Question:
Anaerobic bacteria are being utilized for the destruction of biodegradable organic content of wastewater measured in terms of Chemical Oxygen Demand (COD). The yield coefficient (Y) is 0.06 gram Volatile Suspended Solids (VSS) per gram of COD. Bacterial biomass forms the sludge during the process. The proximate chemical formula of anaerobic bacteria is \(C_6H_7NO_2\).
The COD (in %) converted into bacterial biomass is (rounded off to the nearest integer).
Anaerobic bacteria are being utilized for the destruction of biodegradable organic content of wastewater measured in terms of Chemical Oxygen Demand (COD). The yield coefficient (Y) is 0.06 gram Volatile Suspended Solids (VSS) per gram of COD. Bacterial biomass forms the sludge during the process. The proximate chemical formula of anaerobic bacteria is \(C_6H_7NO_2\).
The COD (in %) converted into bacterial biomass is (rounded off to the nearest integer).
The COD (in %) converted into bacterial biomass is (rounded off to the nearest integer).
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Find the oxygen (COD) equivalent of one mole of the cell formula \(C_6H_7NO_2\) by balancing its oxidation reaction, then multiply that by the yield coefficient to get the fraction of feed COD that ends up in the biomass.
Updated On: Jul 22, 2026
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Verified By Collegedunia
Correct Answer: 9
Solution and Explanation
Step 1: Understand what "COD converted to biomass" means.
When anaerobic bacteria break down the organic matter in wastewater, only part of the feed COD gets used to build new bacterial cells (the biomass, or sludge). Those cells still carry an oxygen demand of their own, because they can be oxidised further to \(CO_2\), \(H_2O\) and \(NH_3\).
So the fraction of COD that ends up in the biomass equals the mass of biomass formed per gram of COD used, multiplied by the COD equivalent of that biomass, that is, the grams of \(O_2\) needed to fully oxidise one gram of cells.
Step 2: Write the oxidation reaction of the bacterial cells.
The cell formula given is \(C_6H_7NO_2\). Oxidising it fully to carbon dioxide, ammonia and water needs oxygen. Balancing carbon, hydrogen, nitrogen and oxygen atoms gives:
\[ C_6H_7NO_2 + 6O_2 \rightarrow 6CO_2 + NH_3 + 2H_2O \] Check the oxygen balance: the left side has \(2 + 6(2) = 14\) oxygen atoms; the right side has \(6(2) + 2(1) = 14\) oxygen atoms. The equation balances, so 6 moles of \(O_2\) are needed per mole of cells.
Step 3: Find the oxygen demand per gram of biomass.
Molecular weight of \(C_6H_7NO_2 = 6(12) + 7(1) + 14 + 2(16) = 72 + 7 + 14 + 32 = 125\).
One mole of biomass (125 g) needs 6 moles of \(O_2\), which weighs \(6 \times 32 = 192\) g. So the COD equivalent of the biomass is:
\[ \frac{192}{125} = 1.536 \text{ g COD per g VSS} \]
Step 4: Use the yield coefficient to find the biomass formed.
The yield coefficient \(Y = 0.06\) g VSS is formed for every gram of COD used up in the process. So for 1 g of COD removed, \(0.06\) g of biomass (VSS) is produced.
Step 5: Convert that biomass back into its COD equivalent.
The COD locked up in the biomass, per gram of COD originally removed, is:
\[ 0.06 \times 1.536 = 0.09216 \text{ g COD per g COD removed} \] As a percentage this is \(0.09216 \times 100 = 9.216\%\).
Final Answer:
Rounded off to the nearest integer, the COD converted into bacterial biomass is about 9%. \[ \boxed{9\%} \]
When anaerobic bacteria break down the organic matter in wastewater, only part of the feed COD gets used to build new bacterial cells (the biomass, or sludge). Those cells still carry an oxygen demand of their own, because they can be oxidised further to \(CO_2\), \(H_2O\) and \(NH_3\).
So the fraction of COD that ends up in the biomass equals the mass of biomass formed per gram of COD used, multiplied by the COD equivalent of that biomass, that is, the grams of \(O_2\) needed to fully oxidise one gram of cells.
Step 2: Write the oxidation reaction of the bacterial cells.
The cell formula given is \(C_6H_7NO_2\). Oxidising it fully to carbon dioxide, ammonia and water needs oxygen. Balancing carbon, hydrogen, nitrogen and oxygen atoms gives:
\[ C_6H_7NO_2 + 6O_2 \rightarrow 6CO_2 + NH_3 + 2H_2O \] Check the oxygen balance: the left side has \(2 + 6(2) = 14\) oxygen atoms; the right side has \(6(2) + 2(1) = 14\) oxygen atoms. The equation balances, so 6 moles of \(O_2\) are needed per mole of cells.
Step 3: Find the oxygen demand per gram of biomass.
Molecular weight of \(C_6H_7NO_2 = 6(12) + 7(1) + 14 + 2(16) = 72 + 7 + 14 + 32 = 125\).
One mole of biomass (125 g) needs 6 moles of \(O_2\), which weighs \(6 \times 32 = 192\) g. So the COD equivalent of the biomass is:
\[ \frac{192}{125} = 1.536 \text{ g COD per g VSS} \]
Step 4: Use the yield coefficient to find the biomass formed.
The yield coefficient \(Y = 0.06\) g VSS is formed for every gram of COD used up in the process. So for 1 g of COD removed, \(0.06\) g of biomass (VSS) is produced.
Step 5: Convert that biomass back into its COD equivalent.
The COD locked up in the biomass, per gram of COD originally removed, is:
\[ 0.06 \times 1.536 = 0.09216 \text{ g COD per g COD removed} \] As a percentage this is \(0.09216 \times 100 = 9.216\%\).
Final Answer:
Rounded off to the nearest integer, the COD converted into bacterial biomass is about 9%. \[ \boxed{9\%} \]
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