Acetylacetone (see Exercise 69, part a), abbreviated acacH,...

Acetylacetone (see Exercise 69, part a), abbreviated acacH, is a bidentate ligand. It loses a proton and coordinates as acac $^{-}$, as shown below: Acetylacetone reacts with an ethanol solution containing a salt of europium to give a compound that is $40.1 \% \mathrm{C}$ and $4.71 \% \mathrm{H}$ by mass. Combustion of $0.286 \mathrm{~g}$ of the compound gives $0.112 \mathrm{~g}$ $\mathrm{Eu}_{2} \mathrm{O}_{3}$. Assuming the compound contains only $\mathrm{C}, \mathrm{H}, \mathrm{O}$, and $\mathrm{Eu}$, determine the formula of the compound formed from the reaction of acetylacetone and the europium salt. (Assume that the compound contains one europium ion.)

Key Concepts

Coordination Chemistry

Coordination chemistry is the branch of chemistry focused on the study of complexes formed between central metal ions and surrounding ligands. It explores the structure, bonding, reactivity, and properties of these coordination compounds, emphasizing the role of ligands in determining complex geometry and stability.

Bidentate Ligands

Bidentate ligands are molecules that have two sites capable of bonding to a metal center. Their dual attachment points lead to the formation of chelate rings, which generally increase the stability of the resulting metal complex due to the chelate effect.

Empirical Formula Determination

Empirical formula determination involves using experimental data such as percent composition and product mass from quantitative analyses to derive the simplest whole-number ratio of elements present in a compound. This process is key in identifying the composition of unknown substances.

Combustion Analysis

Combustion analysis is an analytical technique used to determine the elemental composition of a compound. By burning a sample and measuring the mass of the combustion products, one can calculate the percentages of elements like carbon and hydrogen, which then aids in empirical formula determination.

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions. It provides the quantitative relationship between substances based on the law of conservation of mass, allowing chemists to deduce formula relationships and predict yields in chemical reactions.

Transcript

00:01 This problem allows us to practice an very important skill, which is using the knowledge we know about mass and mass percent in order to calculate the empirical formula of a structure.

00:14 And more importantly, we're going to combine that formula with our concepts of transition metals and ligands.

00:21 The first thing that we should do is find the mass of uropium.

00:24 So we're given, we have 0 .112 grams of our uropium compound.

00:30 And we know that we have 352 grams in our europium compound, and 304 grams of that is just purely european.

00:40 So when we do that calculation, we'll find that the mass of europium is 0 .0967 grams.

00:49 And from here, we can calculate the mass percent of europium.

00:53 We'll take the mass we just found .0967 grams, and we'll divide that by the total mass, and 286 grams and multiply that by 100.

01:05 And we find that our mass percent of europium is 33 .8%.

01:12 And now from here, we can determine the mass percent of oxygen by using the mass percent of the other elements in our empirical formula.

01:23 We'll take 100 minus the fact that we have 33 .8 percent of and we also have plus 41, pardon me, 40 .1 % of carbon plus 4 .71 % of hydrogen.

01:38 When we do this calculation, we'll find that we have 21 .4 % oxygen.

01:45 And this is extremely helpful to know because now we can find the moles of each element if we are assuming we have a 100 gram sample.

01:55 For europium, we'll take 33 .8 grams of european, and we know that in every, one mole of europa, we have 152 grams.

02:04 So we'll find that we have 0 .22 moles of europa.

02:09 We'll do the same thing for all the other elements.

02:13 We have 40 .1 grams of carbon.

02:15 We know that in every one mole of carbon, we have 12 .01 grams.

02:20 So when we do this calculation, we'll have 3 .34 moles of carbon.

02:26 And again, for hydrogen, we start with 4 .71 grams...

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