Question 4 of 18: Oxygen Required for Complete Oxidation from Elemental Composition

Question 4: Oxygen Required for Complete Oxidation from Elemental Composition (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. Mass of material $=30$ kg; dry-weight composition C 52.85%, H 6.48%, O 24.76%, N 15.12%; atomic weights C=12, H=1, O=16, N=14.

Find. Mass of O₂ (kg) required for complete oxidation of the 30 kg sample.

Approach. Convert each element's mass to moles, then apply the general oxidation balance for an organic compound $\text{C}_a\text{H}_b\text{O}_c\text{N}_d$ oxidized to CO₂, H₂O and NH₃: $$\text{C}_a\text{H}_b\text{O}_c\text{N}_d + \left(a+\tfrac{b}{4}-\tfrac{c}{2}-\tfrac{3d}{4}\right)\text{O}_2 \to a\,\text{CO}_2 + \left(\tfrac{b}{2}-\tfrac{3d}{2}\right)\text{H}_2\text{O} + d\,\text{NH}_3$$ which, applied on a per-mole-of-element basis, gives the total moles of O₂ directly without first reducing to an empirical formula.

1. Mass of each element in 30 kg. $$m_C=30(0.5285)=15.855\ \text{kg}, \quad m_H=30(0.0648)=1.944\ \text{kg}$$ $$m_O=30(0.2476)=7.428\ \text{kg}, \quad m_N=30(0.1512)=4.536\ \text{kg}$$
2. Moles of each element. $$n_C=\frac{15{,}855}{12}=1{,}321.3\ \text{mol}, \quad n_H=\frac{1{,}944}{1}=1{,}944.0\ \text{mol}$$ $$n_O=\frac{7{,}428}{16}=464.3\ \text{mol}, \quad n_N=\frac{4{,}536}{14}=324.0\ \text{mol}$$
3. Moles of O₂ required. Applying the oxidation-balance coefficient term-by-term (the "$a,b,c,d$" ratios cancel into the element moles directly): $$n_{O_2}=n_C+\frac{n_H}{4}-\frac{n_O}{2}-\frac{3n_N}{4} = 1{,}321.3+486.0-232.1-243.0 = 1{,}332.1\ \text{mol}$$
4. Mass of O₂. $$\boxed{m_{O_2}=1{,}332.1\ \text{mol}\times 32\ \text{g/mol} = 42{,}628\ \text{g} = 42.6\ \text{kg}}$$