   # Mesitylene is a liquid hydrocarbon Burning 0.115 g of the compound in oxygen gives 0.379 g of CO 2 and 0.1035 g of H 2 O. What is the empirical formula of mesitylene? ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 4, Problem 32PS
Textbook Problem
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## 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 of Solution

Mesitylene is a liquid hydrocarbon.  Burning 0.115g of the compound in oxygen gives 0.379g of CO2and0.1035gofH2O

From the given masses the amount of CO2andH2O can be calculated as follows,

Thus, amount of CO2andH2O isolated from the combustion are,

0.379gCO2×1molCO244.010gCO2=0.008611molCO20.1035gH2O×1molH2O18.015gH2O=0.005745molH2O

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

0.008611molCO2×1molCinunknown1molCO2=0

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