Question 8 of 18: Theoretical Oxygen Demand of Phenol

Question 8: Theoretical Oxygen Demand of Phenol (5 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. 100 g of phenol (C₆H₅OH), the balanced complete-combustion stoichiometry C₆H₅OH + 7 O₂ → 6 CO₂ + 3 H₂O, and atomic weights C = 12, H = 1, O = 16.

Find. The theoretical oxygen demand (ThOD), i.e. the mass of O₂ required to fully oxidize the 100 g phenol sample.

Approach. Convert the phenol mass to moles using its molecular weight, apply the 1:7 phenol-to-O₂ stoichiometric ratio from the balanced equation to get moles of O₂, then convert back to a mass of O₂.

1. Molecular weight of phenol. C₆H₅OH has 6 C, 6 H (5 ring H + 1 hydroxyl H) and 1 O: $$MW_{phenol} = 6(12) + 6(1) + 1(16) = 72 + 6 + 16 = 94\ \text{g/mol}$$
2. Moles of phenol in the 100 g sample. $$n_{phenol} = \dfrac{100\ \text{g}}{94\ \text{g/mol}} = 1.0638\ \text{mol}$$
3. Moles of O₂ required. The balanced equation consumes 7 mol O₂ per mol phenol: $$n_{O_2} = 7 \times 1.0638 = 7.4468\ \text{mol}$$
4. Mass of O₂ (ThOD). With $MW_{O_2} = 2(16) = 32\ \text{g/mol}$: $$\boxed{ThOD = 7.4468 \times 32 = 238.3\ \text{g O}_2 \text{ per 100 g phenol}}$$ Equivalently, phenol has a theoretical oxygen demand of $238.3/100 \approx 2.38\ \text{g O}_2\text{/g phenol}$.