Question

a procedure similar to that described in this experiment can be used to precipitate Cl- as AgCl a sample containing chloride weighing 0.207g was precipitated with Ag+ and the AgCl produced weighed 0.472g.
calculate the mass percent of Cl- in the original sample

Answer 1

Answer on Question #55917 - Chemistry - General chemistry

Question:

a procedure similar to that described in this experiment can be used to precipitate Cl⁻ as AgCl a sample containing chloride weighing 0.207g was precipitated with Ag⁺ and the AgCl produced weighed 0.472g.

calculate the mass percent of Cl⁻ in the original sample

Solution:

The mass of AgCl is 0.472 g. Let's calculate the number of the moles of AgCl (molar mass of AgCl is 143.32 g/mol):

$$n(AgCl) = \frac{m(AgCl)}{M(AgCl)} = \frac{0.472\,g}{143.32\,\frac{g}{mol}} = 3.29 \cdot 10^{-3}\,mol$$

As one can notice from the formula of AgCl, the number of the moles of Cl⁻ and AgCl are equal (quantity of AgCl molecules and quantity of Cl⁻ ions are the same). Then, using the number of the moles of Cl⁻, we can calculate the mass of Cl⁻:

$$n(AgCl) = n(Cl)$$ $$m(Cl) = n(Cl) \cdot M(Cl) = 3.29 \cdot 10^{-3} \cdot 35.45 = 0.117\,g$$

Then, one can calculate the mass percent of the chloride in the sample:

$$\omega(Cl) = \frac{m(Cl)}{m(sample)} \cdot 100\% = \frac{0.117}{0.207} \cdot 100\% = 57.0\%$$

Answer: 57%

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