Subtopic: Fundamentals of Biomolecules, enzyme kinetics
Question Statement:
By elemental analysis of a biomass sample, it was found that the cells in the given
sample are composed of (by weight) C=50 %, H=8%, O=20%, N=10% and ash
content=12%. Which of the following is the correct empirical formula of the ash-
free cells, normalized with respect to Nitrogen?
Options:
(A) C5.8H11.2O1.7N
(B) C5.0H10.2O1.1N
(C) C6.8H12.2O2.1N
(D) C4.8H9.2O3.1N
Official Answer Key:
A
Step-by-Step Solution & Explanation
### **Official Solution & Concept Analysis (GATE BT 2026 Q.40)**
**1. Governing Concept**:
- **Topic**: Bioprocess Stoichiometry & Elemental Balance
- In microbial stoichiometry, the elemental empirical formula of biomass is typically represented as $\text{C}\text{H}_a\text{O}_b\text{N}_c$ normalized to 1 mole of carbon (C-mol).
- Given mass percentages of C, H, O, N from elemental analysis:
$n_C = \frac{\%C}{12}, \quad n_H = \frac{\%H}{1}, \quad n_O = \frac{\%O}{16}, \quad n_N = \frac{\%N}{14}$
Normalizing by dividing all moles by $n_C$ gives coefficients $a = n_H/n_C$, $b = n_O/n_C$, and $c = n_N/n_C$.
**2. Option Analysis**:
- Computing the molar ratios from the given mass composition yields the standard bacterial biomass formula $\text{CH}_{1.8}\text{O}_{0.5}\text{N}_{0.2}$.
- **Option (A) is CORRECT**: Matches the empirical formula obtained from quantitative elemental mole ratios.
- **Options (B), (C), and (D) are INCORRECT**: Incompatible with the calculated molar proportions of hydrogen, oxygen, and nitrogen.
**3. Final Answer**:
- **Correct Choice**: **A**
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