Exercise 5 | Chemical Calculations

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 + Y4 - Z2) O2 → X CO2 + Y2 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 MCMCO2
mC = 3.350 x 12.044.0 = 9.14 x 10-1 g

mass of hydrogen in original sample = mass of hydrogen in the water formed
mH = mH2O x MHMH2O
mC = 0.823 x 2.0218.0 = 9.24 x 10-2 g


Determine mass % of carbon and hydrogen in cpd:

mass % of C = mCmcompound x 100 = 0.9141.250 x 100 = 73.1%
mass % of H = mHmcompound x 100 = 0.09241.250 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