Chapter 4, Problem 32PS

### 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

# Mesitylene is a liquid hydrocarbon Burning 0.115 g of the compound in oxygen gives 0.379 g of CO2 and 0.1035 g of H2O. What is the empirical formula of mesitylene?

Interpretation Introduction

Interpretation:

The empirical formula of mesitylene should be determined.

Concept introduction:

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

Mesitylene is a liquid hydrocarbon. Â Burning 0.115â€‰g of the compound in oxygen gives 0.379â€‰g of CO2â€‰andâ€‰0.1035â€‰gâ€‰ofâ€‰H2O

From the given masses the amount of CO2â€‰andâ€‰H2O can be calculated as follows,

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

â€‚Â 0.379â€‰gâ€‰CO2â€‰Ã—â€‰1â€‰molâ€‰CO244.010â€‰gâ€‰CO2â€‰=â€‰0.008611â€‰â€‰molâ€‰CO20.1035gâ€‰H2Oâ€‰Ã—1â€‰molâ€‰H2O18.015â€‰gâ€‰H2Oâ€‰=â€‰0.005745â€‰molâ€‰H2O

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

â€‚Â 0.008611â€‰molâ€‰CO2â€‰Ã—1â€‰molâ€‰Câ€‰inâ€‰unknown1â€‰molâ€‰CO2â€‰=â€‰0

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