Mass proportion in reactant

1. Jan 9, 2014

marellasunny

1. The problem statement, all variables and given/known data

Fuel general formula: $C_x H_y O_z$

How do I calculate the mass proportion of oxygen in the air participating in the following reaction $\xi_{O_2,air}$?
$$C_x H_y O_z+(x+\frac{y}{4}-\frac{z}{2}).O_2 \rightarrow xCO_2+\frac{y}{2}H_2O$$

2. Relevant equations

Mass proportion of oxygen in fuel:
$$Oxygen=\frac{M_{oxygen}}{M_{fuel}} .z$$
M-molecular weight
z- stoichiometric coefficient
3. The attempt at a solution

The number of moles in the air would be:$(x+\frac{y}{4}-\frac{z}{2})$

So,$$\xi_{O_2,air}=(x+\frac{y}{4}-\frac{z}{2}) * \frac{M_{O2}}{M_{air/fuel?}}$$

Is this correct? Is the mass proportion calculated above EQUAL to the stoichiometric air requirement?

Last edited: Jan 9, 2014
2. Jan 9, 2014

epenguin

Correct? It's a correct calculation of moles of O2 consumed per mole of this fuel.
That is just the first step to getting the answer requested as I understand it, which is about masses. You will need to use atomic masses.

To answer this question it is not necessary to use the fact that oxygen is diatomic. It does no harm and you have done it correctly but it is just an unnecessary complication for the purposes of answering the question.

3. Jan 10, 2014

Staff: Mentor

Take as a basis one mole of fuel. You already showed that 1 mole of fuel requires (x+y/4-z/2) moles of oxygen. How many moles of nitrogen are there for every mole of oxygen in air? How many moles of N2 are carried along for every mole of fuel? On the basis of 1 mole of fuel, what is the seight of the fuel? What is the weight of the oxygen from the air? What is the weight of the nitrogen from the air? What fraction of the total weight is oxygen from the air?

Chet