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)
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
I got
You'll want to work backwards here. This needed
"Ag"^(+)(aq) + "SCN"^(-)(aq) -> "AgSCN"(s)
Therefore, this accounts for
"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
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")