please help with these problems. Much appreciated. Thank you 4. You are given an

Question

please help with these problems. Much appreciated. Thank you

4. You are given an unknown hydrated metal salt containing chloride ion, MCl, nH,O. You dissolve 0.500 g of this salt in water and add excess silver nitrate solution, AgNO, to precipitate the chloride ion as insoluble silver chloride, AgCI. After ound to weigh 0.720g a. 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 0.319 g. What is the mass percent water in the metal salt? The metal cation has a charge of two. What is the molar mass of the metal? What is the identity of the metal? c. d. Determine the complete empirical formula of the hydrated metal salt.

Explanation / Answer

Ans. #4a.        AgNO3(aq) + MCl2(aq) ----------> 2 AgCl(s) + M(NO3­)2(aq)

Stoichiometry: 1 mol MCl2 forms 2 mol AgCl.

Moles of AgCl formed = Mass / Molar mass

                                    = 0.720 g / (143.3209 g)

                                    = 0.005024 mol

# According to stoichiometry of balanced reaction, 1 mol MCl2 forms 2 mol AgCl.

So,

            Moles of Cl- in sample = (½) x Moles of AgCl formed

                                                = ( ½ ) x 0.005024 mol

                                                = 0.002512 mol

            Mass of Cl- in sample = Moles x Molar mass

                                                = 0.002512 mol x (35.4527 g/ mol)

                                                = 0.0891 g

Now,

            Mass % Cl in sample = (Mass of Cl / Mass of sample) x 100

                                                = (0.0891 g / 0.500 g) x 100

                                                = 17.81 %

#4b. Mass of water in sample = Mass of hydrated sample – Mass of dried sample

                                                = 0.500 g – 0.319 g

                                                = 0.181 g

Mass % water in sample = (Mass of water / Mass of hydrated sample) x 100

                                                = (0.181 g / 0.500 g) x 100

                                                = 36.20 %

#4c. Mass of metal = Mass of hydrated sample – Mass of (Cl, from #4a + H2O, from #4b)

                                    = 0.500 g – (0.0891 g + 0.181 g)

                                    = 0.2299 g

1 mol MCl2 consists of 1 mol metal and 2 moles of Cl-.

So,

            Number of moles of metal = ½ x Moles of Cl-                            ; [see #4a]

                                                = ½ x 0.002512 mol

                                                = 0.001256 mol

Now,

            Molar mass of metal = Mass of metal in sample / Moles of metal in sample

                                                = 0.2299 g / (0.001256 mol)

                                                = 183.041 g/ mol

The metal is Tungsten, W (actual molar mass = 183.84 g/ mol).

#d. So, far we have

            Moles of W = 0.001256 mol

            Moles of Cl = 0.002512 mol

            Mass of water = 0.181 g

So,

            Moles of water = 0.181 g / (18.01528 g/ mol) = 0.010047 mol

Now,

Simple stoichiometric molar ratio of constituents is-

            W : Cl : H2O = 0.001256 mol : 0.002512 mol : 0.010047 mol

            Or, W : Cl : H2O = 1 : 2 : 7.9997 = 1 : 2 : 8 (nearest whole number)

That is, there are 1 mol of W, 2 mol of Cl and 8 moles of water of hydration in the hydrated salt.

Hence,

Empirical formula = WCl2.8H2O

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