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A 5.0 g sample of a pesticide was decomposed with metallic sodium in alcohol, and the liberated chloride ion was precipitated with AgCl. Express the results of this analysis in terms of percent DDT (C14H9Cl5, molar mass, 354.72 g/mol)) based on the recovery of 0.1606 g of AgCl

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Answer:

1.59%

Explanation:

First, we convert 0.1606 g of AgCl into moles, using its molar mass:

  • 0.1606 g ÷ 143.32 g/mol = 1.12x10⁻³ mol AgCl

Then we convert the calculated moles of Cl (equal to AgCl moles) into moles of DDT:

  • 1.12x10⁻³ mol Cl * [tex]\frac{1molDDT}{5molCl}[/tex] = 2.24x10⁻⁴ mol DDT

Now we convert 2.24x10⁻⁴ moles of DDT into grams, using its molar mass:

  • 2.24x10⁻⁴ mol DDT * 354.48 g/mol = 0.0794 g

Finally we calculate the percentage of DDT in the sample:

  • 0.0794 g / 5.0 * 100% = 1.59%

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