Mr Cole Chemistry

Combustion Calculations

Combustion Calculations

Intro to Organic Worksheet

Combustion & Gas Volumes

1. A 400 cm3 sample of ethane was mixed with 3000 cm3 of oxygen at 25.0oC. The mixture was ignited.
a) Write a balanced equation for the combustion reaction that occurs.

C2H6 + 3.5O2 → 2CO2 + 3H2O

b) After the reaction, the gases were cooled to 25.0°C. Calculate the volume of gas after the reaction.
  • Reacting Volumes Ratio: 1 (Ethane) : 3.5 (O2) : 2 (CO2)
  • Ethane reacted: 400 cm3
  • Oxygen reacted: 400 × 3.5 = 1400 cm3
  • CO2 produced: 400 × 2 = 800 cm3
  • Oxygen remaining: 3000 – 1400 = 1600 cm3
  • Total Volume: 1600 (O2) + 800 (CO2) = 2400 cm3
  • Note: Water is a liquid at 25oC, so its volume is negligible.
2. A sample of ethanal was heated to 90.0oC, where it had a volume of 150cm3. 500cm3 of oxygen, also at 90.0oC, was added to the ethanal.
a) Write a balanced equation for the combustion reaction that occurs.

CH3CHO + 2.5O2 → 2CO2 + 2H2O

b) Calculate volume after the reaction was complete. The temperature remained at 90.0oC.
  • Ratio: 1 : 2.5 : 2 : (liquid water)
  • Oxygen reacted: 150 × 2.5 = 375 cm3
  • Oxygen remaining: 500 – 375 = 125 cm3
  • CO2 produced: 150 × 2 = 300 cm3
  • Total Volume: 125 + 300 = 425 cm3
  • Note: Water is a liquid at 90oC.
3. A 50.0g sample of ethanol was vaporised by heating it to 100oC. It was mixed with 150 dm3 of oxygen, also at 100oC.
a) Write a balanced equation for the complete combustion of ethanol.

C2H5OH + 3O2 → 2CO2 + 3H2O

b) Calculate the volume after the reaction was complete. The temperature remained at 100°C.
  • Moles Ethanol: 50.0g / 46.0 = 1.087 mol
  • Volume Ethanol (gas): PV=nRT → V = (1.087 × 8.31 × 373) / 100000 = 0.0337 m3 = 33.7 dm3
  • Ratio (all gas at 100oC): 1 : 3 : 2 : 3
  • Oxygen Reacted: 33.7 × 3 = 101.1 dm3
  • Oxygen Remaining: 150 – 101.1 = 48.9 dm3
  • CO2 Produced: 33.7 × 2 = 67.4 dm3
  • H2O Produced (gas): 33.7 × 3 = 101.1 dm3
  • Total Volume: 48.9 + 67.4 + 101.1 = 217.4 dm3
4. 50.0cm3 of methane was kept at 25.0oC and reacted with 800cm3 of air (also at 25.0oC before reacting). After the reaction, the mixture of gases was cooled to 25.0oC.
a) Write the equation for the reaction.

CH4 + 2O2 → CO2 + 2H2O

b) Calculate the volume after the mixture was cooled to 25.0°C.
  • Composition of Air: ~21% Oxygen, ~79% Nitrogen
  • Initial O2: 800 × 0.21 = 168 cm3
  • Initial N2 (unreacted): 800 – 168 = 632 cm3
  • Oxygen Reacted: 50 (CH4) × 2 = 100 cm3
  • Oxygen Remaining: 168 – 100 = 68 cm3
  • CO2 Produced: 50 cm3
  • Total Volume: 68 (O2) + 50 (CO2) + 632 (N2) = 750 cm3
5. 20.0g of cyclohexane was vaporised by heating it to 80oC and mixed with 600dm3 of air (also at 80oC). The pressure was 100kPa.
a) Write the equation for the complete combustion reaction that occurred.

C6H12 + 9O2 → 6CO2 + 6H2O

b) Calculate the volume of gas present after the reaction had occurred.
  • Moles Cyclohexane (C6H12): 20.0g / 84.0 = 0.238 mol
  • Moles O2 Reacted: 0.238 × 9 = 2.142 mol
  • Total Moles in 600dm3 Air: n = PV/RT = (100000 × 0.6) / (8.31 × 353) = 20.45 mol
  • Initial Moles O2 (21%): 4.29 mol
  • Initial Moles N2 (79%): 16.16 mol (This remains unreacted)
  • Final Moles:
    • O2 remaining: 4.29 – 2.142 = 2.148 mol
    • CO2 produced: 0.238 × 6 = 1.428 mol
    • N2 unreacted: 16.16 mol
  • Total Final Moles: 2.148 + 1.428 + 16.16 = 19.736 mol
  • Final Volume: V = nRT/P = (19.736 × 8.31 × 353) / 100000 = 0.579 m3 = 579 dm3

Empirical Formula Calculations

6. 46.0g of a hydrocarbon with an unknown formula reacts with excess oxygen, forming carbon dioxide and water. When cooled to 25oC at a pressure of 101325Pa, the volume of carbon dioxide was 77.4dm3 and the mass of water was 71.4. What was the empirical formula of the hydrocarbon?
  • Moles C (from CO2): n = PV/RT = (101325 × 0.0774) / (8.31 × 298) = 3.17 mol
  • Moles H (from H2O): (71.4 / 18) × 2 = 7.93 mol
  • Ratio C:H: 3.17 : 7.93
  • Simplify: 1 : 2.5 → 2 : 5
  • Empirical Formula: C2H5
7. 89.0g of a hydrocarbon with an unknown formula reacts with excess oxygen, forming carbon dioxide and water. When cooled to 20oC at a pressure of 100kPa, the volume of carbon dioxide was 154.8dm3 and the mass of water was 114.4. What was the empirical formula of the hydrocarbon?
  • Moles C (from CO2): n = (100000 × 0.1548) / (8.31 × 293) = 6.36 mol
  • Moles H (from H2O): (114.4 / 18) × 2 = 12.71 mol
  • Ratio C:H: 6.36 : 12.71 ≈ 1 : 2
  • Empirical Formula: CH2
8. 50.0g of an organic compound containing carbon, hydrogen and oxygen was reacted with 100dm3 of oxygen at 298k and 100kPa (an excess). When the products were cooled back to the original conditions, 64.0dm3 of carbon dioxide and 46.6g of water were formed.
a) What was the empirical formula of the organic compound?
  • Moles C: (100000 × 0.064) / (8.31 × 298) = 2.58 mol. Mass = 31.0g
  • Moles H: (46.6 / 18) × 2 = 5.18 mol. Mass = 5.18g
  • Mass O: 50.0 – 31.0 – 5.18 = 13.82g. Moles O = 13.82 / 16 = 0.864 mol
  • Ratios: C(2.58/0.864) : H(5.18/0.864) : O(1) = 3 : 6 : 1
  • Empirical Formula: C3H6O
b) Given the Mr of 116, what was the molecular formula of the compound?

Empirical Mass (C3H6O) = 58.
116 / 58 = 2.
Molecular Formula: C6H12O2

c) Calculate the volume of oxygen left unreacted at the end of the reaction.
  • Equation: C6H12O2 + 8 😯2 → 6CO2 + 6H2O
  • Moles Compound: 50.0 / 116 = 0.431 mol
  • Moles O2 Reacted: 0.431 × 8 = 3.448 mol
  • Initial Moles O2: (100000 × 0.1) / (8.31 × 298) = 4.038 mol
  • Remaining O2: 4.038 – 3.448 = 0.590 mol
  • Volume Remaining: (0.590 × 8.31 × 298) / 100000 = 0.0146 m3 = 14.6 dm3
Return to Topic