Question

A 7.000-g sample of a pesticide was decomposed with metallic sodium in alcohol, and the liberated chloride ion was precipitated as AgCl. Express the results of this analysis in terms of percent DDT (C14H9Cl5) based on the recovery of 0.2513 g of AgCl. 

Answer 1

Mass Cl

Ar Cl = 35.453 g/mol

Mm AgCl = 143.32 g/mol

mass Cl = 35.453*0.2513/143.32 = 0.06216

Mass DDT in sample

Mm DDT = 354.47 g/mol

There is 5 atom Cl in DDT so :

Mass DDT = 0.1243 g

Then, % mass of DDT in sample:

%DDT = 0.1243/7 = 0.0178 = 1.78%