Oxygen Required for Complete Oxidation: 18-Env-B9 Environmental Chemistry and Microbiology, May 2016, Q6

Question 6: Oxygen Required for Complete Oxidation (10 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. Dry-weight elemental composition C 52.85% / H 6.48% / O 24.78% / N 15.2%; sample mass 50 kg; atomic weights as stated.

Find. Mass of O₂ required for complete oxidation of the 50 kg sample.

Approach. Work on a 100 g basis to convert each element’s mass percentage into moles, then apply the same complete-oxidation stoichiometry as Question 3 ($\text{C}_c\text{H}_h\text{O}_o\text{N}_n\to c\,\text{CO}_2+\dots+n\,\text{NH}_3$, nitrogen released as ammonia rather than further oxidized) to get the oxygen demand per gram of material, then scale to 50 kg.

Moles of each element per 100 g of material. $$n_C=\dfrac{52.85}{12}=4.404,\quad n_H=\dfrac{6.48}{1}=6.480,\quad n_O=\dfrac{24.78}{16}=1.549,\quad n_N=\dfrac{15.2}{14}=1.086\ \text{mol}$$
Oxygen demand from the elemental balance. Treating the material as $\text{C}_c\text{H}_h\text{O}_o\text{N}_n$ oxidized to $\text{CO}_2$, $\text{H}_2\text{O}$ and $\text{NH}_3$ (the same convention as Question 3), the O₂ required per formula unit is $x=c+\tfrac{h-3n}{4}-\tfrac{o}{2}$; applying it directly to the per-100 g mole counts: $$x=4.404+\dfrac{6.480-3(1.086)}{4}-\dfrac{1.549}{2}=4.404+0.806-0.775=4.436\ \text{mol O}_2\ \text{per 100 g}$$
Mass of O₂ per unit mass of material. $$\dfrac{m_{\text{O}_2}}{100\ \text{g material}}=4.436(32)=141.9\ \text{g}\quad\Rightarrow\quad 1.419\ \text{g O}_2/\text{g material}$$
Scale to the 50 kg sample. $$m_{\text{O}_2}=50\ \text{kg}\times1.419=\boxed{71.0\ \text{kg O}_2}$$

Check: the printed percentages sum to 99.3% (52.85+6.48+24.78+15.2), a normal rounding gap in a lab elemental analysis — used as-given rather than re-normalized to 100%, since the gap is well within analytical precision and re-normalizing would not be defensible without knowing which value carries the error. Nitrogen is taken to end as NH₃ (not further nitrified), consistent with the C₅H₄NO₂ convention used in Question 3.

Quantity	Value
O₂ demand per 100 g material	4.436 mol (141.9 g)
O₂ demand ratio	1.419 g O₂/g material
O₂ required for 50 kg material	71.0 kg

Concept & Theory

Topic. Theoretical oxygen demand (ThOD) from an elemental (proximate) analysis.

Key relations. Element mass % $\to$ moles ($n_i=\%_i/A_i$ per 100 g); $x=c+\tfrac{h-3n}{4}-\tfrac{o}{2}$ mol O₂ per formula unit.

Why this works. Because the elemental-balance oxygen demand is linear in the mole counts of C, H, O and N, it can be applied directly to the per-100 g mole inventory without first forcing those moles into a small-integer empirical formula — the two approaches give the identical answer, but skipping the (here, not perfectly clean) integer-ratio step avoids introducing rounding error.

Common pitfall. Ignoring the oxygen already present in the material (the $-o/2$ term) — that oxygen is already available in the molecule and reduces the external O₂ demand; omitting it overstates the answer.

Source. Metcalf & Eddy, Wastewater Engineering, ch. 8 (stoichiometry of biological oxidation).

Question 6 of 20: Oxygen Required for Complete Oxidation

Section 1: Chemistry (8 questions, 50 marks)

Question 6: Oxygen Required for Complete Oxidation (10 marks)