   Chapter 4, Problem 33PS

Chapter
Section
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 0.3093g of the compound is burned in oxygen,  1.0620g of CO2and0.1739gH2O can be isolated.

From the given masses the amount of CO2andH2O can be calculated.

Thus, amount of CO2andH2O isolated from the combustion of naphthalene are,

1.0620gCO2×1molCO244.010gCO2=0.02413molCO20.1739gH2O×1molH2O18.015gH2O=0.00965molH2O

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

0.02413molCO2×1molCinunknown1molCO2=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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