Question

a 5g sample of pesticide was decomposed with metallic sodium in alcohol and the liberate ion was precipitated as AgCl. the amount of AgCl obtained by this method is .1606g. Express the results in terms of percent weight by Cl, b. percent weight of Cl2, and percent weight by weight DDT(C14H9Cl5)(354.47g/mol)

Answer 1

a. 5g sample of pesticide was decomposed with metallic sodium in alcohol and the liberate ion was precipitated as AgCl. the amount of AgCl obtained by this method is .1606g. Express the results in terms of percent weight by Cl, b. percent weight of Cl2, and percent weight by weight DDT(C14H9Cl5)(354.47g/mol)

Solution:

a. Determine the mass of Cl in AgCl according to the equivalent's law:

$$\frac {m (\mathrm {A g C l})}{m (\mathrm {C l})} = \frac {\mathrm {E} (\mathrm {A g C l})}{\mathrm {E} (\mathrm {C l})}$$

Using atomic masses from the periodic table and valence of Ag and Cl we will find the following: $\mathrm{E(AgCl)} = 143.321\mathrm{g / mol}$ ; $\mathrm{E(Cl)} = 35.453\mathrm{g / mol}$ .

Therefore, $m(\mathrm{Cl}) = \frac{m(\mathrm{AgCl}) \cdot \mathrm{E}(\mathrm{Cl})}{E(\mathrm{AgCl})} = \frac{0.1606 \cdot 35.453}{143.321} = 0.0397\mathrm{g}$ .

The percent weight of Cl in pesticide is:

$$\omega (\mathrm {C l}) = \frac {0 . 0 3 9 7}{5} 1 0 0 = 0. 7 9 \%$$

b. The percent weight of $\mathrm{Cl}_2$ in pesticide is:

$$\omega \left(\mathrm {C l} _ {2}\right) = \frac {0 . 0 3 9 7 \cdot 2}{5} 1 0 0 = 1. 5 8 8 \%$$

The percent weight of Cl in DDT $(C_{14}H_9Cl_5)$ is:

$$\omega (\mathrm {C l}) = \frac {3 5 . 4 5 3 \cdot 5}{3 5 4 . 4 7 \cdot 2} 1 0 0 = 2 5. 0 0 4 \%$$

The percent weight of $\mathrm{Cl}_2$ in DDT $(\mathrm{C}_{14}\mathrm{H}_9\mathrm{Cl}_5)$ is:

$$\omega \left(\mathrm {C l} _ {2}\right) = \frac {3 5 . 4 5 3 \cdot 2 \cdot 5}{3 5 4 . 4 7 \cdot 2} 1 0 0 = 5 0. 0 0 8 \%$$

Answer:

a. The percent weight of Cl in pesticide is $0.79\%$ .

b. The percent weight of $\mathrm{Cl}_2$ in pesticide is $1.588\%$ ; the percent weight of Cl in DDT $(\mathrm{C}_{14}\mathrm{H}_{9}\mathrm{Cl}_{5})$ is $25.004\%$ ; the percent weight of $\mathrm{Cl}_2$ in DDT $(\mathrm{C}_{14}\mathrm{H}_{9}\mathrm{Cl}_{5})$ is $50.008\%$ .