Ammonia and potassium iodide solutions are added to an...

Ammonia and potassium iodide solutions are added to an aqueous solution of $\mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3} .$ A solid is isolated (compound $\mathrm{A}$ ), and the following data are collected: i. When $0.105 \mathrm{~g}$ of compound $\mathrm{A}$ was strongly heated in excess $\mathrm{O}_{2}, 0.0203 \mathrm{~g} \mathrm{CrO}_{3}$ was formed. ii. In a second experiment it took $32.93 \mathrm{~mL}$ of $0.100 \mathrm{M} \mathrm{HCl}$ to titrate completely the $\mathrm{NH}_{3}$ present in $0.341 \mathrm{~g}$ compound $\mathrm{A}$. iii. Compound A was found to contain $73.53 \%$ iodine by mass. iv. The freezing point of water was lowered by $0.64^{\circ} \mathrm{C}$ when $0.601$ g compound $A$ was dissolved in $10.00 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}$ $\left(K_{\mathrm{f}}=1.86^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol}\right)$ What is the formula of the compound? What is the structure of the complex ion present? (Hints: $\mathrm{Cr}^{3+}$ is expected to be sixcoordinate, with $\mathrm{NH}_{3}$ and possibly $\mathrm{I}^{-}$ as ligands. The $\mathrm{I}^{-}$ ions will be the counterions if needed.)

Ammonia and potassium iodide solutions are added to an aqueous solution of $\mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3} .$ A solid is isolated (compound $\mathrm{A}$ ), and the following data are collected:
i. When $0.105 \mathrm{~g}$ of compound $\mathrm{A}$ was strongly heated in excess $\mathrm{O}_{2}, 0.0203 \mathrm{~g} \mathrm{CrO}_{3}$ was formed.
ii. In a second experiment it took $32.93 \mathrm{~mL}$ of $0.100 \mathrm{M} \mathrm{HCl}$ to titrate completely the $\mathrm{NH}_{3}$ present in $0.341 \mathrm{~g}$ compound $\mathrm{A}$.
iii. Compound A was found to contain $73.53 \%$ iodine by mass. iv. The freezing point of water was lowered by $0.64^{\circ} \mathrm{C}$ when $0.601$ g compound $A$ was dissolved in $10.00 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}$ $\left(K_{\mathrm{f}}=1.86^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol}\right)$
What is the formula of the compound? What is the structure of the complex ion present? (Hints: $\mathrm{Cr}^{3+}$ is expected to be sixcoordinate, with $\mathrm{NH}_{3}$ and possibly $\mathrm{I}^{-}$ as ligands. The $\mathrm{I}^{-}$ ions will be the counterions if needed.)

Key Concepts

Coordination Complex

A coordination complex is a structure consisting of a metal ion bonded to surrounding molecules or ions known as ligands. In inorganic chemistry, understanding the formation, composition, and reactivity of such complexes is crucial because they exhibit unique geometries and electronic properties that influence their chemical behavior and function.

Coordination Number and Geometry

The coordination number refers to the number of ligand atoms that are bonded directly to the central metal ion. For many transition metals, including chromium(III), a coordination number of six is common, typically resulting in an octahedral geometry. This concept is central to predicting the structure and reactivity of the complex formed in a reaction.

Ligands and Counterions

Ligands are molecules or ions that donate a pair of electrons to the central metal ion, forming coordinate covalent bonds. In the context of coordination complexes, some ions can act as ligands while others serve as counterions to balance the charge of the complex. The distinction between ligands within the coordination sphere and counterions outside it is essential for determining the overall structure and composition of the complex.

Analytical Techniques in Inorganic Chemistry

Determining the empirical formula of a coordination compound often involves a combination of analytical techniques. Techniques such as gravimetric analysis, which measures the mass of a decomposition product, and titrimetric analysis, which quantifies specific components like ammonia, are employed together with physical property measurements to deduce the composition of the complex.

Gravimetric Analysis

Gravimetric analysis involves isolating and weighing a compound or its decomposition product to determine the amount of a particular species in a sample. In the context of coordination compounds, converting the metal ion into a known form (such as an oxide) allows for the calculation of the metal content, which is crucial when reconstructing the overall chemical formula.

Titration in Analytical Chemistry

Titration is an analytical technique where a solution of known concentration is used to determine the concentration of an analyte by reacting with it. In coordination chemistry, titration can be employed to quantify ligands or counterions, such as ammonia, providing key information about the number of ligand molecules present in or outside the coordination sphere.

Colligative Properties

Colligative properties, such as freezing point depression or boiling point elevation, depend on the number of solute particles in a solution rather than their identity. These properties are useful for determining the molar mass of a compound, as they provide insights into the extent of solute dissociation and thereby help in confirming the empirical formula of a coordination compound.

Transcript

00:01 This problem allows us to understand how we would find the empirical formula of a metal complex ion using our knowledge of mass percent and moles.

00:12 And from this point, we can figure out, based on our metal formula, the freezing point of those compounds.

00:22 The way we'll start this problem is we'll find the mass percent of chromium, because remember, the ideal metal that we're dealing with right now is chromium.

00:30 We'll start with 0 .023 grams of our chromium compound, and we'll multiply that by the fact that we have 100 grams of that compound, and 52 grams of that compound is chromium.

00:46 We'll find that the mass of chromium is 0 .0106 grams of chromium.

00:53 So then we can find the mass percent by taking .0106 grams divided by .105 grams, and we'll multiply that by 100, and we'll find that we have 10 .1 % of chromium.

01:10 So now we can, before we find the mass percent of our other elements, we have to find the amount of nh3 or ammonia that we have in our compound.

01:24 So we start with 32 .93 milliliters of hcl, hydrogen, pardon me, hydrochloric acid.

01:32 And we know in every one milliliter, we have 0 .1 millimoles of hcl.

01:37 And based on our molar ratio, we have 1 milamol of ammonia per every 1 milamol of hydrochloric acid.

01:44 And in every 1 millimole of ammonia, we have 17 .04 milligrams of ammonia.

01:51 So we can do the stoichiometry to find that we have 56 .1 milligrams of nh3.

01:59 And now we can find the mass percent of nh3 by taking 56 .1 milligrams, dividing that by 341 milligrams, and multiplying by 100, and we'll find that we have 16 .5 percent of ammonia.

02:17 And now we can find the moles of each element because we have the mass percent of each of our elements.

02:23 We'll start with chromium.

02:25 We have 10 .1 grams of chromium.

02:28 And in every one mole of chromium, we have 52 grams.

02:32 So we'll find that we have 0 .194 moles of chromium.

02:37 We'll do that again for ammonia.

02:40 We start with 16 .5 grams of ammonia.

02:43 We'll multiply that by the fact that in every one mole, we have 17 .03 grams.

02:49 So we'll find that we have 0 .969 moles of ammonia.

02:53 And then we'll do it for iodine.

02:56 We start with 74 .54 grams of iodine.

03:00 Multiply that by in every one mole.

03:03 We have 126 .9 grams.

03:05 So we'll find that we have 0 .5794 moles of iodine.

03:12 And now what we want to do to find our formula is find the molar ratios between these elements, because that will tell us the number of atoms in each element, essentially.

03:25 And the way we're going to do that, we'll take our moles from before and divide it by the lowest number of moles we calculated, which was 0 .194.

03:35 So we'll first start out with chromium.

03:39 We have 0 .194 moles and we'll divide it by itself...

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