Oxygen content from combustion analysis

Worked example

A 0.3424 g sample of a C, H, O compound burns completely to give 0.7377 g of carbon dioxide and 0.3021 g of water. What percentage of its mass is oxygen? Use M(C) = 12.01 g mol⁻¹, M(H) = 1.008 g mol⁻¹, M(CO₂) = 44.01 g mol⁻¹ and M(H₂O) = 18.02 g mol⁻¹.

1. 1. Moles from mass: solve for n(C) = n(CO₂)

   $n = \dfrac{m}{M}$

   = 0.01676 mol

2. 2. Moles from mass: solve for m(C)

   $m = nM$

   = 0.2013 g

3. 3. Moles from mass: solve for n(H₂O)

   $n = \dfrac{m}{M}$

   = 0.01676 mol

4. 4. Mole ratio from the balanced equation: solve for n(H) = 2n(H₂O)

   $n_\mathrm{B} = \dfrac{b}{a}\,n_\mathrm{A}$

   = 0.03353 mol

5. 5. Moles from mass: solve for m(H)

   $m = nM$

   = 0.03380 g

6. 6. Weighted sum of 1·m(sample) − 1·m(C) − 1·m(H) → m(O) by difference

   = 0.1073 g

7. 7. Divide m(O) by difference ÷ m(sample) → O by mass

   = 0.3133

Answer: O by mass = 0.313

Method

All the carbon ends up in CO₂ and all the hydrogen in H₂O. n(C) = n(CO₂) and n(H) = 2n(H₂O); convert each to a mass. Oxygen is what is left: m(O) = m(sample) − m(C) − m(H). The same data give the empirical formula from the mole ratio C : H : O.

Common mistakes

• Each H₂O contains two hydrogen atoms: n(H) = 2 × n(H₂O).

Formulas used

• Moles from mass

  $n = \dfrac{m}{M}$

• Mole ratio from the balanced equation

  $n_\mathrm{B} = \dfrac{b}{a}\,n_\mathrm{A}$

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