4. You are given an unknown hydrated metal salt containing chloride ion, MCI nH,

Question

4. You are given an unknown hydrated metal salt containing chloride ion, MCI 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 filtering, washing, drying, and weighing, the AgCl is found to weigh 0.720 g. What is the mass percent chloride in the metal salt? a. 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

a) AgCl contains 1mol of chloride ions. From the molar mass of AgCl as 143.32g/mol and that of chloride as 35.453g/mol, the moles of AgCl formed is calculated as Moles = mass/molar mass = 0.72/143.32 = 5.0237mmol. This means that there is 5.0237mmol of chloride ion in the salt. This gives the mass of chloride in the sample as 35.453*5.0237/1000 = 0.1781g. Now the mass percent is calculated as (mass of chloride/mass of salt taken)*100 = (0.1781/0.5)*100 = 35.62%.

b) From the molar mass of water as 18.0153g/mol, using the mass difference between the hydrated and anhydrous salts, the mass of water in the sample is calculated as 0.5-0.319 = 0.181g which gives its mass percent as 36.2%.

c) Since we know that the salt has two equivalents of chloride ion and that the moles of chloride ion in the salt is 5.0237mmol, the no.of moles of the metal ion will be 5.0237mmol/2 = 2.51185mmol. Also, since we know that the salt is composed only of the metal ion, chloride ions and water, from the mass percentage of metal ion calculated as 100-(35.62+36.2) = 28.18% the mass of metal ion is obtained as (28.18/100)*0.5 = 0.1409g. Since we know that 0.1409g equals to 2.51185mmol of the metal ion, the molar mass of the metal ion is found as molar mass = mass/moles = 0.1409/2.51185mmol = 56.0941g/mol which is approximately that of iron. Thus the identity of the metal ion is Iron (II).

d) The water of hydration is required for the empirical formula now that we know the identity of our salt as iron (II) chloride. To find this, using the no.of moles of the anydrous salt which is 0.319g and molar mass of the anhydrous salt as 126.751g/mol, the no.of moles of the salt is 0.319/126.751 = 2.5167mmol and using the moles of water in the sample as 0.181/18.0153 = 10.047mmol and taking the ratio of moles of water in the salt to the moles of anhydrous salt, we get 10.047/2.5167 = 3.99 which can be approximated to 4.

Thus the empirical formula of the given salt is FeCl2.4H2O.

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