$}$
