1. Atoms, Molecules, and Stoichiometry
Recap
of Lecture 5

2. Recap Combustion Analysis Actual Yield Theoretical Yield
Percentage Yield =
Actual Yield
Theoretical Yield
X 100%
Combustion Analysis
Values given in terms of mass
Values given in terms of volume
Find the mass of each element
 Find mole ratio and hence empirical formula
Make use of Avogadro’s law to compare volumes

3. Combustion Analysis Example 1: Using mass proportion:
Complete combustion of an unknown organic compound of mass 1.000g containing only carbon, hydrogen, and oxygen, gave 1.500g of carbon dioxide and 0.405g of water What is the empirical formula of the organic compound?
Using mass proportion:
Mass of C in 1.500g CO2 = MC
MCO2
x mCO2
12.0
44.0 x =
= g
Mass of C
Mass of CO2 MC
MCO2 =

4. Combustion Analysis Example 1: Using mass proportion:
Complete combustion of an unknown organic compound of mass 1.000g containing only carbon, hydrogen, and oxygen, gave 1.500g of carbon dioxide and 0.405g of water What is the empirical formula of the organic compound?
Using mass proportion:
Mass of C in 1.500g CO2 = MC
MCO2
x mCO2
12.0
44.0 x =
= g MC
Mass of C = x
Mass of CO2
MCO2

5. Combustion Analysis Example 1: 2 x MH MH2O Mass of H in 0.405g H2O =
Complete combustion of an unknown organic compound of mass 1.000g containing only carbon, hydrogen, and oxygen, gave 1.500g of carbon dioxide and 0.405g of water What is the empirical formula of the organic compound?
2 x MH
MH2O
x mH2O
Mass of H in 0.405g H2O = =
2 x 1.0
18.0
x 0.405
= g
Mass of oxygen in the compound = – – 0.045
= g

6. Compound’s empirical formula is C3H4O3
Combustion Analysis
C H O
Mass ratio
Mole ratio
0.409  16.0
= = =
Simplest whole number ratio
0.0341  0.0341
= 1 = = 1
Compound’s empirical formula is C3H4O3

7. Combustion Analysis Avogadro’s Law: no of moles of gas  volume of gas
Provided that temperature and pressure are kept constant
This means that, Mole ratio is equivalent to the Volume ratio
CXHY + (x + y/4) O2  x CO y/2 H2O
H2O
CO2 O2
CXHY
Mole ratio
Volume ratio
y/2 mol
x mol
(x + y/4) mol
1 mol
y/2 cm3
x cm3
(x + y/4) cm3
1 cm3
10 (y/2) cm3
10x cm3
10(x + y/4) cm3
10 cm3

8. CXHY + (x + y/4) O2  x CO2 + y/2 H2O
Combustion Analysis
Example 1
When 10 cm3 of a gaseous hydrocarbon was completely burnt in excess oxygen, 20 cm3 of carbon dioxide and 30 cm3 of steam was produced (all gas volumes measured under the same conditions).
Calculate the molecular formula of the hydrocarbon.
CXHY + (x + y/4) O2  x CO y/2 H2O
By Avogadro’s law, x 1 =
nCO2
nCxHy
VCO2
VCxHy = 20 10 =
 x = 2

9. Combustion Analysis Example 1 CXHY + (x + y/4) O2  x CO2 + y/2 H2O
When 10 cm3 of a gaseous hydrocarbon was completely burnt in excess oxygen, 20 cm3 of carbon dioxide and 30 cm3 of steam was produced (all gas volumes measured under the same conditions).
Calculate the molecular formula of the hydrocarbon.
CXHY + (x + y/4) O2  x CO y/2 H2O
By Avogadro’s law,
y/2 1 =
nH2O
nCxHy
VH2O
VCxHy = 30 10 =
 y = 6
Molecular formula is C2H6

10. Combustion Analysis 19 Points to note:
The volume of CO2 may be given as a decrease in volume when the residual gases are passed through NaOH or other alkali
The volume of H2O may be given as
Decrease in volume when the residual gases are passed through anhydrous CaCl2 or conc. H2SO4
Decrease in volume when the residual gases are cooled to below 100oC at atm pressure
Oxygen is usually added in excess and not in stoichiometric amount, so there will be excess oxygen as residual gas
Contraction = volume of reactants – volume of products
Expansion = volume of products – volume of reactants

11. CXHY + (x + y/4) O2  x CO2 + y/2 H2O
liquid state
CXHY + (x + y/4) O2  x CO y/2 H2O
Example 2
In an experiment, 10.0 cm3 of a gaseous hydrocarbon was exploded with an excess of oxygen, and a contraction of 35.0 cm3 occurred. The resultant volume was passed through aqueous sodium hydroxide, and the volume was reduced by another 40.0 cm3. Assuming that all volumes were measured at room temperature and pressure, determine the empirical formula of the hydrocarbon.

12. CXHY + (x + y/4) O2  x CO2 + y/2 H2O
liquid state
CXHY + (x + y/4) O2  x CO y/2 H2O
Example 2
In an experiment, 10.0 cm3 of a gaseous hydrocarbon was exploded with an excess of oxygen, and a contraction of 35.0 cm3 occurred. The resultant volume was passed through aqueous sodium hydroxide, and the volume was reduced by another 40.0 cm3. Assuming that all volumes were measured at room temperature and pressure, determine the empirical formula of the hydrocarbon.
Volume of CO2 gas = 40.0 cm3 [absorbed by aq NaOH]
By Avogadro’s Law,
nCO2
nCxHy
VCO2
VCxHy = x 1 40 10 = 
 x = 4

13. CXHY + (x + y/4) O2  x CO2 + y/2 H2O
Example 2
In an experiment, 10.0 cm3 of a gaseous hydrocarbon was exploded with an excess of oxygen, and a contraction of 35.0 cm3 occurred. The resultant volume was passed through aqueous sodium hydroxide, and the volume was reduced by another 40.0 cm3. Assuming that all volumes were measured at room temperature and pressure, determine the empirical formula of the hydrocarbon.
H2O
CO2 O2
CXHY
Mole ratio
Volume ratio
y/2 mol
x mol
(x + y/4) mol
1 mol
0 cm3
10x cm3
10(x + y/4) cm3
10 cm3

14. CXHY + (x + y/4) O2  x CO2 + y/2 H2O
Example 2
In an experiment, 10.0 cm3 of a gaseous hydrocarbon was exploded with an excess of oxygen, and a contraction of 35.0 cm3 occurred. The resultant volume was passed through aqueous sodium hydroxide, and the volume was reduced by another 40.0 cm3. Assuming that all volumes were measured at room temperature and pressure, determine the empirical formula of the hydrocarbon.
H2O
CO2 O2
CXHY
Mole ratio
Volume ratio
y/2 mol
x mol
(x + y/4) mol
1 mol
10 cm3
10(x + y/4) cm3
40 cm3
0 cm3

15. CXHY + (x + y/4) O2  x CO2 + y/2 H2O
Example 2
In an experiment, 10.0 cm3 of a gaseous hydrocarbon was exploded with an excess of oxygen, and a contraction of 35.0 cm3 occurred. The resultant volume was passed through aqueous sodium hydroxide, and the volume was reduced by another 40.0 cm3. Assuming that all volumes were measured at room temperature and pressure, determine the empirical formula of the hydrocarbon.
= Initial volume of gases – final volume of gases
35.0 = { ( x + y/4 ) + excess O2 } – { 40 + excess O2 }
65 = 10(x+y/4)
Since x = 4, solving y = 10
Hence the empirical formula of the hydrocarbon is C4H10

16. CXHY + (x + y/4) O2  x CO2 + y/2 H2O
Example 3
In an experiment, 10.0 cm3 of a gaseous hydrocarbon was exploded with cm3 of oxygen. The volume of the resulting gas mixture was 80.0 cm3, which decreased to 40.0 cm3 when passed through aqueous sodium hydroxide. All volumes were measured at room temperature and pressure. What is the formula of the hydrocarbon?
Volume of CO2 produced = 80.0 – 40.0 = 40.0 cm3
nCO2
nCxHy
VCO2
VCxHy = x 1 40 10 = 
 x = 4 20
Volume of resulting gas mixture = 80.0 cm3
= volume of CO2 + volume of unreacted O2
Hence volume of unreacted O2 = 80.0 – 40.0 = 40.0 cm3

17. CXHY + (x + y/4) O2  x CO2 + y/2 H2O
Example 3
In an experiment, 10.0 cm3 of a gaseous hydrocarbon was exploded with cm3 of oxygen. The volume of the resulting gas mixture was 80.0 cm3, which decreased to 40.0 cm3 when passed through aqueous sodium hydroxide. All volumes were measured at room temperature and pressure. What is the formula of the hydrocarbon?
Therefore volume of O2 used up in the reaction
= – 40.0 = 60.0 cm3
nO2
nCxHy
VO2
VCxHy =
x + y/4 1 60 10 = 
y = 8
The formula for the hydrocarbon is C4H8

18. What have I learnt?
Determine the formula of a hydrocarbon given the combustion analysis data
In terms of mass
In terms of volume

19. End of Lecture 6 (End of part I)
Have a prosperous and happy lunar new year (and do take care of your health!)