1. Empirical Formula
Combustion analysis is one of the most common methods for determining empirical formulas.
A weighed compound is burned in oxygen and its products analyzed by a gas chromatogram.
It is particularly useful for analysis of hydrocarbons.

2. 0.250 g of compound X produces:
Combustion Analysis
Combustion Analysis: the technique of finding the mass composition of an unknown sample (X) by examining the products of its combustion.
X + O2 → CO2 + H2O
0.250 g of compound X produces:
0.686 g CO2 and g H2O

3. Combustion Analysis X + O2 → CO2 + H2O
1. Find the mass of C & H that must have been present in X (multiply masses of products by percent composition of the products).
C: g x g/44.01 g = g C
H: g x (2 x g)/18.02 g = g H

4. Combustion Analysis X + O2 → CO2 + H2O
0.187 g C g H = g total
So compound X must contain only C and H!!
2. Find the number of moles of C and H
C: g x mole/12.01 g = moles C
H: g x mole/1.008 g = moles H

5. Combustion Analysis X + O2 → CO2 + H2O
3. Find the RELATIVE number of moles of C and H in whole number units (divide by smallest number of moles)
C: / = 1
H: / = 4
If these numbers are fractions, multiply each by the same whole number.

6. Combustion Analysis X + O2 → CO2 + H2O
3. Write the Empirical Formula (use the relative numbers as subscripts)
CH4

7. Combustion Analysis Summary Find the mass of C and H in the sample.
Find the actual number of moles of C and H in the sample.
Find the relative number of moles of C and H in whole numbers.
Write the empirical formula for the unknown compound.

8. Combustion Analysis NOTE!
In step # 1 always check to see if the total mass of C and H adds up to the total mass of X combusted. If the combined mass of C and H is less than the mass of X, then the remainder is an unknown element (unless instructed otherwise).
If a third element is known, calculate the mass of that element by subtraction (at the end of step 1), and include the element in the remaining steps.

9. Combustion Analysis
Combustion Analysis provides the Empirical Formula. If a second technique provides the molecular weight, then the molecular formula may be deduced.
Calculate the empirical formula weight.
Find the number of “formula units” by dividing the known molecular weight by the formula weight.
Multiply the number of atoms in the empirical formula by the number of formula units.

10. Combustion Analysis
The molecular weight of glucose is 180 g/mole and its empirical formula is CH2O. Deduce the molecular formula.
Formula weight for CH2O is g/mole
# of formula units = 180/30.03 = 6
Molecular formula = C6H12O6