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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