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: $$\begin{array}{lll}{\text { Mass }} & {0.2022 \mathrm{g}} & {0.2224 \mathrm{g}} \\ {\text { Volume }} & {22.6 \mathrm{cm}^{3}} & {26.0 \mathrm{cm}^{3}} \\ {\text { Temperature }} & {13^{\circ} \mathrm{C}} & {17^{\circ} \mathrm{C}} \\ {\text { Pressure }} & {765.2 \mathrm{mm} \mathrm{Hg}} & {764.6 \mathrm{mm}}\end{array}$$ 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

Convert the data to appropriate units

We need to convert the given data to the appropriate units to use in the Ideal Gas Law equation. 1. Volume: Convert cm³ to L 2. Temperature: Convert Celsius to Kelvin 3. Pressure: Convert mmHg to atm

02

Calculate the number of moles using the Ideal Gas Law equation

Using the converted data, plug the values into the Ideal Gas Law equation and solve for the number of moles (n) for both experiments.

03

Calculate the molar mass of the gaseous product from both experiments

Divide the mass (in g) by the number of moles (in mol) to get the molar mass of the gaseous product for both experiments.

04

Deduce the valency of beryllium

Using the molar mass of the gaseous product, deduce if beryllium is divalent or trivalent. For divalent formula, \(\mathrm{Be}\left(\mathrm{C}_{5} \mathrm{H}_{7} \mathrm{O}_2\right)_2\), molar mass: \(1 \times M_{\mathrm{Be}} + 2 \times M_{\mathrm{C_5H_7O_2}}\) For trivalent formula, \(\mathrm{Be}\left(\mathrm{C}_{5} \mathrm{H}_{7} \mathrm{O}_2\right)_3\), molar mass: \(1 \times M_{\mathrm{Be}} + 3 \times M_{\mathrm{C_5H_7O_2}}\) Where: - \(M_{\mathrm{Be}}\) is the molar mass of beryllium - \(M_{\mathrm{C_5H_7O_2}}\) is the molar mass of the anion \(\mathrm{C_5H_7O_2^-}\) Compare the calculated molar mass of the gaseous product with the divalent and trivalent formulas and determine which one is closer to the actual data. This will help confirm if beryllium is divalent or trivalent.

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