A 5.000 g sample of a pesticide was decomposed with metallic sodium in alcohol, and the liberated...
Question:
A {eq}\rm \displaystyle 5.000\ g {/eq} sample of a pesticide was decomposed with metallic sodium in alcohol, and the liberated chloride ion was precipitated as {eq}\rm AgCl {/eq}. Express the results of this analysis in terms of percent DDT {eq}\rm \displaystyle (C_{14}H_9Cl_5,\ 354.50\ g/mol) {/eq}, one of the components, based on the recovery of {eq}\rm 0.1606\ g {/eq} of {eq}\rm AgCl {/eq}.
Gravimetric Analysis:
The gravimetric analysis makes use of the masses of the involved chemical species to determine the amount of the target compound. This is usually done by precipitating out the target compound for easier quantification. One of the setbacks of this is when incomplete precipitation occurs which means that the calculated mass of the target compound will be lower than the true value.
Answer and Explanation: 1
The mass of the recovered AgCl will be converted to the mass of DDT using the number of chloride ions. AgCl has 1 chloride ion while DDT has 5. As such, the gravimetric factor is {eq}\rm 1\ MM\ AgCl = 5\ MM\ DDT {/eq}. Using this factor, the mass of DDT can be calculated,
- MM AgCl = 143.32 g/mol
{eq}\rm mass\ DDT = 0.1606\ g\ AgCl \times \dfrac{5 \times 354.50\ g/mol\ DDT}{1 \times 143.32\ g/mol\ AgCl} = 1.986\ g\ DDT {/eq}
Using the mass of DDT, its percentage in the sample can be calculated.
{eq}\rm \%\ DDT = \dfrac{1.986\ g}{5.000\ g} \times 100 = \boxed{\mathbf{39.72 \%}} {/eq}
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