# Consider a sample of a hydrocarbon (a compound consisting of only carbon and hydrogen) at 0.959 atm and 298 K. Upon combusting the entire sample in oxygen, you collect a mixture of gaseous carbon dioxide and water vapor at 1.51 atm and 375 K. This mixture has a density of 1.391 g/L and occupies a volume four times as large as that of the pure hydrocarbon. Determine the molecular formula of the hydrocarbon.

### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

Chapter
Section

### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 8, Problem 153CP
Textbook Problem
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## Consider a sample of a hydrocarbon (a compound consisting of only carbon and hydrogen) at 0.959 atm and 298 K. Upon combusting the entire sample in oxygen, you collect a mixture of gaseous carbon dioxide and water vapor at 1.51 atm and 375 K. This mixture has a density of 1.391 g/L and occupies a volume four times as large as that of the pure hydrocarbon. Determine the molecular formula of the hydrocarbon.

Interpretation Introduction

Interpretation: The formula of hydrocarbon in the given reaction of combustion of this hydrocarbon with O2 at 0.959atm and 298K is needed to be identified, if the combustion producing a mixture of CO2 and H2O having a density of 1.391g/L and occupies a volume of 4 times larger than hydrocarbon.

Concept introduction:

• Number of moles of a substance,
• According to ideal gas equation,

Numberofmoles=Pressure×VolumeR×Temperature

• Mass of a substance from its density and volume is,

Mass=Density×Volume

Mass of a substance from its number of moles is,

Number of moles×Molecularmass in grams=takenMass

• Balance equation of a reaction is written according to law of conservation of mass.
• Mole ratio between the reactants of a reaction is depends upon the coefficients of reactants in a balanced chemical equation.

### Explanation of Solution

Explanation

• To find: the balanced chemical equation of the given reaction of combustion of hydrocarbon compound.

CxHy+O2xCO2+y2H2O

Let’s take the hydrocarbon as CxHy .

The combustion of CxHy in O2 is producing CO2 and H2O gases.

Therefore,

The chemical equation of the reaction given is,

CxHy+O2CO2+H2O

This equation is not balanced, Balanced equation is the chemical equation of a reaction which is written according to law of conservation of mass the mass of reactant molecules should conserve in product molecules.

The balanced chemical equation of the given reaction is,

Balance the molecules in both sides of arrow in order of atoms other than hydrogen, hydrogen atoms and then oxygen atoms.

Therefore, the balanced chemical equation of the given reaction of combustion of hydrocarbon compound will be,

CxHy+O2xCO2+y2H2O

Hence,

The balanced chemical equation for the given reaction of combustion of hydrocarbon compound is, CxHy+O2xCO2+y2H2O .

The volume of hydrocarbon is taken as 1 L.

To find: the number of moles of reacted CxHy in the given reaction.

The number of moles of CxHy is 0.0392mol .

The pressure of reacted CxHy in the given reaction is given as 0.959atm .

The volume of CxHy is 1L .

The temperature is given as 298K .

According to ideal gas equation is,

Numberofmoles=Pressure×VolumeR×Temperature

Therefore, the number of moles of CxHy ,

nCxHy=0.959atm×1L0.08206Latm/Kmol×298K=0.0392mol

Hence, the number of moles of CxHy is in given reaction is 0.0392mol .

The volume of mixture of products is taken as 4 L (since the volume of hydrocarbon is taken as 1 L).

To find: the number of moles and mass of produced mixture of CO2 and H2O having a density of 1.391g/L in the given reaction.

The number of moles of produced mixture is 0.196mol .

The mass of produced mixture is 5.564g .

The pressure of produced mixture in the given reaction is given as 1.51atm .

The volume of produced mixture is 4L .

The temperature is given as 375K .

According to ideal gas equation is,

Numberofmoles=Pressure×VolumeR×Temperature

Therefore, the number of moles of produced mixture is,

nmixture=1.51atm×4L0.08206Latm/Kmol×375K=0.196mol

Hence, the number of moles of produced mixture is in given reaction is 0.196mol .

The density of produced mixture is given as 1.391g/L

Mass of a substance from its density and volume is,

Mass=Density×Volume

Therefore,

The mass of produced mixture is,

Mass=1

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