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
Apr 11, 2018

I got 25.75%25.75% "w/w Cl"w/w Cl.


You'll want to work backwards here. This needed "19.93 mL"19.93 mL of "0.1553 M KSCN"0.1553 M KSCN, which accounts for "0.003095 mols SCN"^(-)0.003095 mols SCN. This reacted with the excess "AgNO"_3AgNO3 which gave

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

Therefore, this accounts for "0.003095 mols excess AgNO"_30.003095 mols excess AgNO3. From the total "AgNO"_3AgNO3, the reacted "AgNO"_3AgNO3 can be found.

"0.1619 M AgNO"_3 xx "0.04062 L" = "0.006576 mols AgNO"_30.1619 M AgNO3×0.04062 L=0.006576 mols AgNO3 total

This means that

"0.006576 - 0.003095 = 0.003481 mols"0.006576 - 0.003095 = 0.003481 mols of "AgNO"_3AgNO3 reacted

to form "AgCl"AgCl, and thus, "0.003481 mols Cl"^(-)0.003481 mols Cl reacted with that "Ag"^(+)Ag+ from "AgNO"_3AgNO3. Knowing the mols of "Cl"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")