Chapter 22, Problem 148CP

### Chemistry

9th Edition
Steven S. Zumdahl
ISBN: 9781133611097

Chapter
Section

### Chemistry

9th Edition
Steven S. Zumdahl
ISBN: 9781133611097
Textbook Problem

# Consider a sample of a hydrocarbon 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 and name it.

Interpretation Introduction

Interpretation: The molecular formula of the hydrocarbon and its name is to be stated.

Concept introduction: The molecular formula of an element defines the total number of atoms present in a molecule. The number of moles is defined as the ratio of mass with the molecular mass of an element. The mass of an element is the amount of the substance present in an element.

To determine: The name and molecular formula of the hydrocarbon.

Explanation

Explanation

Given

The pressure of the hydrocarbon sample is 0.959 atm .

The pressure after combusting the sample is 1.51 atm .

The temperature of the hydrocarbon sample is 298 K .

The temperature after combusting the sample is 375 K .

The density of the density of the mixture is 1.391 g/L .

The mixture occupies a volume four times as large as that of pure hydrocarbon.

The hydrocarbon is considered to be CxHy . The balanced chemical equation for the combustion of CxHy is,

CxHy+O2xCO2+(y2)H2O .

The initial volume of the reactants is considered to be 1 L .

The mixture occupies a volume four times as large as that of pure hydrocarbon

So, the volume of products is 4×1 L=4 L .

Mass of products is calculated by using the formula,

Mass=Density×volume .

Substitute the value of density and volume in the above formula to calculate the mass of the product.

Mass=Density×volume=1.391 g/L×4 L=5.564 g

The number of moles of reactant and product is calculated by using the expression,

Number of moles(n)=PVRT

Where,

• P is the pressure.
• V is the volume.
• R is the universal gas constant (0.0821 L atm mol1 k1) .
• T is the temperature.

For CxHy ,

Substitute the value of P, V, R and T in the above expression to calculate the moles of CxHy .

Moles(n) of CxHy=PVRT=0.959 atm×1 L0.0821 L atm mol1 K1×298 K=0.039198 mol

For product,

Substitute the value of P, V, R and T in the above expression to calculate the moles of product.

Moles(n) of product=PVRT=1.51 atm×4 L0.0821 L atm mol1 K1×375 K=0.196184 mol

According to balanced combustion equation, 1 mole of CxHy produces x moles of CO2 and y2 moles of H2O

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