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

According to the problem statement, if you did everything right,
$$\begin{align}
\text{AFR} &amp;= 37.54 \\
\text{AFR}_{\text{stoich}} &amp;= \frac{\nu_{\text{oxygen}}}{\nu_{\text{fuel}}}\times 4.76 \\
\ce{C4H10 + $6.5$\,O2 &amp;-&gt; 4 CO2 + 5 H2O} \\
\text{AFR}_{\text{stoich}} &amp;= 4.76 \times 6.5 = 30.94 \\
\text{Equivalence ratio} &amp;= \frac {37.54}{30.94}=1.21
\end{align}$$
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: 
$$\text{FAR equivalence ratio} = \frac{1}{\text{AFR equivalence ratio}} = \frac{1}{1.21}=0.82$$