GATE 2025 CE (CE2) – Question 49
A compound has a general formula C$_a$H$_b$O$_c$N$_d$ and molecular weight 187. A 935 mg/l solution of the compound is prepared in distilled deionized water. The Total Organic Carbon (TOC) is measured as 360 mg/l (as C). The Chemical Oxygen Demand (COD) and the Total Kjeldahl Nitrogen (TKN) are determined as 600 mg/l (as O$_2$) and 140 mg/l (as N), respectively (as per the chemical equation given below). Which of the following options is/are CORRECT?
$C_aH_bO_cN_d + \frac{(4a + b - 2c - 3d)}{4}O_2 \to aCO_2 + \frac{(b - 3d)}{2}H_2O + dNH_3$
Atomic weight: C(12), H(1), O(16), N(14)
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Correct answer: (A) a = 6; (B) b = 7; (C) c = 5
Explanation
The concentration of the compound is $\frac{935}{187} = 5$ mmol/l. The carbon is $\frac{360}{12} = 30$ mmol/l, so $a = \frac{30}{5} = 6$. The nitrogen is $\frac{140}{14} = 10$ mmol/l, so $d = 2$. The molecular weight gives $72 + b + 16c + 28 = 187$, that is $b + 16c = 87$. The COD is $\frac{600}{32} = 18.75$ mmol O₂/l, which is $3.75$ mol per mol of compound, so $\frac{4a + b - 2c - 3d}{4} = 3.75$, that is $24 + b - 2c - 6 = 15$ and $b - 2c = -3$. Solving, $87 - 18c = -3$ gives $c = 5$ and $b = 7$. So $a = 6$, $b = 7$, $c = 5$ and $d = 2$ (not 3).