Question

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.)

Short answer

To determine the formula of the compound formed from the reaction of acetylacetone and the europium salt, we first find the mass percentages of O and Eu by subtracting the given percentages of C and H from 100%. Then, we calculate the mass of each element in the given sample and determine the moles of each element using their respective molar masses. Finally, we divide the moles of each element by the smallest number of moles (which is the number of moles of Eu) to find a whole number ratio and obtain the empirical formula of the compound as EuC_xH_yO_z, where x, y, and z are the whole number ratios.

Step-by-step solution

Calculate the mass percentages of oxygen and europium

First, we will find the mass percentages of O and Eu by subtracting the given percentages of C and H from 100%. Percentage of O + Percentage of Eu = 100% - (Percentage of C + Percentage of H) Percentage of O + Percentage of Eu = 100% - (40.1% + 4.71%)

Calculate the mass of each element in the given sample

Now we will calculate the mass of each element (C, H, O, and Eu) in the given sample of 0.286 g of compound. 1. Mass of C = 0.286 g × (40.1/100) = \(0.286 \times 0.401\) g 2. Mass of H = 0.286 g × (4.71/100) = \(0.286 \times 0.0471\) g 3. Since combustion of the 0.286 g compound gives 0.112 g Eu2O3, the mass of oxygen in Eu2O3 can be calculated as: Mass of Eu = 2 × molar mass of Eu = 2 × 151.964 g/mol (from the periodic table) Mass of O in Eu2O3 = 3 × molar mass of O = 3 × 16 g/mol Thus, Mass of Eu2O3 = Mass of 2Eu + Mass of 3O 0.112 g = (2 × 151.964) + (3 × 16) ⇒ Mass of Eu = 0.0655 g 4. We can now calculate the mass of O in the sample as the difference between the total mass and the mass of other elements. Mass of O = 0.286 g - (Mass of C + Mass of H + Mass of Eu)

Calculate the moles of each element in the given sample

Now we will calculate the moles of each element in the given sample using their respective molar masses. 1. Moles of C = Mass of C / Molar mass of C = (0.286 × 0.401) / 12.01 g/mol 2. Moles of H = Mass of H / Molar mass of H = (0.286 × 0.0471) / 1.008 g/mol 3. Moles of O = Mass of O / Molar mass of O (calculated from mass of O in step 2) 4. Moles of Eu = Mass of Eu / Molar mass of Eu = 0.0655 g / 151.964 g/mol

Determine the formula of the compound

To determine the formula of the compound, we will divide the moles of each element by the smallest number of moles (which is the number of moles of Eu) to find a whole number ratio. 1. Moles of C / Moles of Eu 2. Moles of H / Moles of Eu 3. Moles of O / Moles of Eu 4. Moles of Eu / Moles of Eu Round the ratios to the nearest whole numbers to get the empirical formula of the compound formed from the reaction of acetylacetone and the europium salt. Finally, the formula of the compound will be EuC_xH_yO_z, where x, y, and z are the whole number ratios obtained in Step 4.

Key concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that deals with the quantitative relationship between reactants and products in a chemical reaction. It's like a recipe for chemistry, where you need specific proportions of ingredients (reactants) to make the desired dish (product). In stoichiometry, we use the chemical equation and the mole concept to determine how much of each substance is involved. When solving a stoichiometric problem, we need to follow a systematic approach:

• Determine the balanced chemical equation for the reaction.
• Convert all given information into moles (using molar masses from the periodic table).
• Use the mole ratio from the balanced equation to find the number of moles of the unknown substance.
• Convert the moles of the unknown back into the desired units (grams, liters, molecules, etc.).

In the given exercise, stoichiometry helps us deduce the chemical formula of a compound formed by acetylacetone and a europium salt. By calculating the moles of each element present and comparing them, we can determine the empirical formula, which reflects the simplest whole-number ratio of atoms in a compound. It's essential to maintain a clear view of each step, ensuring proportions and units are consistent for a precise answer.

Combustion Analysis

Combustion analysis is an analytical technique used to determine the elemental composition of a substance through its complete burning. During the combustion process, carbon and hydrogen in the compound are converted into carbon dioxide and water, respectively. By measuring the masses of CO₂ and H₂O produced, we can back-calculate to find the amounts of carbon and hydrogen in the original sample.Most combustion analyses will also consider the production of oxides from other non-metals or metal salts in the sample. This extra information is crucial for compounds containing elements beyond C and H, such as oxygen or metals, like in the exercise provided. Here's how you would typically proceed:

• Burn the sample in a controlled environment to ensure complete combustion.
• Measure the masses of the products (such as CO₂, H₂O, and metal oxides).
• Calculate the masses of C, H, and other elements by stoichiometry.
• Convert these masses to moles and then to percentages to determine the empirical formula.

In our exercise, the combustion of the europium compound produces Eu₂O₃. Through meticulous measurement and calculation, the exercise leads us to the mass of europium in the compound, allowing us to deduce the other components' proportions and ultimately, the compound's empirical formula.

Bidentate Ligand Coordination

Bidentate ligands are versatile molecules that can form two bonds with a central metal atom or ion in coordination chemistry. They operate like two-handed clamps, grabbing onto the metal at two different points. When a ligand coordinates with a metal atom in this way, it forms what we call a chelate complex, which tends to be more stable than complexes with monodentate ligands (ones that attach at only one point).In the context of our exercise, acetylacetone (acacH) acts as a bidentate ligand. Once it loses a proton (H⁺), it transforms into acac^- and coordinates with europium (Eu³⁺). To visualize this, imagine acacH as a creature with two arms; upon losing a proton, those arms become free to hold onto the europium ion simultaneously. This mode of coordination is significant because it affects the stoichiometry of the complex. For instance, the number of acac^- ligands surrounding one europium ion affects the compound's composition and, therefore, its empirical formula. Understanding the role of bidentate ligands enables us to piece together the molecular puzzle and unveil the structure of the complex.