Chapter 22, Problem 4TE

There are five atoms in the methane molecule, ${\text{CH}}_{\text{4}}$ . Out of these five is a carbon atom, which is $1/5\times 100=20\%$ carbon. What is the percentage of carbon in ethane, ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$ ? Propane, ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}$ ? Butane, ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}$ ? Do heavier hydrocarbons tend to produce more or less carbon dioxide upon combustion compared to lighter hydrocarbons? Why?

To determine

Carbon percentage present in ethane ${\text{(C}}_{\text{2}}{\text{H}}_{\text{6}}\text{)}$, Propane ${\text{(C}}_{\text{3}}{\text{H}}_8\text{)}$ and butane ${\text{(C}}_{\text{4}}{\text{H}}_{10}\text{)}$. Comparison of combustion of heavier hydrocarbons with lighter hydrocarbons.

Answer to Problem 4TE

The percentage of carbon in ethane propane butane is $\text{25}$ ,$\text{27}\text{.27}$ , and $\text{28}\text{.57}$ respectively. And yes, there will be higher emission of carbon dioxide during combustion of heavier combustion.

Explanation of Solution

In Ethane, number of carbons =2

So, the percentage of carbon $=\frac{2}{8}\times 100=25\%.$

In propane, number of carbons=3

So, the percentage of carbon $=\frac{3}{11}\times 100=27.27\%.$

In butane, number of carbons=4

So, the percentage of carbon $=\frac{4}{14}\times 100=28.57\%.$

The equation for the general formula of combustion of hydrocarbons is stated below-

From above we can conclude that amount of carbon dioxide is not constant, it depends upon the carbon number of hydrocarbons. So heavier be the hydrocarbon more the number of carbons and more will be carbon dioxide.

Conclusion:

Therefore, the percentage of carbon in ethane propane butane is $\text{25}$ ,$\text{27}\text{.27}$ and $\text{28}\text{.57}$ respectively. And yes, there will be higher emission of carbon dioxide during combustion of heavier combustion.