AP Chemistyr salt lab question?

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M+ is an unknown metal cation with a +1 charge. A student dissolves the chloride of the unknown metal, MCl, in enough water to make 100.0 mL of solution. The student then mixes the solution with excess AgNO3 solution, causing AgCl to precipitate. The student collects the precipitate by filtration, dries it, and records the data shown below. (The molar mass of AgCl is 143 g / mol.)

Mass of unknown chloride, MCl=0.74 g

Mass of filter paper=0.80 g

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Mass of filter paper plus AgCl precipitate=2.23 g

  1. What is the identity of the metal chloride?

a. NaCl

b. KCl

c. CuCl

d. LiCl

1 Answer

  • Mass of AgCl = (mass of filter paper + AgCl) – (mass of filter paper) = 2.3 g – 0.80 g = 1.5 g AgCl

    1.5 g AgCl x (1 mole AgCl / 143 g AgCl) = 0.010 moles Ag

    Since there is 1 Ag+ in the formula AgCl and 1 Cl- in the formula AgCl, then moles Ag+ = moles AgCl = 0.010 and moles Cl- = moles AgCl = 0.010.

    Since the formula of the unknown chloride is MCl, 0.010 moles Cl- x (1 mole M+ / 1 mole Cl-) = 0.010 moles M+

    0.010 moles Cl- x (35.5 g Cl- / 1 mole Cl-) = 0.37 g Cl-

    Mass M+ = Mass of MCl – mass of Cl- = 0.74 g – 0.37 g = 0.37 g.

    0.37 g M+ / 0.010 moles M+ = 37 g/mole. Potassium ion (K+) has a molar mass of 39 g/mole (pretty close to 37).

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