Question

You are given an unknown hydrated metal salt containing chloride ion MCl2 times nH2O

a. you dissolve 0.500 g of this salt in water and add excess silver nitrate slutionAgNO3 to precipitate the chloride ion as insoluble silver chloride, AgCl. After filtering washing-drying and weighing the AgCl is found to weigh 0.720g. what is the mass percent chloride in the metal salt?

b. a second 0.500 g sample is dehydrated to remove water of hydration. After drying the sample is found to weigh 0.319g what is the mass percent chloride in the metal salt?

c. The metal cation has a charge of two. what is the molar mass of the metal?

d. determine the complete empirical formula of the hydrated metal salt

Answer 1

a) 0.500 g of MCl₂.nH₂O was treated with AgNO₃ to give AgCl. The mass of AgCl was found to be 0.720 g.

Molar mass of AgCl = 143.32 g/mol.

                

Mole(s) of AgCl obtained = (0.720 g)/(143.32 g/mol) = 0.0050237 mole.

We know that 1 mole AgCl = 1 mole Cl atoms. Since the Cl atoms came from MCl₂.nH₂O, hence, the hydrated salt must contain 0.0050237 mole Cl atoms.

Atomic mass of Cl = 35.453 g/mol; therefore, the mass of Cl in the hydrated salt = (0.0050237 mole)*(35.453 g/mol) = 0.1781 g.

Mass percentage Cl in the salt = (0.1781 g)/(0.500 g)*100 = 35.62% (ans).

b) Mass of water lost = (0.500 – 0.319) g = 0.181 g.

Molar mass of H₂O = (2*1.008 + 1*15.9994) g/mol = 18.0154 g/mol.

Mole(s) of water in the hydrated salt = (0.181 g)/(18.0154 g/mol) = 0.01005 mole.

Mass percentage water in the salt = (0.181 g)/(0.500 g)*100 = 36.20% (ans).

c) Mass of metal in 0.500 g hydrated salt = [0.500 – (0.1781 + 0.181)] g = 0.1409 g.

We see that 0.1781 g Cl combines with 0.1409 g metal; therefore, by the law of equivalents, 35.453 (equivalent weight of Cl) will combine with (0.1409 g metal)*(35.453 g Cl/0.1781 g Cl) = 28.04788 g metal. This is the equivalent weight of the metal.

Since the metal has valence 2, hence, the molecular weight of the metal is (equivalent weight*2) = (28.04788*2) g/mol = 56.09576 g/mol ≈ 59.096 g/mol. This is the molar mass of the metal (ans).

d) Percentage of metal in the salt = (0.1409 g)/(0.500 g)*100 = 28.18%.

100 g of the hydrated salt contains 35.62 g Cl atoms, 36.20 g H₂O and 28.18 g metal. Divide by the atomic/molar masses to find the simplest ratio of the moles of each.

Metal:Chlorine:Water = (28.18 g/56.096 g/mol):(35.62 g/35.453 g/mol):(36.20/18.0154 g/mol) = 0.5023:1.0047:2.0094 = (0.5023/0.5023):(1.0047/0.5023):(2.0094/0.5023) = 1:2:4.

The empirical formula of the hydrated metal salt is MCl₂.4H₂O (ans).