How much larger is the contribution of Justine's father's car (48 gal of 92 octane gas / month) to global warming than Justine's cow?

Question

Justine, a precocious 9 year old and the daughter of a farmer, has learned that cows generate a lot of methane (CH₄), a greenhouse gas that leads to global warming. The EPA estimates that the 100 million cattle in the United States release approximately 6 million metric tons (1 metric ton = 1000 kg) of methane per year. Methane is roughly 6 times as efficient at trapping solar energy in the earth's atmosphere than is the same mass of CO₂ (g). At the same time, Justine's father's truck burns about 40.8 gallons of 92 octane gasoline (92% C₈H₁₈ and 8% C₇H₁₆ by weight, respectively) per month. Gasoline has a density of 0.73 g/mL. Hydrocarbon combustion reactions for unsaturated alkanes like C₈H₁₈ and C₇H₁₆ are of the form:

C_nH_2n+2 + (3n + 1)/2 O₂ --> n CO₂ + (n + 1) H₂O.

Justine wants to calculate the relative effects of the family cow (Ol' Bessie) and her father's driving habits on the environment. How much larger is the contribution of her father's driving to global warming than that of Justine's cow? Note: 1 gallon = 3.78541 L.

Answer 1

When you compare two things, you need to put them in the same units. Here, we'll use CO₂ equivalents.

In chemistry, we usually use factor labelling to go from one extensive (how much stuff) variable to another. Let's start with 1 cow and 1 truck and see how much CO₂ equivalents they produce in the same time. (since they are already giving the truck value in months (not a great unit as it varies in length of time) we will specify one month)

1 cow( 6 x 10⁹ kg Meth/10⁸ cows-year)(1yr/12 mos)(6 CO₂ kg/1 CH₄ kg) = kg/month CO₂

1 truck(40.8 gal Gas/truck-month)(3.785 liter/ga)(1000 ml/liter)(.92 g Oct/100 g Gas)(1000 g/kg) times

the last term which is the kg CO₂/kg Oct. We could go to moles and use the stoichiometry, but because you are given the reaction and all the C appears as CO₂, we can connect grams to grams with the molar masses:

There will be 8 x M_CO2 g of CO₂/ M_Oct or 8(44.01) g CO₂/114.23 g oct (note that because this is a mass ratio, it is true for kg as well. This will also yield kg/mo CO₂

Please consider a tutor. Take care.