Chapter 22, Problem 162CP

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.

Answer to Problem 162CP

Answer

The formula of the hydrocarbon is ${\text{C}}_2{\text{H}}_6$ and the name of the hydrocarbon is ethane.

Explanation of Solution

Explanation

Given

The pressure of the hydrocarbon sample is $0.95\text{9 atm}$.

The pressure after combusting the sample is $\text{1}\text{.51 atm}$.

The temperature of the hydrocarbon sample is $2\text{98 K}$.

The temperature after combusting the sample is $\text{375 K}$.

The density of the density of the mixture is $1.39\text{1 g/L}$.

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

The hydrocarbon is considered to be ${\text{C}}_x{\text{H}}_y$. The balanced chemical equation for the combustion of ${\text{C}}_x{\text{H}}_y$ is,

${\text{C}}_x{\text{H}}_y+{\text{O}}_{\text{2}}\underset{}{\to }x{\text{CO}}_{\text{2}}+\left(\frac{y}{2}\right){\text{H}}_{\text{2}}\text{O}$.

The initial volume of the reactants is considered to be $\text{1 L}$.

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

So, the volume of products is $4\times \text{1 L}=\text{4 L}$.

Mass of products is calculated by using the formula,

$\text{Mass}=\text{Density}\times \text{volume}$.

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

$\begin{array}{c}\text{Mass}=\text{Density}\times \text{volume}\\ =1.39\text{1 g/L}\times \text{4 L}\\ =5.56\text{4 g}\end{array}$

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

$\text{Number of moles}\left(\text{n}\right)=\frac{\text{PV}}{\text{RT}}$

Where,

• P is the pressure.
• V is the volume.
• R is the universal gas constant $\left(0.08{\text{21 L atm mol}}^{-1}{\text{ k}}^{-1}\right)$.
• T is the temperature.

For ${\text{C}}_x{\text{H}}_y$,

Substitute the value of P, V, R and T in the above expression to calculate the moles of ${\text{C}}_x{\text{H}}_y$.

$\begin{array}{c}\text{Moles}\left(\text{n}\right){\text{ of C}}_x{\text{H}}_y=\frac{\text{PV}}{\text{RT}}\\ =\frac{0.9\text{59 atm}\times \text{1 L}}{0.08{\text{21 L atm mol}}^{-1}{\text{ K}}^{-1}\times 29\text{8 K}}\\ =0.0391\text{98 mol}\end{array}$

For product,

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

$\begin{array}{c}\text{Moles}\left(\text{n}\right)\text{ of product}=\frac{\text{PV}}{\text{RT}}\\ =\frac{\text{1}\text{.51 atm}\times 4\text{ L}}{0.08{\text{21 L atm mol}}^{-1}{\text{ K}}^{-1}\times 375\text{ K}}\\ =0.196184\text{ mol}\end{array}$

According to balanced combustion equation, 1 mole of ${\text{C}}_x{\text{H}}_y$ produces $x$ moles of ${\text{CO}}_{\text{2}}$ and $\frac{y}{2}$ moles of ${\text{H}}_{\text{2}}\text{O}$.

So, the equation is written as,

$0.0391\text{98}x+0.0391\text{98}\frac{y}{2}=0.196184$.

Simplify the above equation,

$0.0391\text{98}x+0.019599\text{y}=0.196184$.

$2x+y=10.00998$ (1)

Mass of ${\text{CO}}_{\text{2}}$ and ${\text{H}}_{\text{2}}\text{O}$ is calculated by using the formula,

$\text{Mass}=\text{Number of moles}\times \text{Molar mass}$

Molar mass of ${\text{CO}}_{\text{2}}$ is $4\text{4 g/mol}$.

Molar mass of ${\text{H}}_{\text{2}}\text{O}$ is $\text{18 g/mol}$.

For ${\text{CO}}_{\text{2}}$,

Substitute the value of number of moles and molar mass in the above formula to calculate the mass of ${\text{CO}}_{\text{2}}$.

$\begin{array}{c}\text{Mass}=\text{Number of moles}\times \text{Molar mass}\\ =0.039198x\text{ mol}\times 4\text{4 g/mol}\\ =1.725085x\end{array}$

For ${\text{H}}_{\text{2}}\text{O}$,

Substitute the value of number of moles and molar mass in the above formula to calculate the mass of ${\text{H}}_{\text{2}}\text{O}$.

$\begin{array}{c}\text{Mass}=\text{Number of moles}\times \text{Molar mass}\\ =0.039198\left(\frac{y}{2}\right)\text{ mol}\times 18\text{ g/mol}\\ =0.352778y\end{array}$

The total mass of product is $5.56\text{4 g}$.

Therefore,

$1.725085x+0.35317\text{y}=5.564$

Simplify the above equation,

$4.884573x+\text{y}=15.75445$ (2)

Subtract equation (1) from equation (2).

$\begin{array}{c}2.884573x=5.754446\\ x=1.994904\approx 2\end{array}$.

Substituting $x$ in equation (1) the value of $y$ is calculated as,

$\begin{array}{c}2x+y=10\\ 2\left(2\right)+y=10\\ y=6\end{array}$

Therefore the formula of the hydrocarbon is ${\text{C}}_x{\text{H}}_y={\text{C}}_2{\text{H}}_6$ and the name of the hydrocarbon is ethane.

Conclusion

Conclusion

The formula of the hydrocarbon is ${\text{C}}_2{\text{H}}_6$ and the name of the hydrocarbon is ethane.