Combustion analysis or combustion reaction involving hydrocarbons ( compounds composed only of carbon and hydrogen) have the following equation (unbalanced)
Hydrocarbon + O2 ⇒ CO2 + H2O
What we are given:
20g Hydrocarbon + O2 ⇒ 67.6g CO2 + 13.8g H2O
The empirical formula of an compound is defined as the lowest ratio of elements within that compound.
For example:
Glucose; C6H12O6 , in its lowest ratio, the empirical formula for glucose would be CH2O
Erythrulose; C4H8O4 in its lowest ratio, the empirical formula is the same as glucose: CH2O.
Step 1
The first step in finding the empirical formula of our hydrocarbon is to convert the grams of CO2 and H2O into mols
C = 12.1g
O = 16.00g * 2 = 32.0g
67.6g CO2 * 1mol44.1g = 1.53 mol CO2
H = 1.01g * 2 = 2.02g
O = 16.00g
13.8g H2O 1mol18.02g = .766 mol H2O
We will now factor the amount of O2 that was used in the reaction.
With the Law of Conservation of Mass in mind, we are given that 20g of Hydrocarbon reacts with O2 to form 67.6g CO2 and 13.8g of H2O or a total of 81.4g.
The equation 81.4g (H2O+CO2) - 20g Hydrocarbon = 61.4g O2
O = 16.00g, but oxygen is a diatomic, so we multiply by 2 to get 32.00g
61.4g O2 * 1molO232.00g = 1.92 mol O2
Step 2
Now we multiply our three mol ratios by the smallest to get a whole number ratio between them
CO2 ; 1.53mol.766mol = 2.00 mol CO2
H2O ; .766mol.766mol = 1.00 mol H2O
O2 ; 1.92mol.766mol = 2.51 mol O2
Since O2 is at 2.51 mol, we multiply everything by 2
2.00 mol CO2 * 2 = 4.00 mol CO2
1.00 mol H2O * 2 = 2.00 mol H2O
2.51 mol O2 * 2 = 5.02 mol O2 ⇒ 5.00 mol O2
Step 3
going back to our equation, we can balance it with the mol ratios
Hydrocarbon + 5 O2 ⇒ 4 CO2 + 2 H2O
Left hand side: 10 O
Right hand side: 10 O, 4C, 4H
For the equation to be balanced, the Hydrocarbon must have 4 C and 4 H. The empirical formula is the lowest ratio of the elements (C,H), thus it is C1H1 or just CH
