   Chapter 4, Problem 34PS

Chapter
Section
Textbook Problem

Azulene is a beautiful blue hydrocarbon. If 0.106 g of the compound is burned in oxygen, 0.364 g of CO2 and 0.0596 g of H2O are isolated.(a) What is the empirical formula of azulene?(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 formula of Azulene has to be determined.

Concept introduction:

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

Explanation

Given that 0.106g of the compound is burned in oxygen, 0.364g of CO2and0.0596gH2O can be isolated.

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

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

0.364gCO2×1molCO244.010gCO2=0.00827molCO20.0596gH2O×1molH2O18.015gH2O=0.00330molH2O

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

0.00827molCO2×1molCinunknown1molCO2=0

(b)

Interpretation Introduction

Interpretation:

Molecular formula of Azulene has to be determined.

Concept Introduction:

• Equation for finding Molecular formula from the empirical formula,

MolarmassEmpiricalformula mass × Empirical formula

• The molar mass of an element or compound is the mass in grams of 1 mole of that substance, and it is expressed in the unit of grams per mol (g/mol).

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