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?