Empirical formula of hydrocarbon by combustion analysis?

1) The elements making up the unknown substance almost always include carbon and hydrogen. Oxygen is often involved and nitrogen is involved sometimes. Other elements can be involved, but problems with C and H tend to predominate followed by C, H and O and then by C, H, O and N.

2) We must know the mass of the unknown substance before burning it.

3) All the carbon in the sample winds up as CO2 and all the hydrogen in the sample winds up as H2O.

4) If oxygen is part of the unknown compound, then its oxygen winds up incorporated into the oxides. The mass of oxygen in the sample will almost always be determined by subtraction.

5) Often the N is determined via a second experiment and this introduces a bit of complexity to the problem. Nitrogen dioxide is the usual product when nitrogen is involved. Sometimes the nitrogen product is N2, sometimes NH3.

6) Sometimes the problem asks you for the empirical formula and sometimes for the molecular formula (or both). Two points: (a) you have to know the molar mass to get to the molecular formula and (b) you have to calculate the empirical formla first, even if the question doesn't ask for it. A few lines below is a link that goes to a file that discusses how to go from empirical to molecular formulas.

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g

answer: a chemical reaction

explanation:

im not wrong