Chapter 4.4, Problem 4.5CYU

### 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

# A 0.1342-g sample of a compound composed of C, H, and O was burned in oxygen and 0.240 g of CO2 and 0.0982 g of H2O was isolated. What is the empirical formula of the compound? If the experimentally determined molar mass is 74.1 g/mol, what is the molecular formula of the compound?

Interpretation Introduction

Interpretation:

The empirical formula and molecular formula of the given compound 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

From the masses the of CO2â€‰andâ€‰H2O amount of carbon and hydrogen present in the sample can be determined as follows,

Thus, amount of Câ€‰andâ€‰H isolated from the combustion are,

â€‚Â 0.240â€‰gâ€‰CO2â€‰Ã—â€‰1â€‰molâ€‰CO244.010â€‰gâ€‰CO2â€‰Ã—â€‰â€‰1â€‰molâ€‰C1â€‰molâ€‰â€‹CO2Ã—12.01â€‰gâ€‰C1â€‰molâ€‰Câ€‰=â€‰0.06549â€‰gâ€‰C0.0982â€‰gâ€‰H2Oâ€‰Ã—1â€‰molâ€‰H2O18.015â€‰gâ€‰H2Oâ€‰â€‰Ã—â€‰â€‰2â€‰molâ€‰H1â€‰molâ€‰â€‹H2OÃ—1.008â€‰gâ€‰H1â€‰molâ€‰H=â€‰0.01099â€‹â€‹â€‹â€‰gâ€‰â€‰molÂ H

The mass of oxygen in the sample can be determined from the mass of original sample and the masses of carbon and oxygen in the sample as follows,

â€‚Â Massâ€‰ofâ€‰sampleâ€‰=â€‰0.1342â€‰gâ€‰=â€‰0.06549â€‰gâ€‰Câ€‰+â€‰0.01099â€‰gâ€‰Hâ€‰+â€‰xâ€‰gâ€‰O

â€‚Â Massâ€‰ofâ€‰Oâ€‰=â€‰0.05772â€‰gâ€‰O

Amount of Carbon, Hydrogen and Oxygen can be calculated from its mass and molar mass as follows,

Amountâ€‰ofâ€‰carbonâ€‰=â€‰0.06549â€‰gâ€‹â€‰â€‰Câ€‰â€‰Ã—1â€‰molâ€‰C12.01â€‰gâ€‰Câ€‰â€‰=â€‰0

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