Question

A newspaper article about the danger of global warming from the accumulation of green house gases such as carbon dioxide states that ‘reducing driving your car by 20 miles a week would prevent the release of over thousand pounds of \({\bf{C}}{{\bf{O}}_{\bf{2}}}\) per year into the atmosphere.” Is this a reasonable statement? Assume that gasoline is octane \(\left( {{{\bf{C}}_{\bf{8}}}{{\bf{H}}_{{\bf{18}}}}} \right)\) and that is burnt completely to \({\bf{C}}{{\bf{O}}_{\bf{2}}}{\rm{ }}{\bf{and}}{\rm{ }}{{\bf{H}}_{\bf{2}}}{\bf{O}}\) in the engine of your car. Facts (Reasonable guesses) about your cars gas milage, the density of octane and other factors also needed.

Short answer

Yes, the newspaper statement is quite reasonable.

In a newspaper article it is written that if one reduced car driving 20 miles in 1 week it will stop the Emission of over thousand lb. of\({{\mathop{\rm CO}\nolimits} _2}\)each year. We need to find out whether the newspaper is right taking into account octane as a gasoline.

Step-by-step solution

Step 1:

Balanced Equation of Octane burning

When, octane burns completely it generates\({{\mathop{\rm CO}\nolimits} _2}\)and water

• Balanced equation can be written as

\(2{C_8}{H_{18}} + 25{O_2} \to 16C{O_2} + 18{H_2}O\)

Step 2:

Density of Octane

The density of octane is expected to be close to the density of hydrocarbon denzene. Assume it to be\(0.8gc{m^{ - 3}}\)

• Density can be written as

\(0.8gm{l^{ - 1}}\left( {1ml = 1c{m^{ - 3}}} \right)\)

Step 3:

Total reduction in gasoline consumption

Let us assume that the car drives 20 miles in 1 gallon .

• There are 52 weeks in one year

Thus, consumption of gasoline by the car will be reduced by the amount calculated as follows if driving is reduced by 20 miles per week

\(\dfrac{{20miles}}{{1week}} \times \dfrac{{52weeks}}{{1year}} \times \dfrac{{1gallon}}{{20miles}} = 52gallonsperyear\)

This amount of gasoline should be converted to mass of gasoline in grams.

• Total volume of gasoline, 1 Gallon = 3.7854118l

\(\begin{aligned}{}52Gallon = 52 \times 3.7854118l\\ = 196.8414136l\end{aligned}\)

Step 4:

Molar mass of\({{\mathop{\rm C}\nolimits} _8}{H_{18}}\)C scale

\({\mathop{\rm Density}\nolimits} = \dfrac{{mass}}{{Volume}} \Rightarrow Mass = Density \times Volume\)

• Putting values in the upper formula

\(\begin{aligned}{}Mass = \dfrac{{0.8g}}{{ml}} \times 196.8414136l \times \dfrac{{1000ml}}{l}\\ = 157476.1309g\end{aligned}\)

• The relative molar mass of \(\left( {{C_8}{H_{18}}} \right)\) on the C scale

8(Atomic mass of c) + 18(Atomic mass of H)

\(\begin{aligned}{} = 8 \times 12.011 + 18 \times 1.0079\\ = 96.088 + 18.1422\\ = 114.2302\end{aligned}\)

Hence, the relative molecular mass of \(\left( {{C_8}{H_{18}}} \right)\)on the c scale 18 114.2302. The mass of 1 mole of \(\left( {{C_8}{H_{18}}} \right)\)18114.2302g

Step 5:

Number of moles of\(\left( {{C_8}{H_{18}}} \right){\rm{ }}and{\rm{ }}C{O_2}\)

Number of moles of\(\left( {{{\rm{C}}_{\rm{8}}}{{\rm{H}}_{{\rm{18}}}}} \right)\)can be calculated as follows:

\(\begin{aligned}{}{}^n{C_8}{H_{18}} = 157473.1309g{\rm{ }}{C_8}{H_{18}} \times \dfrac{{1mol{C_8}{H_{18}}}}{{114.2302g{C_8}{H_{18}}}}\\ = 1378.5595mol{C_8}{H_{18}}\end{aligned}\)

• According to equation we know,

2 moles of octanes generate 16 moles of carbon dioxide amount of carbon dioxide produce from 1378.5595 moles of\({{\mathop{\rm C}\nolimits} _8}{H_{18}}\)are

\(\begin{aligned}{l}1378.5595mol{C_8}{H_{18}} \times \dfrac{{16molesC{O_2}}}{{2moles{C_8}{H_{18}}}}\\ = \dfrac{{22056.952}}{{2mol}}C{O_2} = 11028.476molC{O_2}\end{aligned}\)

Step 6:

Molar mass of Carbon dioxide on C scale

Relative molar mass of Carbon dioxide on C scale

\(\begin{aligned}{} = 1\left( {atomicmassofC} \right) + 2\left( {AtomicMassOfO} \right)\\ = 1 \times 12.011 + 2 \times \left( {15.999} \right)\\ = 12.011 + 31.998 = 44.009\end{aligned}\)

Step 7:

Reasonability of the statement

Hence, relative molar mass of\({{\mathop{\rm CO}\nolimits} _2}\)on C scale is 44.009. the mass of 1 mole of\({{\mathop{\rm CO}\nolimits} _2}\)is 44.009

• The mass of 11028.476 mol of\(C{O_2}\)is

\(\begin{aligned}{}MassC{O_2} = 11028.476molC{O_2} \times \dfrac{{44.009gC{O_2}}}{{1molC{O_2}}}\\ = 485352.200284gC{O_2}\end{aligned}\)

• Converting mass of\({{\mathop{\rm CO}\nolimits} _2}\)into lb.

\(\begin{aligned}{}453.63 = 1lb\\1g = \dfrac{1}{{453.6g}}lb\\485352.200284g = \dfrac{{485352.200284}}{{453.6}}lb\\ = 1070.000442lb\end{aligned}\)

Hence, the newspaper statement is quite reasonable