A 0.4793-g sample of a chlorocarbon compound was analyzed by burning it in oxygen and collecting the evolved gases in a solution of NaOH. After neutralizing, the sample was treated with 40.62 mL of a 0.1619 M AgNO3 solution?

This precipitated the chloride (Cl-) out as AgCl and left an excess of AgNO3. The excess AgNO3 was titrated with 0.1553 M KSCN and required 19.93 mL to reach the endpoint in a Volhard titration.Calculate the % w/w Cl– (35.45 g/mol) in the sample.

Reactions: Cl– + Ag+ → AgCl(s) 

              Ag+ + SCN–  →   AgSCN(s)

1 Answer
Truong-Son N.
Apr 11, 2018

I got 25.75% "w/w Cl".

You'll want to work backwards here. This needed "19.93 mL" of "0.1553 M KSCN", which accounts for "0.003095 mols SCN"^(-). This reacted with the excess "AgNO"_3 which gave

"Ag"^(+)(aq) + "SCN"^(-)(aq) -> "AgSCN"(s)

Therefore, this accounts for "0.003095 mols excess AgNO"_3. From the total "AgNO"_3, the reacted "AgNO"_3 can be found.

"0.1619 M AgNO"_3 xx "0.04062 L" = "0.006576 mols AgNO"_3 total

This means that

"0.006576 - 0.003095 = 0.003481 mols" of "AgNO"_3 reacted

to form "AgCl", and thus, "0.003481 mols Cl"^(-) reacted with that "Ag"^(+) from "AgNO"_3. Knowing the mols of "Cl" atoms that came from the chlorocarbon, the mass that came from it is

0.003481 cancel"mols Cl" xx "35.453 g Cl"/cancel"1 mol Cl" = "0.1234 g Cl"

Therefore,

"0.1234 g Cl"/"0.4793 g chlorocarbon compound" xx 100%

= color(blue)(25.75% "w/w Cl")

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