# Is it possible to obtain the amount of air required for a combustion analysis?

I'm not very sure if the required amount of oxygen (I assume) can be found by only knowing the amounts of the produced subtances after a combustion.

To illustrate this I found this problem:

$$1.14\,kg$$ of octane $$C_{8}H_{18}$$ is burned with a certain amount of air. The resulting products from the combustion are known to have the following percentages by volume: $$CO_{2} = 41.18\%$$ ; $$CO = 5.8%$$ ; $$H_2O_{(vapor)} = 52.94\%$$. Find the weight of required air in kilograms. Assume the composition of the air in this combustion is $$21\%$$ of $$O_{2}$$ and $$79\%$$ of $$N_{2}$$ by volume and its average molecular weight is $$28.8\,\frac{g}{mol}$$.

The alternatives given are as follows:

$$\begin{array}{ll} 1.&9.0\,kg\\ 2.&16.4\,kg\\ 3.&12.5\,kg\\ 4.&45.1\,kg\\ 5.&120\,kg\\ \end{array}$$

What I think to approach this problem was to use the given percentages to get the number of moles using the initial moles of octane.

Using these moles I can obtain the grams of oxygen produced. Since it is given the percentage which is of oxygen in the air I could use this to find the required mass of the air to get that amount of air.

This is shown as follows: (for the sake of brevity I'm ommiting units but they are consistent.)

$$\textrm{FW of octane} = 114$$

$$n_octane=\frac{1140}{114}=10$$

Then moles of each product:

For $$CO_{2} = 41.18\%$$

$$\frac{41.18}{100}\times 10 = 4.118$$

Then in this amount of $$CO_{2}$$ there must be these grams of oxygen:

$$4.118 \times \frac{32\,g\O}{1\,mol\,CO_{2}}= 131.776 \,g$$

For $$CO = 5.8\%$$

$$\frac{5.8}{100}\times 10 = 0.058$$

$$0.058 \times \frac{16\,g\O}{1\,mol\,CO}= 0.928 \,g$$

For $$H_2O_{(vapor)} = 52.94\%$$.

$$\frac{52.94}{100}\times 10 = 5.294$$

$$5.294 \times \frac{16\,g\O}{1\,mol\,H_2O}= 84.704 \,g$$

$$131.776+0.928+84.704=217.408\,g\,O$$

Then using the known percentage of oxygen which is $$21\%$$

$$\frac{21}{100}x=217.408$$

$$x=1035.28 \,g$$

Which is approximately 1.035 kg. but this answer doesn't get any close to the specified alternatives. Did I overlooked something. Or could it be that this approach is not applicable?. In this given situation is it possible to find the required mass of air?.

For the given problem there isn't much $$\ce{CO}$$ created. Given that the question is multiple choice an exact amount isn't needed. The problem only needs to be solve close enough to determine which choice is the answer. So assume complete combustion and then guess that the answer will be about 5% high. If octane, $$\ce{C8H18}$$, is completely burned with oxygen then the reaction is:

$$\ce{C8H18 + 12.5O2 -> 8CO2 + 9H2O}$$

since the problem states that 1.14 kg of octane was burned, the number of moles is:

$$\dfrac{1140}{114} = 10.0$$

Thus for complete oxidation:

• 125 moles of oxygen would be needed.
• Air is only 21% oxygen so 125/0.21 = 595 "moles" of air are needed.
• At 28.8 g/"mole" that would be 17.1 kg of air.

In reality there is about 5% less oxygen required since there is some $$\ce{CO}$$ produced, so $$17.1*0.95 = 16.3\ \pu{ kg}$$ is a good guess.

I'll leave finding the exact solution to the OP.

• The reason why i did not attempted to use a complete combustion equation was due the fact that there appears other products and im referring to carbon monoxide and who knows what else as the sum of percentages given isnt $100\%$. This confusion is tied to how did you obtained that there is $5\%$ less oxygen required. Whats the justification for this? In the problem states carbon monoxide is $5.8\%$ is it because of this? Why was i given the average molecular weight? It was not needed right? – Chris Steinbeck Bell Jan 3 at 12:46
• But could be found by mutiplying the molar fraction to their respective formula weights and adding these quantities together. But again this information was unused. So as the percentage of nytrogen. – Chris Steinbeck Bell Jan 3 at 12:49
• @ChrisSteinbeckBell - Given $CO_{2} = 41.18\%$ ; $CO = 5.8\%$ and assuming ideal gas behavior, then normalize on the carbon atom to obtain the equivalent of 8 carbons. – MaxW Jan 3 at 19:49
• @ChrisSteinbeckBell - The 5% correction was just a guess. The $\ce{CO}$ is about 10% of the $\ce{CO2}$ and contains 1/2 the oxygen. So about a 5% correction. // I used a slide rule for about a decade. You have to estimate calculations since 2.3, 23 and 230 are all on the same spot on the slide rule. Given the answers in this case just assuming complete combustion make the problem simple and gives an answer close enough to choose between the multiple choice answers. – MaxW Jan 3 at 19:56
• @ChrisSteinbeckBell - $8 * (5.8/(5.8+41.18)) = 0.9877$ atoms of $\ce{CO}$ per molecule of octane. Thus there must be 7.0123 atoms of $\ce{CO2}$. I'll leave it to you to figure out how many oxygen atoms are needed. – MaxW Jan 3 at 20:09