Combustion analysis | Fundamentals of Chemical Reactions
General Chemistry 2 - Combustion analysis
Combustion analysis is used to determine the formula of a compound containing carbon, hydrogen and oxygen.
1.250 g of this compound is burned in a combustion analysis apparatus. The mass of CO2 produced is 3.350 g and the mass of H2O is 0.823 g.
Determine the empirical formula of the compound.
Chemical equation:
CxHyOz + O2 → CO2 + H2O (unbalanced)
CxHyOz + (X + - ) O2 → X CO2 + H2O (balanced)
Determine the mass of carbon and hydrogen in the original sample:
mass of carbon in original sample = mass of carbon in the carbon dioxide formed
mC = mCO2 x
mC = 3.350 x = 9.14 x 10-1 g
mass of hydrogen in original sample = mass of hydrogen in the water formed
mH = mH2O x
mC = 0.823 x = 9.24 x 10-2 g
Determine mass % of carbon and hydrogen in cpd:
mass % of C = x 100 = x 100 = 73.1%
mass % of H = x 100 = x 100 = 7.39 %
Determine mass % of oxygen in cpd:
mass % of O = 100 - mass % of C - mass % of H = 19.5 %
Determine the empirical formula:
In a 100 g sample: 73.1 g of C, 7.39 g of H and 19.5 g of O
⇒ 6.09 mol of C, 7.32 mol of H and 1.219 mol of O
⇒ For 1 mole of O, we have 5 moles of C and 6 moles of H.
The empirical formula of the compound is C5H6O