Chapter 4, Problem 33PS

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Chapter
Section

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

# Naphthalene is a hydrocarbon that once was used in mothballs. If 0.3093 g of the compound is burned in oxygen, 1.0620 g of CO2 and 0.1739 g of H2O are isolated.(a) What is the empirical formula of naphthalene?(b) If a separate experiment gave 128.2 g/mol as the molar mass of the compound, what is its molecular formula?

(a)

Interpretation Introduction

Interpretation:

The empirical and molecular formula of naphthalene should be determined.

Concept introduction:

Empirical formula of a compound represents the smallest whole number relative ratio of elements in that compound.

Explanation

If 0.3093â€‰g of the compound is burned in oxygen, Â 1.0620g of CO2â€‰andâ€‰0.1739â€‰gâ€‰H2O can be isolated.

From the given masses the amount of CO2â€‰andâ€‰H2O can be calculated.

Thus, amount of CO2â€‰andâ€‰H2O isolated from the combustion of naphthalene are,

â€‚Â 1.0620gâ€‰CO2â€‰Ã—â€‰1â€‰molâ€‰CO244.010â€‰gâ€‰CO2â€‰=â€‰0.02413â€‰â€‰molâ€‰CO20.1739â€‰gâ€‰H2Oâ€‰Ã—1â€‰molâ€‰H2O18.015â€‰gâ€‰H2Oâ€‰=â€‰0.00965â€‰molâ€‰H2O

For every mole of CO2 isolated, 1 mol of C must have been present in naphthalene.Â  So,

â€‚Â 0.02413â€‰molâ€‰CO2â€‰Ã—1â€‰molâ€‰Câ€‰inâ€‰unknown1â€‰molâ€‰CO2â€‰=â€‰0

(b)

Interpretation Introduction

Interpretation:

The molecular formula of naphthalene has to be determined.

Concept Introduction:

Molecular formula of a compound is given as,

MolarmassEmpiricalformula mass × Empirical formula

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