Determining Formulas by Combustion Analysis
Used mostly for compounds of formulas: CxHy or CxHyOz
Complete combustion:
Fuel + O2  CO2 + H2O
All carbon atoms go to CO2 (1 mole C per mole CO2).
All hydrogen atoms go to H2O (2 moles H per mole H2O).
If O is in the compound it is distributed between the CO2
and H2O. The O from O2 is also distributed between CO2
and H2O.
Procedure
1. Weigh sample.
2. Burn the sample.
3. Collect and weigh water produced.
4. Collect and weigh carbon dioxide produced.
See fig. 4.9
Calculations
1. Find mass and moles of C from mass of CO2.
2. Find mass and moles of H from mass of H2O.
3. If present find mass and moles O by subtraction from
the original sample mass.
4. Use mole ratio to determine empirical formula.
5. Molecular formula can be found if molar mass is
known from other data.
Example (This is the correct version; exercise 4.9; pg. 164)
Unknown formula CxHyOz.
Sample mass: 0.1342 g
Mass of recovered CO2: 0.240 g
Mass of recovered H2O: 0.0982 g
Molar mass: 74.1 g/mol
1. Find mass and moles of C.
(Molar masses: CO2 =44.0098 g/mol; C = 12.011 g/mol)
Convert grams CO2 to moles CO2.
 1molCO2
0.240 gCO2 
 44.0098gCO2

  5.45  10  3 molCO2


Since 1 mol CO2 has 1 mol C We have 5.54x10-3 moles of C.
Convert moles C tog rams C.
 12.011gC 
5.45  10  3 molC 
  0.0655 gC
 1molC 
2. Find mass and moles of H.
(Molar mass of H2O = 18.0152 g/mol; H = 1.0079 g/mol)
Convert grams water to moles water.
 1molH 2 O 
  5.45 10 3 molH 2 O
0.0982 gH 2 O
 18.0152 gH 2 O 
Notice that there are 2 mol H for every 1 mol H2O.
 2molH 
  1.09 10  2 molH
5.45 10 3 molH 2 O
 1molH 2 O 
 1.0079 gH 
1.09 10  2 molH 
  0.0110 gH
 1molH 
Convert mol H to g H.
3. Find mass and moles O.
Subtract masses of C and H from the original sample mass.
0.1342 g – 0.0655 g – 0.0110 g = 0.0577 g O
Convert g O to mol O.
 1molO 
  0.00361molO
0.0577 gO
 15.9994 gO 
4. Find mole ratio and empirical formula.
Divide each moles (C, H and O) by the lowest of the
three and convert to a whole number ratio.
C
H
3
2
5.53 10 mol
 1.5
0.00361mol
1.09 10 mol
3
0.00361mol
O
0.00361mol
1
0.00361mol
Multiply through by 2 to get whole numbers, so empirical
Formula is:
C2H6O2
5. Find molecular formula.
Add atomic weights to find the empirical weight (EW).
EW = (2 x 12.011u) + (6 x 1.0079u) + (2 x 15.9994u)
= 74.1u
= 74.1 g/mol
The given molar mass (MW) is also 74.1 g/mol, so the
empirical formula is the same as the molecular
formula.
MW
1
EW
Molecular formula is: C2H6O2