Chapter 4, Problem 4.87PAE

4.93 A mixture of methane (CH₄) and propane (C₃H₈) has a total mass of 29.84 g. When the mixture is burned completely in excess oxygen, the CO₂ and H₂O products have a combined mass of 142.97 g. Calculate the mass of methane in the original mixture.

Interpretation Introduction

Interpretation: The mass of methane in the original mixture should be found out if $C{O}_2$ and $H_{2}O$ products result in a combined mass of 142.97 g when a mixture of methane ( $C{H}_4$ ) and propane ( $C_3{H}_8$ ) with a total mass of 29.84 g is burned completely in the presence of excess oxygen.

Concept introduction:

Burning of methane and propane in excess oxygen is a combustion reaction. When hydrocarbons burn out, that results in water and carbon dioxide gas. To complete combustion, there should be enough oxygen supply. Otherwise half combustion can occur. As a result of half combustion carbon monoxide and carbon particles are released to the environment.

Answer to Problem 4.87PAE

Solution:

$\begin{array}{l}Massofmethaneinoriginalsample\text{ is}12.53g\hfill \\ \hfill \end{array}$

Explanation of Solution

Completecombustion (Balance equation) reaction for both $C{H}_4$ and $C_3{H}_8$ can be written as follows.

$\begin{array}{l}C{H}_4+2O_2 \to C{O}_2+2H_{2}O\hfill \\ \hfill \end{array}$

$C_3{H}_8+5O_2\to 3C{O}_2+4H_{2}O$

Let’s take initial weight of propane as X, then initial weight of methane should be (29.84-X).

$n=\frac{m}{M}\left(n=noofmoles,m=weightofthesample,M=molarmass\right)$

Using above equation, number of moles of each molecules present can be found out.

$\begin{array}{l}In\text{ initial sample,}\\ \\ Numberofmolesofmethane=\frac{(29.84-X)}{16}\\ Numberofmoleso\end{array}$

$Nunber\text{ of moles of propane= }\frac{X}{44}$

After the combustion of methane,

$\begin{array}{l}Numb{\text{er of moles of CO}}_{2}formed=\frac{(29.84-X)}{16}\\ Number{\text{ of moles of H}}_{2}O\text{ formed= 2}\times \frac{(29.84-X)}{16}\end{array}$

After the combustion of propane,

$\begin{array}{l}N{\text{umber of moles of CO}}_{2}formed=3\times \frac{X}{44}\\ Number{\text{ of moles of H}}_{2}O\text{ = 4}\times \frac{X}{44}\end{array}$

So final products weight can be obtained as follows,

$\begin{array}{l}(Molar{\text{ mass of CO}}_2\times Total{\text{ number of moles of CO}}_2)+(Molar{\text{ mass of H}}_{2}O\times Total{\text{ number of moles of H}}_{2}O)=142.97\\ 44\times (\raisebox{1ex}{$(29.84-X)$}\!\left/ \!\raisebox{-1ex}{$16$}\right.+\frac{3x}{44})\text{ + 18}\times \text{(}\frac{2\times (29.84-X)}{16}\text{ + }\frac{4X}{44}\text{ ) = 142}\text{.97 }\\ X=17.31\end{array}$

So the mass of methane in original sample = 29.84 g -17.31 g

                            = 12.53 g

Conclusion

When the mas of a hydrocarbon mixture is known along with the mass of the product mix formed when it has undergone combustion reaction, the mass of individual hydrocarbons can be found out if the corresponding molar masses of the hydrocarbons are known.