A 0.1156-g sample of a chlorocarbon compound was analyzed by burning it in oxyge

Question

A 0.1156-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 26.00 mL of a 0.1141 M AgNO3 solution. This precipitated the chloride (Cl-) out as AgCl and left an excess of AgNO3. The excess AgNO3 was titrated with 0.1048 M KSCN and required 23.85 mL to reach the endpoint in a Volhard titration. Calculate the % w/w Cl– (35.45 g/mol) in the sample. Give your answer to 2 places after the decimal point. Reactions: Cl– + Ag+ ? AgCl(s) Reaction 1 Ag+ + SCN– ? AgSCN(s) Reaction 1

Explanation / Answer

% w/w Cl– in the chlorocarbon compound = 14.47 %

The excess AgNO3 = 0.1048 M, 23.85 mL KSCN = 0.1048 x 23.85 /1000 = 0.002495 moles

No. of moles of AgNO3 in 26.00 mL of a 0.1141 M = 0.1141 x 26.00 / 1000 = 0.002967 moles

No. of moles of AgCl = 0.002967 - 0.002495 = 0.000472 moles

So the no. of moles Cl– (35.45 g/mol) in 0.000472 moles of AgCl = 0.000472 moles

0.000472 moles of Cl- (35.45 g/mol) = 35.45 x 0.000472 = 0.016732 g

% w/w Cl– in the chlorocarbon compound = (0.016732 g / 0.1156 g) x 100 = 14.47 %

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