Compounds containing ruthenium(II) and bipyridine, C10 H8 N2, have received considerable intere

step 1 Compounds containing ruthenium and bipyridine. Bipyridine is this carbon compound. They have rec

Compounds containing ruthenium(II) and bipyridine, C10 H8...

Compounds containing ruthenium(II) and bipyridine, $\mathrm{C}_{10} \mathrm{H}_{8} \mathrm{~N}_{2},$ have received considerable interest because of their role in systems that convert solar energy to electricity. The compound $\left[\mathrm{Ru}\left(\mathrm{C}_{10} \mathrm{H}_{8} \mathrm{~N}_{2}\right)_{3}\right]$$\mathrm{Cl}_{2}$ is synthesized by reacting $\mathrm{RuCl}_{3} \cdot 3 \mathrm{H}_{2} \mathrm{O}(s)$ with three molar equivalents of $\mathrm{C}_{10} \mathrm{H}_{8} \mathrm{~N}_{2}(s),$ along with an excess of triethylamine, $\mathrm{N}\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{3}(l),$ to convert ruthenium(III) to ruthenium(II). The density of triethylamine is $0.73 \mathrm{~g} / \mathrm{mL},$ and typically eight molar equivalents are used in the synthesis. (a) Assuming that you start with $6.5 \mathrm{~g}$ of $\mathrm{RuCl}_{3} \cdot 3 \mathrm{H}_{2} \mathrm{O},$ how many grams of $\mathrm{C}_{10} \mathrm{H}_{8} \mathrm{~N}_{2}$ and what volume of $\mathrm{N}\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{3}$ should be used in the reaction? (b) Given that the yield of this reaction is 91 percent, how many grams of $\left[\mathrm{Ru}\left(\mathrm{C}_{10} \mathrm{H}_{8} \mathrm{~N}_{2}\right)_{3}\right] \mathrm{Cl}_{2}$ will be obtained?

Compounds containing ruthenium(II) and bipyridine, $\mathrm{C}_{10} \mathrm{H}_{8} \mathrm{~N}_{2},$ have received considerable interest because of their role in systems that convert solar energy to electricity. The compound $\left[\mathrm{Ru}\left(\mathrm{C}_{10} \mathrm{H}_{8} \mathrm{~N}_{2}\right)_{3}\right]$$\mathrm{Cl}_{2}$ is synthesized by reacting $\mathrm{RuCl}_{3} \cdot 3 \mathrm{H}_{2} \mathrm{O}(s)$ with
three molar equivalents of $\mathrm{C}_{10} \mathrm{H}_{8} \mathrm{~N}_{2}(s),$ along with an excess of triethylamine, $\mathrm{N}\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{3}(l),$ to convert ruthenium(III) to ruthenium(II). The density of triethylamine is $0.73 \mathrm{~g} / \mathrm{mL},$ and typically eight molar equivalents are used in the synthesis.
(a) Assuming that you start with $6.5 \mathrm{~g}$ of $\mathrm{RuCl}_{3} \cdot 3 \mathrm{H}_{2} \mathrm{O},$ how many grams of $\mathrm{C}_{10} \mathrm{H}_{8} \mathrm{~N}_{2}$ and what volume of $\mathrm{N}\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{3}$
should be used in the reaction?
(b) Given that the yield of this reaction is 91 percent, how many grams of $\left[\mathrm{Ru}\left(\mathrm{C}_{10} \mathrm{H}_{8} \mathrm{~N}_{2}\right)_{3}\right] \mathrm{Cl}_{2}$ will be obtained?

Key Concepts

Stoichiometry

Stoichiometry involves using the mole concept to relate quantities of reactants and products in a chemical reaction. It requires converting masses to moles using molar masses, understanding the molar ratios of reactants as specified by the balanced chemical equation, and then determining how much of each substance is required or produced. This concept is crucial to any synthetic procedure, as it ensures that the correct proportions of chemicals are used to drive the reaction efficiently.

Coordination Complex Synthesis

Coordination complex synthesis deals with the formation of compounds where a central metal atom is bonded to surrounding ligands. This discipline covers ligand selection, stoichiometry in the formation of these complexes, and the necessary conditions to facilitate their formation. Understanding the principles behind the stabilization of different oxidation states of the central metal ion and the electronic and steric properties of the ligands is foundational in designing and synthesizing coordination compounds.

Redox Reactions

Redox reactions, or oxidation-reduction reactions, involve the transfer of electrons between species, leading to changes in the oxidation states of the elements involved. In such reactions, one species is reduced (gains electrons) while the other is oxidized (loses electrons). This concept is key when a metal ion must change its oxidation state to form the desired coordination compound, as it impacts the reactivity and formation of the product.

Percent Yield Calculation

Percent yield is a measure of the efficiency of a chemical reaction, calculated as the ratio of actual product obtained to the theoretical maximum product expected, multiplied by 100. This concept is essential for evaluating and optimizing chemical syntheses, as it reflects practical limitations like incomplete reactions, side reactions, or losses during workup. Understanding percent yield helps in refining reaction conditions to improve overall productivity.

Density and Volume Conversion

The concept of density links the mass of a substance to its volume, expressed as mass per unit volume. This relationship is particularly useful when dealing with liquids in chemical reactions, where volumes are measured but conversions to mass may be necessary for stoichiometric calculations. Accurate density and volume conversions ensure that the quantities of reagents are correctly accounted for in the reaction setup.

Transcript

00:01 Compounds containing ruthenium and bipyridine.

00:03 Bipyridine is this carbon compound.

00:06 They have received considerable interest because of their role in systems that convert solar energy to electricity.

00:12 The compound of ruthenium and chlorine and bipyridine is synthesized by reacting this ruthenium compound with three molar equivalents of the bipyridine, so that's why we have this three along with an excess of triethylamine.

00:28 Triethylamine is this liquid compound here with the nitrogen.

00:33 So to convert ruthenium to ruthenium -3 to ruthenium -2, the density of the triethylamine is 0 .73 grams per milliliter and typically eight molar equivalents, so that's why we have an eight here, are used in the synthesis.

00:50 So for part a, now we're assuming that we're how many grams of the bipyridine and the volume of the triethylamine should be used in the reaction.

01:05 So this is a limiting reactive problem.

01:07 We're going to be starting with 6 .5 grams of our ruthenium hydrated.

01:13 So i'm just going to call it ruthenium and then water.

01:17 I'm shortening it just because it's going to be very, very long.

01:20 So from here, we're going to convert it to moles, one mole over the molecular weight.

01:26 Now i went ahead and calculated these earlier, so the molecular weight we're going to be using here is 261 .448 grams.

01:36 Now this is moles of our hydrated compound.

01:40 We want to go to grams of our bipyridine, so we're going to say one mole of our ruthenium compound will give or will need three moles of our carbon compound.

01:54 So i'm just going to shorten that to the c10 compound.

01:57 We need to go to grams, so let's put one mole of our c10, our bipyridine on the bottom, and the molecular weight for that is 156 .184 grams.

02:09 So let's do some math.

02:12 We have 6 .5 divided by 261 .448.

02:18 From there, we're going to multiply it by 3 and then multiply it by 156 .184, and this is going to be 11 .6489, or approximately 11 .6 grams of our bipyridine.

02:35 So let's write that in.

02:38 Now we have to go to the density or the volume of our triethylamine.

02:43 So we're going to start with 6 .5 grams of our ruthenium hydrate.

02:49 We're first two steps are the same.

02:58 Next, we have one mole of our hydrated compound to eight moles of our nc2h5, our triethylamine.

03:12 So i'm just going to say tri -e.

03:15 Let's erase that and let's make it a little bit neater.

03:20 So we're going to label this as tri -e.

03:23 Now we're in moles, so let's go to grams since that's what our density is in, and we're looking for volume.

03:32 So one mole of our triethylamine...