# Determining the Equivalence Ratio

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

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?

• I have improved the formatting of your post using $\LaTeX$. For more information on how to do this yourself please see here and here. – bon Jul 10 '15 at 20:14

\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: $$\text{FAR equivalence ratio} = \frac{1}{\text{AFR equivalence ratio}} = \frac{1}{1.21}=0.82$$

• where'd you get the 4.76? – james Jul 11 '15 at 15:42
• In air, for every mole of Oxygen you have 3.76 mole of Nitrogen or 4.76 mole of total (Oxygen+Nitrogen) (neglecting other trace compounds) – Osman Mamun Jul 11 '15 at 17:09
• $\frac{79}{21}=3.76$ – Osman Mamun Jul 11 '15 at 17:10

I followed this 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