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One of the chemical controversies of the nineteenth century concerned the element beryllium (Be). Berzelius originally claimed that beryllium was a trivalent element (forming Be 3+ ions) and that it gave an oxide with the formula Be 2 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 Be 2+ ions) and that it gave an oxide with the formula Be 2 O 3 . 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 H 7 O 2 − and measured the density of the gaseous product. Combes’s data for two different experiments are as follows: I II Mass 0.2022 g 0.2224 g Volume 22.6 cm 3 26.0 cm 3 Temperature 13°C 17°C Pressure 765.2 mm Hg 764.6 mm If beryllium is a divalent metal, the molecular formula of the product will be Be(C 5 H 7 O 2 ) 2 ; if it is trivalent, the formula will be Be(C 5 H 7 O 2 ) 3 . Show how Combes’s data help to confirm that beryllium is a divalent metal.

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Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

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BuyFindarrow_forward

Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 8, Problem 138AE
Textbook Problem
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One of the chemical controversies of the nineteenth century concerned the element beryllium (Be). Berzelius originally claimed that beryllium was a trivalent element (forming Be3+ ions) and that it gave an oxide with the formula Be2O3. This resulted in a calculated atomic mass of 13.5 for beryllium. In formulating his periodic table, Mendeleev proposed that beryllium was divalent (forming Be2+ ions) and that it gave an oxide with the formula Be2O3. This assumption gives an atomic mass of 9.0. In 1894, A. Combes (Comptes Rendus 1894, p. 1221) reacted beryllium with the anion C5H7O2and measured the density of the gaseous product. Combes’s data for two different experiments are as follows:

  I II
Mass 0.2022 g 0.2224 g
Volume 22.6 cm3 26.0 cm3
Temperature 13°C 17°C
Pressure 765.2 mm Hg 764.6 mm

If beryllium is a divalent metal, the molecular formula of the product will be Be(C5H7O2)2; if it is trivalent, the formula will be Be(C5H7O2)3. Show how Combes’s data help to confirm that beryllium is a divalent metal.

Interpretation Introduction

Interpretation: From the given data, it should be confirmed that beryllium is a divalent element.

Concept introduction:

  • Ideal gas equation in terms of Density,

Density=Pressure×Molecular massR×Temperature

Explanation of Solution

Explanation

  • To find: The molar mass of data set 1

Molarmass=209g/mol

Ideal gas equation in terms of Density,

Density=Pressure×Molecular massR×TemperatureMolecular mass=Density×R×TemperaturePressureSince,Density=MassVolumeMolecular mass=Mass×R×Temperature(T)Pressure(P)×Volume(V)

Here,Mass=0.2022gR=0.08206LatmKmolT=13°C=286KK=°C+273=13°C+273=286KP=765.2mmHg=765.2torr=1.006atm1atm=760torr

765.0×1atm760=1.006atmV=22.6cm3=22.6×10-3Lsince,1L=1000cm3Molecular mass=0.20220.08206LatmKmol×286K1.006atm×22.6×10-3L=209g/mol

  • To find: The molar mass of data set 2

Molarmass=202g/mol

Density=Pressure×Molecular massR×TemperatureMolecular mass=Density×R×TemperaturePressureSince,Density=MassVolumeMolecular mass=Mass×R×Temperature(T)Pressure(P)×Volume(V)Here,Mass=0

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Chapter 8 Solutions

Chemistry: An Atoms First Approach
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