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Chapter 4: Problem 128

A 5.012 -g sample of an iron chloride hydrate was dried in an oven. The mass of the anhydrous compound was \(3.195 \mathrm{~g}\). The compound was then dissolved in water and reacted with an excess of \(\mathrm{AgNO}_{3} .\) The AgCl precipitate formed weighed 7.225 g. What is the formula of the original compound?

Short Answer

Expert verified
The formula of the original compound is FeCl2

Step by step solution

01

Finding the amount of Fe

Evaluating the amount of Fe in the compound can be done by subtracting the weight of the anhydrous compound from the weight of the hydrate. Given the weight of the hydrate as \(5.012 \mathrm{~g}\) and the anhydrous compound as \(3.195 \mathrm{~g}\), the weight of Fe can then be found as \(5.012 \mathrm{~g} - 3.195 \mathrm{~g} = 1.817 \mathrm{~g}\)
02

Finding the amount of Cl

Since the AgCl precipitate formed in reaction contains Cl, its amount can be used to find the weight of Cl in the original compound. The molecular weight of AgCl is \(143.32 \mathrm{~g/mol}\). Having this information, the weight of Cl can then be found by multiplying the mass of the AgCl precipitate with the atomic weight ratio of Cl to AgCl. Thus, the weight of Cl is \((7.225 \mathrm{~g}) \times \frac{35.45 \mathrm{~g/mol}}{143.32 \mathrm{~g/mol}} = 1.799 \mathrm{~g}\)
03

Finding the amount of H2O

The weight of water is the difference between the weight of the anhydrous compound and the weight of Fe and Cl combined. Thus, the weight of water is \(3.195 \mathrm{~g} - (1.817 \mathrm{~g} + 1.799 \mathrm{~g}) = -0.421 \mathrm{~g}\). There is no water attached to the FeCl present.
04

Finding the formula of the compound

With the obtained weights of Fe and Cl, it is now possible to determine the formula of the anhydrous compound. The formula of the compound can be found to be FeCl2 based on their proportions in mass, since the mass of Fe and Cl are roughly in the 1:1 ratio.

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