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?

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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)

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Answer 1 · 2018-04-11T08:37:07

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 ${AgNO}_{3}$ $"AgNO"_3$ which gave

${Ag}^{+} (a q) + {SCN}^{-} (a q) \to AgSCN (s)$ $"Ag"^(+)(aq) + "SCN"^(-)(aq) -> "AgSCN"(s)$

Therefore, this accounts for ${0.003095 mols excess AgNO}_{3}$ $"0.003095 mols excess AgNO"_3$ . From the total ${AgNO}_{3}$ $"AgNO"_3$ , the reacted ${AgNO}_{3}$ $"AgNO"_3$ can be found.

${0.1619 M AgNO}_{3} \times 0.04062 L = {0.006576 mols AgNO}_{3}$ $"0.1619 M AgNO"_3 xx "0.04062 L" = "0.006576 mols AgNO"_3$ total

This means that

$0.006576 - 0.003095 = 0.003481 mols$ $"0.006576 - 0.003095 = 0.003481 mols"$ of ${AgNO}_{3}$ $"AgNO"_3$ reacted

to form $AgCl$ $"AgCl"$ , and thus, ${0.003481 mols Cl}^{-}$ $"0.003481 mols Cl"^(-)$ reacted with that ${Ag}^{+}$ $"Ag"^(+)$ from ${AgNO}_{3}$ $"AgNO"_3$ . 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,

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

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

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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?

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