Question

A mixture consisting of only chromium(II) chloride (CrCl₂, 122.90 g/mol) and iron(III) chloride (FeCl₃, 162.20 g/mol) weighs 2.3738 g. When the mixture is dissolved in water and an excess of silver acetate is added, all the chloride ions associated with the original mixture are precipitated as insoluble silver chloride (AgCl, 143.32 g/mol). The mass of the silver chloride is found to be 5.7101 g. Calculate the mass percentage of chromium(II) chloride in the original mixture.

Answer 1

Let the mixture contain x g chromium (II) chloride and y g iron (III) chloride.

Therefore,

x + y = 2.3738 g ……(1)

The molar masses are given.

Mole(s) of chromium (II) chloride = (mass of chromium chloride)/(molar mass of chromium chloride) = (x g)/(122.90 g/mol) = x/122.90 mole.

Mole(s) of iron (III) chloride = (mass of iron chlroride)/(molar mass of iron chloride) = (y g)/(162.20 g/mol) = y/162.20 mole.

Write the balanced chemical equations for the formation of the chlorides.

CrCl₂ (aq) + 2 CH₃COOAg (aq) --------> 2 AgCl (s) + Cr(CH₃COO)₂ (aq) ……..(2)

FeCl₃ (aq) + 3 CH₃COOAg (aq) --------> 3 AgCl (s) + Fe(CH₃COO)₃ (aq) ……..(3)

As per the stoichiometric equations (2) and (3),

Mole(s) AgCl from chromium chloride = 2x/122.90 mole (1 mole CrCl₂ gives 2 moles AgCl).

Mole(s) AgCl from iron chloride = 3y/162.20 mole (1 mole FeCl₃ gives 3 moles AgCl).

Total number of moles of AgCl = (2x/122.90 + 3y/162.20) mole.

Mass of AgCl obtained = (moles of AgCl)*(molar mass of AgCl)

= (2x/122.90 + 3y/162.20) mole*(143.32 g/mol)

= 143.32*(2x/122.90 + 3y/162.20) g.

As per the problem,

143.32*(2x/122.90 + 3y/162.20) = 5.7101

=====> 286.64x/122.90 + 429.96y/162.20 = 5.7101

=====> 2.332x + 2.651y = 5.7101

=====> 2.332x + 2.651*(2.3738 – x) = 5.7101 (from (1) above)

=====> 2.332x + 6.29294 – 2.651x = 5.7101

=====> 6.29294 – 0.319x = 5.7101

=====> 6.29294 – 5.7101 = 0.319x

=====> 0.319x = 0.58284

=====> x = 0.58284/0.319

=====> x = 1.82708

=====> x ≈ 1.8271

The mass of chromium chloride in the mixture is 1.8271 g and the mass percentage of chromium chloride = (mass of CrCl₂)/(mass of mixture)*100

= (1.8721 g)/(2.3738 g)*100

= 76.9494%

≈ 76.95% (ans).