Question

Determine the total organic carbon, total inorganic carbon and total carbon in a wastewater that contains a mixture of 50 mg/L of methyl alcohol (CH3OH), 150 mg/L of methyl ethyl ketone (CH3COC2H5) and 200 mg/L of bicarbonate (HCO3-).

Answer 1

Let's consider amount of waste in 1L of water

so in 1 L of water = 50 mg of CH3OH is there

now molar mass of CH3OH = 12+3 + 16 + 1 = 32 g/mol

so in 32 gm of CH3OH, 12 g is carbon as molar mass of carbon in 12g/mol

so in 32000 mg of CH3OH mass of Carbon = 12000 mg

therefore in 50 mg, mass of carbon = (12000/32000) X 50 = 18.75 mg

now methyl ethyl ketone (CH3COC2H5), molar mass = 72 g/mol

mass of C in one mole of methylethyl ketone = 2 X 12 = 24 g

so mass of C in 150 mg methyl ethyl ketone = (24/72) X 150 = 50 mg

so amount of organic C in 1 L of waste wate = 18.75 + 50 = 68.75 mg

Now molar mass of HCO₃^- = 61 g/mol

mass of carbon in 1 mol of HCO₃^- =  12 g

therefore mass of C in 200 mg of HCO₃^- = (12/61) X 200 = 39.34 mg

So amount of inorganic carbon in 1L waste water = 39.34 mg

And, total carbon in a L of wastewater = 68.75 mg + 39.34 mg = 108.1 mg