Determining the Equivalence Ratio

Determine the equivalence ratio of the following mixture: 4g of Butane, $\ce{C4H10}$, and 75 grams of air.


Answer is 0.82


Ok so I try converting everything into mols


$4~\mathrm{g}~(\ce{C4H10}) \times \frac{1~\mathrm{mol}}{58~\mathrm{g}} = \frac{4}{58}~\mathrm{mol}$


$75~\mathrm{g~(Air)} \times \frac{1~\mathrm{mol}}{28.966~\mathrm{g}} = \frac{75}{28.966}~\mathrm{mol}$


Air-fuel ratio = $\frac{\frac{75}{28.966}}{\frac{4}{58}} = 37.54$ 


This is just a guess that the air-fuel ratio is some sort of equivalence ratio.


I saw that equivalence ratio = actual air-to-fuel ratio divided by the theoretical. I don't know how to get either actual or theoretical? Is the data in the problem sufficient enough to get both?


I also see a formula in the reviewer: Fuel + Air = Products of combustion and the appropriate reaction, but it only works for finding the exact amount of mols like: $\ce{C4H10 + aO2 + 3.76aN2 = bCO2 + CH2O + 3.76aN2}$


This is for a mechanical engineering board exam review; unfortunately, university only tapped briefly on the chemistry side of fuels.


Any hint on what I should be doing? Point me to the right direction?




$$\begin{align}
\text{AFR (Air to Fuel ratio)} &= \frac{n\_{\text{air}}}{n\_{\text{fuel}}} \\
\text{Equivalence ratio} &= \frac{\text{AFR}}{\text{AFR}\_{\text{stoich}}} \\
\end{align}$$


According to the problem statement, if you did everything right,
$$\begin{align}
\text{AFR} &= 37.54 \\
\text{AFR}\_{\text{stoich}} &= \frac{\nu\_{\text{oxygen}}}{\nu\_{\text{fuel}}}\times 4.76 \\
\ce{C4H10 + $6.5$\,O2 &-> 4 CO2 + 5 H2O} \\
\text{AFR}\_{\text{stoich}} &= 4.76 \times 6.5 = 30.94 \\
\text{Equivalence ratio} &= \frac {37.54}{30.94}=1.21
\end{align}$$
Now there are two types of equivalence ratio. One is air to fuel and another is fuel to air. I suspect in the answer they meant fuel to air equivalence ratio: 
$$\text{FAR equivalence ratio} = \frac{1}{\text{AFR equivalence ratio}} = \frac{1}{1.21}=0.82$$


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I followed [this](http://fluid.wme.pwr.wroc.pl/~spalanie/dydaktyka/combustion_MiBM/fund/Stoichiometry.pdf) procedure. Note that "oxygen" here refers to ${\rm O}\_{2}$.


They define the equivalence ratio as: $${\rm F/A\over (F/A)\_s}$$ where F/A is your fuel to oxygen ratio (moles of fuel divided by moles of oxygen), and (F/A)$\_{\rm s}$ is the "stoichiometric ratio" defined as the number of moles of oxygen needed to combust 1 mole of fuel.


For the numerator, you have 4/58 moles of fuel (F) and [0.21 $\times$ (75/29)] moles of oxygen (A) (0.21 because 21% of air is oxygen). Your numerator (F/A) is equal to 0.126.


For the denominator (F/A)$\_{\rm s}$, we know that for each mole of fuel ${\rm C\_mH\_n}$ you need ${\rm (m + n/4)}$ moles of oxygen (see linked text for that). For butane (your fuel) that means you need 6.5 moles of oxygen to combust 1 mole of fuel. Your denominator is 0.153 (that's just 1/6.5).


Dividing 0.126/0.153 = 0.82


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