A major component of gasoline is octane.

Question

A major component of gasoline is octane C8H18. When octane is burned in air, it chemically reacts with oxygen gas O2 to produce carbon dioxide CO2 and water H2O. What mass of water is produced by the reaction of 1.4g of octane?

Answer 1

We can do this by a mass calculation as we assume all of the H in octane becomes water.

1.4 grams octane (18 g H/ 114 g octane)(18 g H₂O/2 g H) = 1.99 g H₂O

Please consider a tutor. Take care.

Answer 2

The first step we need to take is to balance the combustion reaction. Heptane is particulary challenging, since the heptane molecule brings 8 carbons and 16 hydrogens that need to be accounted for on the product side. The issue quickly becomes mired down in the fact that oxygen only comes as a diatomic molecule. There is always an even number of O atoms. But the product side contains a molecule that has an odd number of O atoms. This/even/odd relationship often result in large coefficents, as we will see.

The basic reaction is C8H18 + O2 = CO2 + H2O. Start balancing by assigning a coefficient of 1 to the most complex molecule, which is the heptane in this case.

1C8H18 + O2 = CO2 + H2O

1 heptane molecule brings 8 C and 18 H atoms. Find homes for these atoms by changing the coefficients of the product side. The only place for C is the CO2, so assign it a coefficient of 8:

1C8H18 + O2 = 8CO2 + H2O

The only place for the 18 H atoms is the H2O, so assign the H2O a coefficient of 9:

1C8H18 + O2 = 8CO2 + 9H2O

The only element left is Oxygen, O. From the product side, we can count a total of (16 + 9) = 25 O atoms. We can add a coefficient for the reactant O2, but there are no whole number that will result in 25 O atoms. We can't have 1/2 of an O2 atom. For the moment, let's give it a coefficient of 12.5, just to get started:

1C8H18 + 12.5O2 = 8CO2 + 9H2O

The equation is now balanced, illegally. To fix the issue, simply multiply all coefficients by 2 in order to make the O coefficient a whole number:

2C8H18 + 25O2 = 16CO2 + 18H2O

The equation is now balanced, again, but legally now. Looks ugly, but it is correct.

Reactant Product

C 16 16

H 36 36

O 50 50 (32+18)

The question asks for the mass of water formed from 1.4 grams of octane. [I would have preferred a larger amount for all the time this is taking]. The balanced equation tells us that 2 moles of heptane will produce 18 moles of water. That sounds encouraging. But let's see how much mass of water this actually produces. The molar ratio of water to heptane is 18/2 or 9/1 (moles water/moles heptane)

Convert the 1.4 grams of octane into moles octane, by dividing the mass by the molar mass of octane (114 g/mole).

(1.4g octane)/(114 g/mole) = 0.0123 moles heptane [ouch]

Multiply by the molar ratio from above to find moles water produced:

(0.0123 moles heptane)*(9 moles water)/(1 mole heptane) = 0.11 moles H2O

Multiply moles H2O produced by the molar mass of H2O (18g/mole) = 1.99 grams.

We should expect 1.99 grams of water when 1.4 grams of octane is combusted. Fun, but not worth the effort.