   Chapter 11, Problem 132IP

Chapter
Section
Textbook Problem

# 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.

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.

Explanation

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

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