Question

A 0.7498-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 48.85 mL of a 0.2298 M AgNO₃ solution. This precipitated the chloride (Cl^-) out as AgCl and left an excess of AgNO₃. The excess AgNO₃ was titrated with 0.1165 M KSCN and required 20.82 mL to reach the endpoint in a Volhard titration.

Calculate the % w/w Cl^– (35.45 g/mol) in the sample. Provide your answer to 2 places after the decimal point and without units.

Reactions: Cl^– + Ag⁺ → AgCl(s) Reaction 1

Ag⁺ + SCN^–  →   AgSCN(s) Reaction 2

Answer 1

Here is the Solution:

We need to calculate the % (w/w) ci- in the sample. • Total moles of Ag= 0.04885x 0.2298 mollk = 0.0112257 moles, · Moles of excess Agt Stoichiometry of reaction Agt + SCN- Ag SCN (s) is 1:1 moles of Ag* = moles of SCN- = 0.02082 x x 0.1165 mol K = 0.00 242553 moles • No. of moles of Agt ion which reacts with ci ion, ci + Agt D Ag Cl (6) Stoichiometry of reaction I is dit moles of ci = moles of Agt = Total moles Agt - excess moles Ag? = (0.0112257 -0.00242553) mol = 0.00 880017 moles. • mass of cl = No. of moles x molar mass = 0.0088 00 17 mot x 35.45 g/mol = 0.311966 g - X 100 % Given weight of sample is : 0.7498 g mass of ai % (w/w) c1' = weight of sample % (w/w) ci = 0.311966 g xe 0.7498 8 % (w/w) ci = 41.6065 % % (w/w) C1 = 41.61 % X 100%

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