Q.A sample of 0.5 g of an organic compound was treated according to Kjeldahl's method. The ammonia evolved was absorbed in 50 mL of 0.5 M H2SO4. The remaining acid after neutralisation by ammonia consumed 80 mL of 0.5 M NaOH. The percentage of nitrogen in the organic compound is

Once an element's presence is confirmed qualitatively, its percentage BY WEIGHT is found by a dedicated quantitative method. Oxygen alone has no direct method -- its percentage is always found by DIFFERENCE (100% minus every other measured element).

Carbon and hydrogen are estimated together by combustion: a known weight $w$ of sample is burnt completely in excess oxygen, and the masses of water ($x$ g) and $CO_2$ ($y$ g) formed are weighed via an absorption train (conc. $H_2SO_4$-soaked pumice for water, then KOH bulbs for $CO_2$). Since 18 g $H_2O$ contains 2 g H, $\%H = \frac{2}{18}\times\frac{x}{w}\times100$; since 44 g $CO_2$ contains 12 g C, $\%C = \frac{12}{44}\times\frac{y}{w}\times100$.

Sulphur (Carius method): the sample is heated with fuming $HNO_3$ in a sealed tube, oxidising S to $H_2SO_4$, which is precipitated with excess $BaCl_2$ as $BaSO_4$ (mass $x$ from sample mass $w$). Since 233 g $BaSO_4$ contains 32 g S, $\%S = \frac{32}{233}\times\frac{x}{w}\times100$.

Halogens (Carius method): heated with fuming $HNO_3$ + $AgNO_3$; the halogen precipitates directly as $AgX$. $\%Cl = \frac{35.5}{143.5}\times\frac{a}{w}\times100$; $\%Br = \frac{80}{188}\times\frac{b}{w}\times100$; $\%I = \frac{127}{235}\times\frac{c}{w}\times100$ (a/b/c = mass of AgCl/AgBr/AgI).

Phosphorus (Carius method): P oxidises to phosphoric acid, precipitated either as ammonium phosphomolybdate, $(NH_4)_3PO_4{\cdot}12MoO_3$ (1877 g/mol; $\%P = \frac{31}{1877}\times\frac{x}{w}\times100$), or, after igniting a magnesium-ammonium phosphate precipitate, as magnesium pyrophosphate, $Mg_2P_2O_7$ (222 g/mol; $\%P = \frac{62}{222}\times\frac{y}{w}\times100$). …