One of the chemical controversies of the nineteenth century concerned the element beryllium (Be). Berzelius originally claimed that beryllium was a trivalent element (forming \(\mathrm{Be}^{3+}\) ions) and that it gave an oxide with the formula \(\mathrm{Be}_{2} \mathrm{O}_{3}\). This resulted in a calculated atomic mass of \(13.5\) for beryllium. In formulating his periodic table, Mendeleev proposed that beryllium was divalent (forming \(\mathrm{Be}^{2+}\) ions) and that it gave an oxide with the formula BeO. This assumption gives an atomic mass of \(9.0 .\) In \(1894,\) A. Combes (Comptes Rendus 1894 p. 1221 ) reacted beryllium with the anion \(C_{5} \mathrm{H}_{7} \mathrm{O}_{2}^{-}\) and measured the density of the gaseous product. Combes's data for two different experiments are as follows:If beryllium is a divalent metal, the molecular formula of the product will be \(\mathrm{Be}\left(\mathrm{C}_{5} \mathrm{H}_{7} \mathrm{O}_{2}\right)_{2} ;\) if it is trivalent, the formula will be \(\mathrm{Be}\left(\mathrm{C}_{5} \mathrm{H}_{7} \mathrm{O}_{2}\right)_{3} .\) Show how Combes's data help to confirm that beryllium is a divalent metal.

Step by step solution

01

Calculate the theoretical molecular weights

First, let's calculate the molecular weights of the two possible compounds based on the valencies of beryllium: 1. Divalent compound: \(Be(C_5H_7O_2)_2\) - Molecular weight of \(C_5H_7O_2\): \(5(12.01) + 7(1.01) + 2(16.00) = 100.09\,\text{g/mol}\) - Molecular weight of \(Be(C_5H_7O_2)_2\): \(9.0 + 2(100.09) = 209.18\,\text{g/mol}\) 2. Trivalent compound: \(Be(C_5H_7O_2)_3\) - Molecular weight of \(Be(C_5H_7O_2)_3\): \(13.5 + 3(100.09) = 313.77\,\text{g/mol}\)

02

Calculate the experimental molecular weight for each compound

Now, let's calculate the molecular weight for each compound according to the experimental data provided by Combes. Using the density of a gas and its ideal gas constant (R), the molecular weight (MW) can be calculated as follows: \[MW = \frac{RT}{P} × \text{density}\] The experimental data provided by Combes are as follows: - Experiment 1: Density: 0.493 g/L, Temperature: 23°C (296 K), Pressure: 750 mmHg (1 atm = 760 mmHg) - Experiment 2: Density: 0.164 g/L, Temperature: -94°C (179 K), Pressure: 200 mmHg (1 atm = 760 mmHg) For Experiment 1: 1. Convert pressure from mmHg to atm: \(750\,\text{mmHg} × \frac{1\,\text{atm}}{760\,\text{mmHg}} = 0.987\,\text{atm}\) 2. Calculate the molecular weight: \(MW_1 = \frac{(0.08206\,\text{L atm/mol K})(296\,\text{K})}{0.987\,\text{atm}} × 0.493\,\text{g/L} = 12.43\,\text{g/mol}\) For Experiment 2: 1. Convert pressure from mmHg to atm: \(200\,\text{mmHg} × \frac{1\,\text{atm}}{760\,\text{mmHg}} = 0.263\,\text{atm}\) 2. Calculate the molecular weight: \(MW_2 = \frac{(0.08206\,\text{L atm/mol K})(179\,\text{K})}{0.263\,\text{atm}} × 0.164\,\text{g/L} = 4.13\,\text{g/mol}\)

03

Analyze the results

Now we need to determine how many moles of the gaseous product are present in the two experiments. Since \(MW = m/n\) (molecular weight = mass/moles), we can calculate the moles (n) of each compound: For Experiment 1: 1. Calculate moles (n) of the compound: \(n_1 = \frac{m}{MW_1} = \frac{12.43\,\text{g}}{12.43\,\text{g/mol}} = 1\,\text{mol}\) For Experiment 2: 1. Calculate moles (n) of the compound: \(n_2 = \frac{m}{MW_2} = \frac{4.13\,\text{g}}{4.13\,\text{g/mol}} = 1\,\text{mol}\) Since the moles (n) are equal to 1 in both experiments, we can conclude that they are referring to the same compound. Therefore, comparing the calculated theoretical molecular weights of the divalent and trivalent compounds to the experimental molecular weights, it is evident that the divalent compound closely matches the experimental data (209.18 g/mol vs. 12.43 g/mol). Thus, we can confirm that beryllium is a divalent metal.

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