Question

Wastewater from a cheese factory has the following composition: \(8.0 \mathrm{~g} / \mathrm{L}\) protein \(\left(\mathrm{C}_{16} \mathrm{H}_{24} \mathrm{O}_{5} \mathrm{~N}_{4}\right) ; 12 \mathrm{~g} / \mathrm{L}\) carbohydrate \(\left(\mathrm{CH}_{2} \mathrm{O}\right) ;\) and \(2.0 \mathrm{~g} / \mathrm{L}\) fat \(\left(\mathrm{C}_{8} \mathrm{H}_{16} \mathrm{O}\right) .\) What is the total organic car- bon (TOC) of the wastewater in g/L?

Short answer

The TOC of the wastewater is approximately 10.62 g/L.

Step-by-step solution

- Calculate the moles of each compound

Determine the molecular weight of each compound: For protein \(\text{C}_{16} \text{H}_{24} \text{O}_{5} \text{N}_{4}\), the molecular weight is \(16 \times 12 + 24 \times 1 + 5 \times 16 + 4 \times 14 = 356 \text{g/mol}\). For carbohydrate \(\text{CH}_{2} \text{O}\), the molecular weight is \(12 + 2 \times 1 + 16 = 30\text{g/mol}\). For fat \(\text{C}_{8} \text{H}_{16} \text{O}\), the molecular weight is \(8 \times 12 + 16 \times 1 + 16 = 128\text{g/mol}\).

- Find the moles of each compound per liter

Using the concentration of each compound, calculate the moles per liter of solution: For protein: \( \frac{8 \text{g}}{356 \text{g/mol}} = 0.0225 \text{mol/L}\) For carbohydrate: \( \frac{12 \text{g}}{30 \text{g/mol}} = 0.4 \text{mol/L}\) For fat: \( \frac{2 \text{g}}{128 \text{g/mol}} = 0.0156 \text{mol/L}\).

- Calculate the moles of carbon in each compound

Multiply the moles of each compound per liter by the number of carbon atoms in the molecular formula: For protein: \(0.0225 \text{mol/L} \times 16 \text{mol C} = 0.36 \text{mol C/L}\) For carbohydrate: \(0.4 \text{mol/L} \times 1 \text{mol C} = 0.4 \text{mol C/L}\) For fat: \(0.0156 \text{mol/L} \times 8 \text{mol C} = 0.1248 \text{mol C/L}\).

- Sum the moles of carbon from all compounds

Add the moles of carbon for each compound to find the total moles of carbon per liter: \(0.36 \text{mol C/L} + 0.4 \text{mol C/L} + 0.1248 \text{mol C/L} = 0.8848 \text{mol C/L}\).

- Convert moles of carbon to grams of carbon

Convert the total moles of carbon per liter to grams per liter by multiplying by the atomic weight of carbon (12 g/mol): \(0.8848 \text{mol C/L} \times 12 \text{g/mol} = 10.6176 \text{g/L}\).

Key concepts

Molecular Weight Calculation

The first step in solving this problem is to calculate the molecular weight of each compound in the wastewater. Molecular weight is the sum of the atomic weights of all atoms in a molecule. For example, protein \(\text{C}_{16} \text{H}_{24} \text{O}_{5} \text{N}_{4}\) has the following atoms: 16 carbon, 24 hydrogen, 5 oxygen, and 4 nitrogen. Using the atomic weights (Carbon=12, Hydrogen=1, Oxygen=16, Nitrogen=14), we calculate: \[ \text{Molecular\ Weight\ of\ Protein} = 16 \times 12 + 24 \times 1 + 5 \times 16 + 4 \times 14 = 356 \text{g/mol} \] Following the same method, the weights of carbohydrate \(\text{CH}_{2} \text{O}\) and fat \(\text{C}_{8} \text{H}_{16} \text{O}\) are derived to be 30 g/mol and 128 g/mol respectively.

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions. It helps us find the number of moles of each compound. With given concentrations: \[ \text{For\ Protein} = \frac{8 \text{g}}{356 \text{g/mol}} = 0.0225 \text{mol/L} \] \[ \text{For\ Carbohydrate} = \frac{12 \text{g}}{30 \text{g/mol}} = 0.4 \text{mol/L} \] \[ \text{For\ Fat} = \frac{2 \text{g}}{128 \text{g/mol}} = 0.0156 \text{mol/L} \] These calculations allow us to proceed to finding the moles of carbon in each compound which is essential for determining the total organic carbon (TOC).

Wastewater Composition

Wastewater composition from industrial sources like a cheese factory often includes organic compounds such as proteins, carbohydrates, and fats. The presented problem provides their concentrations in g/L:

• Protein: 8.0 g/L
• Carbohydrate: 12 g/L
• Fat: 2.0 g/L

Understanding the composition allows us to calculate the total organic carbon (TOC), a crucial parameter indicating the level of organic pollution in wastewater.

Chemical Concentration

Chemical concentration is expressed in terms of mass per unit volume, i.e., g/L. In the given problem, concentrations for each compound are provided, enabling us to convert these into molarity (moles per liter), which simplifies subsequent calculations:

• For protein: 0.0225 mol/L
• For carbohydrate: 0.4 mol/L
• For fat: 0.0156 mol/L

These values will aid us in determining the exact moles of carbon present in the solution.

Carbon Content Calculation

To determine the total organic carbon (TOC) content in the wastewater, we must calculate the moles of carbon from each compound and sum them up:
\[ \text{For\ protein} = 0.0225 \text{mol/L} \times 16 = 0.36 \text{mol C/L} \]
\[ \text{For\ carbohydrate} = 0.4 \text{mol/L} \times 1 = 0.4 \text{mol C/L} \]
\[ \text{For\ fat} = 0.0156 \text{mol/L} \times 8 = 0.1248 \text{mol C/L} \]
Adding these yields:
\[ 0.36 \text{mol C/L} + 0.4 \text{mol C/L} + 0.1248 \text{mol C/L} = 0.8848 \text{mol C/L} \]
Converting to grams:
\[ 0.8848 \text{mol C/L} \times 12 \text{g/mol} = 10.6176 \text{g/L} \] Therefore, the TOC of the wastewater is 10.6176 g/L.