   Chapter 21, Problem 101MP

Chapter
Section
Textbook Problem

# There are three salts that contain complex ions of chromium and have the molecular formula CrCl3 · 6H2O. Treating 0.27 g of the first salt with a strong dehydrating agent resulted in a mass loss of 0.036 g. Treating 270 mg of the second salt with the same dehydrating agent resulted in a mass loss of 18 mg. The third salt did not lose any mass when treated with the same dehydrating agent. Addition of excess aqueous silver nitrate to 100.0-mL portions of 0.100 M solutions of each salt resulted in the formation of different masses of silver chloride; one solution yielded 1430 mg AgCl; another, 2870 mg AgCl; the third, 4300 mg AgCl. Two of the salts are green and one is violet.Suggest probable structural formulas for these salts, defending your answer on the basis of the preceding observations. State which salt is most likely to be violet. Would a study of the magnetic properties of the salts be helpful in determining the structural formulas? Explain.

Interpretation Introduction

Interpretation: The probable structural formulas for the salts on the basis of the given observations are to be stated. The salt that is most likely to be violet and whether a study of the magnetic properties of the salts be useful in determining the structural formulas are to be explained.

Concept introduction: Electronic configuration is used to describe the distribution of the electrons in the orbitals of an atom. Structure of an atom can be defined by its electronic configuration. It can also be used to denote an atom which is ionized to a cation or anion formed by the loss or gain of electrons in their respective orbitals.

To determine: The formula of the compound A and the structure of complex ion present on the basis of given information.

Explanation

Explanation

The molar mass of the complex CrCl3.6H2O =266.35g/mol .

The given mass of CrCl3.6H2O is 0.27g .

The number of moles of CrCl3.6H2O is calculated by the formula,

GivenmassMolarmass

Substitute the values of given mass and molar mass in the above equation.

0.27g266.35g/mol=1×103mol

Therefore, 1×103mol of CrCl3.6H2O have reacted.

Molar mass of water =18g/mol

Mass lost is 0.036g .

The number of moles of water of hydration is calculated by the formula,

Moles of hydration=MasslostMolarmassofwater

Substitute the values of mass lost and molar mass of water in the above equation.

Moles of hydration=0.036g18g/mol=2×103mol

The number of water molecules released is calculated by the formula,

Numberofwatermolecules=NumberofmolesofwaterofhydrationNumberofmolesofCrCl3.6H2O

Substitute the values of number of moles of water of hydration and number of moles of CrCl3.6H2O in the above equation.

Numberofwatermolecules=2×1031×103=2

Since two water molecules are released, therefore they are present as counter ions. Out of six water ligands, if two are counter ions , the other four water molecules must be attached to central metal as ligands. The complex is octahedral, thus it can accommodate only six ligands. If four water molecules are attached as ligand , then only two more chlorine atoms would be present as ligands. The one remaining chlorine molecule is present as counter ion.

Hence, the formula of first compound is [Cr(H2O)4Cl2].2H2O.Cl .

The molar mass of the complex CrCl3.6H2O =266.35g/mol .

The given mass of second compound having formula CrCl3.6H2O is 270mg .

The conversion of mg to g is done as,

1mg=103g

Therefore, the conversion of 270mg to g is done as,

270mg=270×103g=0.27g

Therefore, the mass of second compound is 0.27g .

The number of moles of second compound is calculated by the formula,

GivenmassMolarmass

Substitute the values of given mass and molar mass in the above equation

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