GATE 2026 BT Q.40: solution and answer

Official answer: ASource: Official answer key, GATE 2026 BT (IIT Guwahati)

1. Removing the ash leaves $100-12=88\text{g}$ of organic matter per $100\text{g}$ of biomass (and indeed $50+8+20+10=88$). On an ash-free basis the percentages become $\text{C}=50/0.88=56.8\%$, $\text{H}=9.09\%$, $\text{O}=22.7\%$, $\text{N}=11.4\%$. Every element is divided by the same factor $0.88$, so the ratios between them do not change, and we can work directly with the given masses.

2. Moles of each element in $100\text{g}$ of the sample (mass $÷$ atomic mass, with $\text{C}=12$, $\text{H}=1$, $\text{O}=16$, $\text{N}=14{\text{g mol}}^{-1}$): $\text{C}:\frac{50}{12}=4.167$, $\text{H}:\frac{8}{1}=8.00$, $\text{O}:\frac{20}{16}=1.25$, $\text{N}:\frac{10}{14}=0.714\text{mol}$.

3. Normalise to one N atom by dividing each by the moles of N: $$\text{C}:\frac{50/12}{10/14}=\frac{700}{120}=5.83,\text{H}:\frac{8}{10/14}=11.2,\text{O}:\frac{20/16}{10/14}=\frac{280}{160}=1.75,\text{N}:1$$

4. The empirical formula of the ash-free cells is ${\text{C}}_{5.83}{\text{H}}_{11.2}{\text{O}}_{1.75}\text{N}$. Written to one decimal place, as the options are, this is ${\text{C}}_{5.8}{\text{H}}_{11.2}{\text{O}}_{1.7}\text{N}$. (The printed option cuts O off at $1.7$ rather than rounding $1.75$ up to $1.8$; no option shows $1.8$, so this does not change the match.)

5. Checking the others: B, C and D each have a C:N ratio ($5.0$, $6.8$, $4.8$) and an O:N ratio ($1.1$, $2.1$, $3.1$) that differ from $5.83$ and $1.75$, so only option (A) is consistent with the data.