Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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 \(\mathrm{C}_{5} \mathrm{H}_{7} \mathrm{O}_{2}^{-}\) 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 \(\mathrm{Be}\left(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{O}_{2}\right)_{2} ;\) 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.

Short Answer

Expert verified
Based on Combes's data and the comparison of theoretical and experimental densities, it is confirmed that beryllium is a divalent metal, forming Be²⁺ ions and the oxide BeO, as also proposed by Mendeleev.

Step by step solution

01

Determine the molar mass of the potential molecular formulas

First, find the molar mass of the molecules for divalent and trivalent metal. For divalent metal (Be(C₅H₇O₂)₂): The molar mass can be calculated by adding up the molar masses of beryllium, carbon, hydrogen, and oxygen multiplied by their respective counts: Molar mass = 1(9.0) + 10(12.01) + 14(1.01) + 4(16.00) = 9.0 + 120.1 + 14.14 + 64 = 207.24 g/mol For trivalent metal (Be(C₅H₇O₂)₃): Molar mass = 1(9.0) + 15(12.01) + 21(1.01) + 6(16.00) = 9.0 + 180.15 + 21.21 + 96 = 306.36 g/mol
02

Calculate the theoretical density of the gaseous product for both scenarios

To calculate the theoretical density of the gaseous product, we can use the ideal gas law (PV = nRT). Rearrange the formula to solve for density(m/v): PV = nRT => PV = (mass/molar mass)RT => density = (P * molar mass) / (RT) Since we want to compare the theoretical densities to Combes's experiments, it's important to note that the experimental conditions were at standard temperature (273.15 K) and pressure (1 atm = 101325 Pa). For divalent metal (Be(C₅H₇O₂)₂): Density = (101325 * 207.24) / (8.314 * 273.15) = 21,189,775 / 2,273.087 = 9.321 g/L For trivalent metal (Be(C₅H₇O₂)₃): Density = (101325 * 306.36) / (8.314 * 273.15) = 31,091,615 / 2,273.087 = 13.681 g/L
03

Compare the theoretical densities to Combes's experimental data

Now we compare the theoretical densities calculated above to the data provided by Combes's experiments. Given data from experiments: Density(Divalent) = 9.321 g/L (Theoretical) Density(Trivalent) = 13.681 g/L (Theoretical) Experimental Density of the product: 9.169 g/L and 9.170 g/L Clearly, the experimental densities align more closely with the theoretical density of the divalent metal (9.321 g/L). Conclusion: Based on Combes's data, beryllium is confirmed as a divalent metal, forming Be²⁺ ions and the oxide BeO, as also proposed by Mendeleev.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Controversies
Chemistry, as a field, is not without its debates and historical controversies. One such debate involved the classification of beryllium in the 19th century. Johan August Arfwedson discovered beryllium in the mineral beryl in 1828, and Jöns Jakob Berzelius identified beryllium as an element. Initially, Berzelius thought beryllium was trivalent, meaning it would form ions with a +3 charge, resulting in a molar mass that we now know to be incorrect.

Such controversies are part of the scientific process. Disputations like these lead to further research and experimentation, which ultimately contribute to our understanding of the elements. A pivotal experiment by A. Combes in 1894 provided data that eventually settled this controversy by confirming beryllium as a divalent metal, thus supporting Mendeleev's periodic table. The meticulous approach demonstrating how empirical evidence aligns with theoretical predictions is a testament to the rigor and subtlety that is intrinsic to the practice of chemistry.

Through challenges and reassessments, scientific knowledge advances. Hence, such controversies, instead of hindering, serve to refine and strengthen scientific concepts over time.
Periodic Table
The periodic table is one of the cornerstones of chemistry, providing a comprehensive framework for categorizing all known elements. Dmitri Mendeleev, often credited with its creation, recognized the need for organization within the chemical elements based on their properties.

Mendeleev's periodic table was consequential in predicting the existence and properties of new elements. His proposal that beryllium was divalent and formed BeO contradicted earlier assumptions by Berzelius. Mendeleev's foresight was rooted in recognizing patterns within the groupings of elements and the valence of elements, which is crucial for their bonding behavior and compound formation.

This historical glimpse into Mendeleev's prediction underscores the periodic table's significance as a prediction tool and not just a classification system. Understanding an element's place in the table allows chemists to infer a wealth of information about its chemical behavior.
Molecular Formulas
Molecular formulas communicate the types and numbers of atoms present in a compound, reflecting its composition. They are essential not only for understanding what a substance is but how it interacts with other chemicals.

The controversy over beryllium highlights the importance of accurately determining molecular formulas. The molecular formula impacts the calculated molar mass which in turn plays a significant role in specific experiments and calculations in chemistry, like determining the density of a gaseous product, as seen in Combes's work.

Applying Molecular Formulas

When Combes experimented with beryllium, he demonstrated through the densities of gaseous compounds that beryllium's accurate molecular formula involved divalent ions (\(Be^{2+}\)). Understanding the correct molecular formula is critical in chemistry as it allows for accurate predictions and explanations of experimental results, just like Combes's results supported Mendeleev's arrangement in the periodic table and the divalent nature of beryllium.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A 20.0-L nickel container was charged with \(0.500\) atm of xenon gas and \(1.50\) atm of fluorine gas at \(400 .{ }^{\circ} \mathrm{C}\). The xenon and fluorine react to form xenon tetrafluoride. What mass of xenon tetrafluoride can be produced assuming \(100 \%\) yield?

Which noble gas has the smallest density at STP? Explain.

Consider two different containers, each filled with 2 moles of \(\mathrm{Ne}(\mathrm{g}) .\) One of the containers is rigid and has constant volume. The other container is flexible (like a balloon) and is capable of changing its volume to keep the external pressure and internal pressure equal to each other. If you raise the temperature in both containers, what happens to the pressure and density of the gas inside each container? Assume a constant external pressure.

One way of separating oxygen isotopes is by gaseous diffusion of carbon monoxide. The gaseous diffusion process behaves like an effusion process. Calculate the relative rates of effusion of \({ }^{12} \mathrm{C}^{16} \mathrm{O},{ }^{12} \mathrm{C}^{17} \mathrm{O}\), and \({ }^{12} \mathrm{C}^{18} \mathrm{O}\). Name some advantages and disad- vantages of separating oxygen isotopes by gaseous diffusion of carbon dioxide instead of carbon monoxide.

A sealed balloon is filled with \(1.00 \mathrm{~L}\) helium at \(23^{\circ} \mathrm{C}\) and \(1.00\) atm. The balloon rises to a point in the atmosphere where the pressure is 220 . torr and the temperature is \(-31^{\circ} \mathrm{C}\). What is the change in volume of the balloon as it ascends from \(1.00\) atm to a pressure of 220 . torr?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free