How many moles of CO2 are emitted into the atmosphere when 20.6g C8H18 is burned?

Question

Consider the combustion reaction for octane  which is a primary component of gasoline.

2C8H18 + 25O2 -> 16CO2 + 18H2O

How many moles of CO2 are emitted into the atmosphere when 20.6g C8H18 is burned?

Answer 1

To understand how to solve this problem, we must first look at what we're given and what we need to find. We are given the chemical equation for the combustion of octane, we have the mass of C8H18 (octane) used, and we want to know the moles of CO2 released into the atmosphere.

To find the moles of CO2 released, we are going to have to do some stoichiometry. We cannot directly convert mass of C8H18 to moles of CO2; however, we can convert the mass of C8H18 to moles of C8H18 and use the molar ratio to convert to moles of CO2.

We get the molar mass of octane by adding the individual molar masses of all the atoms in the molecule [(8x12)+(18x1)]=114g/mol. Using the molar mass of octane, we can convert the grams of C8H18 to moles of C8H18:

(20.6g C8H18)(1mole C8H18/114g C8H18)=moles of C8H18 (I intentionally didn't solve this because I am going to solve it in a later equation, to better account for significant figures)

Now that we have the number of moles of octane, we can convert this to moles of CO2. This is achieved by using the molar ratios found in the balanced chemical equation. We can see that for every 2 moles of C8H18 used, we have 16 moles of CO2 produced. Adding this to our above equation we get:

(20.6g C8H18)(1mole C8H18/114g C8H18)(16moles CO2/2 moles C8H18)=1.45 moles CO2

P.S. When doing stoichiometry I always like to remember that when setting up the equation, I always want to put what I want over what I want to convert from. In the example above, started with grams and wanted moles so I multiplied by moles over grams so the grams cancel out.