Chemistry: An Atoms First Approach
Chemistry: An Atoms First Approach
2nd Edition
ISBN: 9781305079243
Author: Steven S. Zumdahl, Susan A. Zumdahl
Publisher: Cengage Learning
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Textbook Question
Chapter 10, Problem 132IP

Anthraquinone contains only carbon, hydrogen, and oxygen. When 4.80 mg anthraquinone is burned, 14.2 mg CO2 and 1.65 mg H2O are produced. The freezing point of camphor is lowered by 22.3°C when 1.32 g anthraquinone is dissolved in 11.4 g camphor. Determine the empirical and molecular formulas of anthraquinone.

Expert Solution & Answer
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Interpretation Introduction

Interpretation: The empirical formula and molecular formula of Anthraquinone has to be determined.

Concept Introduction:

The ratio of the elements present in a compound and not the arrangement of the atoms are called as Empirical formula

Molecular formula is the representation of sum of number of atoms and molecules, not their arrangement in structure.

Answer to Problem 132IP

The empirical formula of Anthraquinone is C7H4O

The molecular formula of Anthraquinone is C14H8O2

Explanation of Solution

Record the data

Mass of Carbon dioxide                = 14mg

Mass of Water                              = 1.65mg

Freezing point of Camphor            = 22.3°C

Mass of Anthraquinone burned      = 4.80mg

Mass of Anthraquinone Dissolved = 1.32g

Mass of Camphor                         = 11.4g

To calculate the mass percent of Carbon, Hydrogen and Oxygen

Atomic mass of Carbon             = 12.01mg

Molar mass of Carbon dioxide   = 44.01mg

Atomic mass of Hydrogen         = 2.016mg

Molar mass of Water                 = 18.02mg

Mass of Carbon = 14.2mgCO2×12.01mgC44.01mgCO2=3.88mg

Mass percentage of Carbon = 3.88mg4.80mg×100

                                                     = 80.8%

Mass of Hydrogen = 1.65mgH22.016mgH18.02mgH2O=0.185mg

Mass percentage of Hydrogen = 0.185mg4.80mg×100

                                              = 3.85%

Mass percentage of Oxygen = 100.00-(80.8+3.85)

                                          = 15.4%

Mass percentage of Carbon            = 80.8%

Mass percentage of Water               = 3.85%

Mass percentage of Oxygen          = 15.4%

To calculate the empirical formula

Mass percentage of Carbon      = 80.8%

Mass percentage of Hydrogen   = 3.85%

Mass percentage of Oxygen     = 15.4%

Out of 100.00g ,

80.8g1mol12.01g=6.73molC=6.730.963=6.9973.85g1mol1.008g=3.82molH=3.820.963=3.97415.4g1mol16.00g=0.963molO=0.9630.963=1.00 ~

Therefore, the empirical formula is C7H4O

Record the given info

Freezing point of Camphor            = 22.3°C

Molal freezing point depression constant = 40.°C/molal

Mass of Anthraquinone burned      = 4.80mg

Mass of Anthraquinone Dissolved = 1.32g

Mass of Camphor                         = 11.4g

To calculate the mass of Anthraquinone (m),

ΔTf=Kfmm=ΔTfKf=22.3°C40.°C/molal=0.56molal

Molal of Anthraquinone = 0.56molal

Moles of Anthraquinone = 0.0114kgcamphor×0.56molanthraquinonekgcamphor

                                     = 6.4×10-3mol

Moles of Anthraquinone = 6.4×10-3mol

To calculate the molar mass

Moles of Anthraquinone = 6.4×10-3mol

Mass of Anthraquinone Dissolved = 1.32g

Molar mass of Anthraquinone = 1.3g6.4×10-3mol

                                              = 210gmol-1

To determine the Molecular formula of Anthraquinone

Empirical formula mass = 104gmol-1

Molar mass of Anthraquinone = 210gmol-1

Molar mass of Anthraquinone is twice the empirical mass of Anthraquinone, therefore the molecular mass of Anthraquinone is C14H8O2

Molecular Mass of Anthraquinone = C14H8O2

Conclusion

The moles of individual elements were calculated by using the mass percentages to their molar masses. The moles of the individual elements were divided by smallest ratio of moles and approximated to determine the empirical formula. The empirical formula was found to be C7H4O .

The molecular mass of Anthraquinone is calculated by using the molar mass and empirical formula mass. The molecular mass of Anthraquinone was C14H8O2

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Chapter 10 Solutions

Chemistry: An Atoms First Approach

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    Solutions: Crash Course Chemistry #27; Author: Crash Course;https://www.youtube.com/watch?v=9h2f1Bjr0p4;License: Standard YouTube License, CC-BY