Combustion Analysis Example

Overview

Analyze a compound containing carbon, hydrogen, and oxygen.
Objective: Determine the molecular formula from combustion analysis.
Given:
- Mass of compound: 4.18 grams
- Collected: 2.5 grams of water, 6.13 grams of CO2
- Molar mass of the compound: 180.156 g/mol

Use the mass of collected CO2 (6.13 grams)
Molar Mass of CO2: 44.01 g/mol
Convert grams of CO2 to moles:
- [ \text{Moles of CO2} = \frac{6.13 \text{ g}}{44.01 \text{ g/mol}} = 0.139 \text{ moles} ]
Stoichiometry: 1 mole of CO2 = 1 mole of C
Moles of Carbon: 0.139 moles

Use the mass of collected H2O (2.50 grams)
Molar Mass of H2O: 18.02 g/mol
Convert grams of H2O to moles:
- [ \text{Moles of H2O} = \frac{2.50 \text{ g}}{18.02 \text{ g/mol}} = 0.139 \text{ moles} ]
Stoichiometry: 1 mole of H2O = 2 moles of H
Moles of Hydrogen: 0.277 moles

Calculate the mass of carbon and hydrogen in the compound:
- Mass of Carbon:
  - [ \text{Mass of C} = 0.139 \text{ moles} \times 12.011 \text{ g/mol} = 1.67 \text{ g} ]
- Mass of Hydrogen:
  - [ \text{Mass of H} = 0.277 \text{ moles} \times 1.0079 \text{ g/mol} = 0.279 \text{ g} ]
Subtract to find mass of oxygen:
- [ \text{Mass of O} = 4.18 \text{ g (sample)} - 1.67 \text{ g (C)} - 0.279 \text{ g (H)} = 2.231 \text{ g} ]
Molar Mass of O: 16.00 g/mol
Calculate moles of oxygen:
- [ \text{Moles of O} = \frac{2.231 \text{ g}}{16.00 \text{ g/mol}} = 0.139 \text{ moles} ]

Given molar mass of compound: 180.156 g/mol
Empirical formula mass (CH$_2$O):
- [ 12.011 + (2 \times 1.0079) + 16.00 = 30.03 \text{ g/mol} ]
Divide molar mass by empirical formula mass:
- [ \frac{180.156 \text{ g/mol}}{30.03 \text{ g/mol}} = 5 ]
Molecular Formula: C$5$H${10}$O$_5$