Estimation Of Nitrogen (duma's Method)

Duma's method oxidises the organic compound over heated copper oxide so that all bound nitrogen is liberated as free nitrogen gas, which is collected over potassium hydroxide and its volume measured. At STP, 22400 mL of N_2 corresponds to 28 g of nitrogen (the molar mass of N_2). The mass of nitrogen in the sample is therefore (28 / 22400) × 112 = 0.14 g. Percentage of nitrogen = (0.14 / 0.5) × 100 = 28%. Option 14% mistakenly omits dividing by the sample mass and instead reports the nitrogen mass scaled incorrectly. Option 56% would result from using 14 g per 22400 mL, i.e. treating the gas as atomic nitrogen rather than diatomic N_2. Option 22.4% confuses the molar volume figure with a percentage. The working applies the NCERT relation %N = (28 × V_N2)/(22400 × mass) × 100, where the volume is corrected to standard temperature and pressure before substitution. Unlike Kjeldahl's method, Duma's method works for all nitrogen-containing compounds, including those with nitrogen in rings or in nitro groups, because oxidative combustion liberates every form of nitrogen as the element. Plausibility check: 112 mL is one two-hundredth of the molar volume, giving 0.14 g nitrogen, a magnitude consistent with a nitrogen-rich compound such as a small amine.