Combustion analysis or combustion reaction involving hydrocarbons ( compounds composed only of carbon and hydrogen) have the following equation (unbalanced)
Hydrocarbon + O_2 => CO_2 + H_2O
What we are given:
20g Hydrocarbon + O_2 => 67.6g CO_2 + 13.8g H_2O
The empirical formula of an compound is defined as the lowest ratio of elements within that compound.
For example:
Glucose; C_6H_12O_6 , in its lowest ratio, the empirical formula for glucose would be CH_2O
Erythrulose; C_4H_8O_4 in its lowest ratio, the empirical formula is the same as glucose: CH_2O.
Step 1
The first step in finding the empirical formula of our hydrocarbon is to convert the grams of CO_2 and H_2O into mols
C = 12.1g
O = 16.00g * 2 = 32.0g
67.6g CO_2 * (1mol)/(44.1g) = 1.53 mol CO_2
H = 1.01g * 2 = 2.02g
O = 16.00g
13.8g H_2O (1mol)/(18.02g) = .766 mol H_2O
We will now factor the amount of O_2 that was used in the reaction.
With the Law of Conservation of Mass in mind, we are given that 20g of Hydrocarbon reacts with O_2 to form 67.6g CO_2 and 13.8g of H_2O or a total of 81.4g.
The equation 81.4g (H2_O + CO_2) - 20g Hydrocarbon = 61.4g O_2
O = 16.00g, but oxygen is a diatomic, so we multiply by 2 to get 32.00g
61.4g O_2 * (1mol O_2)/(32.00g) = 1.92 mol O_2
Step 2
Now we multiply our three mol ratios by the smallest to get a whole number ratio between them
CO_2 ; (1.53 mol)/(.766mol) = 2.00 mol CO_2
H_2O ; (.766mol)/(.766mol) = 1.00 mol H_2O
O_2 ; (1.92mol)/(.766mol) = 2.51 mol O_2
Since O_2 is at 2.51 mol, we multiply everything by 2
2.00 mol CO_2 * 2 = 4.00 mol CO_2
1.00 mol H_2O * 2 = 2.00 mol H_2O
2.51 mol O_2 * 2 = 5.02 mol O_2 => 5.00 mol O_2
Step 3
going back to our equation, we can balance it with the mol ratios
Hydrocarbon + 5 O_2 => 4 CO_2 + 2 H_2O
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 C_1H_1 or just CH
