Nitrogen and halogen estimation

A chemist who has already spotted nitrogen or a halogen in an unknown compound qualitatively still cannot write a molecular formula — for that, the exact mass percentage of each element is needed. Three classical gravimetric and volumetric methods supply those numbers: Dumas' method and Kjeldahl's method, both for nitrogen, and Carius' method, for halogens (and, with small modifications, sulphur and phosphorus). Each converts the element of interest into a single measurable product — a gas volume, a titrated acid, or a weighed precipitate — and each rests on the same idea: a known mass of compound goes in, one clean quantity comes out, and simple stoichiometry links the two. This article walks through the principle, the formula, and a worked calculation for each method, so that a volume, titre, or precipitate mass can be turned into a percentage composition with confidence.

The core insight

Nitrogen is estimated as collected $\text{N}_2$ gas (Dumas) or as titrated ammonia (Kjeldahl), while halogens (and sulphur) are estimated as a weighed silver halide or barium sulphate precipitate after Carius decomposition — each method converting the element into one measurable quantity linked to mass percentage by simple stoichiometry.

The core idea

Dumas' method (nitrogen). The compound is heated strongly with excess copper(II) oxide in a $\text{CO}_2$ atmosphere. Carbon and hydrogen are oxidised to $\text{CO}_2$ and $\text{H}_2\text{O}$ , while nitrogen forms a mixture of free $\text{N}_2$ and nitrogen oxides. Passing this mixture over hot copper gauze reduces the oxides to more $\text{N}_2$ ; passing it over concentrated KOH solution then absorbs the $\text{CO}_2$ and water vapour, leaving pure nitrogen whose volume is read off. Since 22,400 mL of any gas at STP is one mole (28 g of $\text{N}_2$ ), the volume converts directly to a mass fraction:

\%\,\text{N} = \dfrac{28 \times V_{\text{N}_2}(\text{STP, L})}{22.4 \times m}\times 100

where: $m$ is the mass of compound taken (g), and $V_{\text{N}_2}$ is the volume of collected nitrogen corrected to STP (L).

Kjeldahl's method (nitrogen). The compound is boiled with concentrated $\text{H}_2\text{SO}_4$ (often with a $\text{CuSO}_4$ or $\text{K}_2\text{SO}_4$ catalyst), converting its nitrogen to ammonium sulphate. Excess NaOH then liberates ammonia:

(\text{NH}_4)_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{NH}_3\uparrow + 2\text{H}_2\text{O}

The ammonia is absorbed in a known excess of standard acid, and the unreacted acid is found by back-titration with standard alkali; the acid actually neutralised by the ammonia gives the nitrogen mass:

\%\,\text{N} = \dfrac{1.4\,N\,V}{m}

where: $N$ is the normality and $V$ the volume (mL) of standard acid neutralised by the liberated ammonia, and $m$ is the compound's mass (g); the constant $1.4$ comes from $\dfrac{14}{1000}\times100$ . Kjeldahl's method fails whenever nitrogen sits in a ring (pyridine) or as a nitro/azo/diazo group, since these do not convert to ammonium sulphate under the reaction conditions — Dumas' method is used instead.

Carius' method (halogens, sulphur, phosphorus). The compound is heated in a sealed Carius tube with fuming nitric acid and (for halogens) silver nitrate. The fierce oxidation completely decomposes the compound, precipitating the halogen as silver halide ( $\text{AgCl}$ , $\text{AgBr}$ or $\text{AgI}$ ), which is filtered, dried and weighed:

\%\,\text{X} = \dfrac{\text{atomic mass of X}}{\text{molar mass of AgX}}\times\dfrac{\text{mass of AgX}}{m}\times 100

Sulphur is similarly oxidised to sulphate and precipitated (with $\text{BaCl}_2$ ) as barium sulphate:

\%\,\text{S} = \dfrac{32}{233}\times\dfrac{\text{mass of BaSO}_4}{m}\times 100

Phosphorus is oxidised and precipitated analogously (commonly as magnesium ammonium phosphate) and estimated by the same mass-ratio logic.

Figure

Flow diagram of Dumas method showing compound heated with copper oxide producing gases that pass over hot copper gauze then through potassium hydroxide solution leaving pure nitrogen collected in a graduated tube

Derivation ATurning a measured gas volume into percentage nitrogen

The goal of this derivation is to show, from nothing more than the definition of a mole and the molar volume of a gas at STP, exactly why the Dumas percentage-nitrogen formula has the shape $\%\text{N} = \dfrac{28\,V_{\text{N}_2}}{22.4\,m}\times100$ , so that the formula is remembered as a piece of reasoning rather than a fact to memorise.

Step 1. Start from the defining fact of molar volume: at STP, one mole of any gas occupies $22.4$ L (or equivalently $22{,}400$ mL). This holds for $\text{N}_2$ exactly as it does for any other gas, because it is a statement about the number of gas particles, not about their chemical identity.

Step 2. Convert the measured volume $V_{\text{N}_2}$ (at STP, in litres) into moles of nitrogen gas using that molar volume: moles of $\text{N}_2 = \dfrac{V_{\text{N}_2}}{22.4}$ .

Step 3. Convert moles of $\text{N}_2$ into a mass. Since one mole of $\text{N}_2$ has mass $28$ g (two nitrogen atoms of atomic mass $14$ each), the mass of nitrogen collected is mass $= 28\times\dfrac{V_{\text{N}_2}}{22.4}$ grams.

Step 4. Express this mass as a percentage of the original compound's mass $m$ by dividing and multiplying by $100$ : $\%\text{N} = \dfrac{\text{mass of N}}{m}\times100 = \dfrac{28\,V_{\text{N}_2}}{22.4\,m}\times100$ , which is exactly the working formula — every symbol in it now traces back to the mole concept and nothing else. Check: the units work out cleanly — grams over grams cancels to a pure number, which $\times100$ turns into a percentage, confirming the formula is dimensionally consistent.

Worked example

Question: In a Dumas' method estimation, 0.30 g of an organic compound gave 50 mL of nitrogen gas collected at STP. Calculate the percentage of nitrogen, and then find the number of nitrogen atoms per molecule if the compound's molar mass is 134 g/mol.

Step 1. Note the given data: $m = 0.30$ g, $V_{\text{N}_2} = 50$ mL $= 0.050$ L (already at STP).

Step 2. Apply the Dumas formula: $\%\,\text{N} = \dfrac{28\times V_{\text{N}_2}(\text{L})}{22.4\times m}\times 100 = \dfrac{28\times 0.050}{22.4\times 0.30}\times 100 = \dfrac{1.40}{6.72}\times 100 = 20.83\%$ .

Step 3. Extend to atoms per molecule. Mass of nitrogen per mole of compound $= \dfrac{20.83}{100}\times134 = 27.9$ g, and dividing by the atomic mass of nitrogen ( $14$ ) gives $\dfrac{27.9}{14} \approx 2$ nitrogen atoms per molecule.

The compound contains approximately 20.83% nitrogen by mass, consistent with 2 nitrogen atoms per molecule of molar mass 134 g/mol.

Competitive stretch

0.15 g of an organic chlorine compound, treated by Carius' method, gave 0.2451 g of AgCl (Ag = 108, Cl = 35.5, molar mass AgCl = 143.5). Step 1. $\%\,\text{Cl} = \dfrac{35.5}{143.5}\times\dfrac{0.2451}{0.15}\times 100$ . Step 2. $\dfrac{35.5}{143.5}=0.2474$ and $\dfrac{0.2451}{0.15}=1.634$ . Step 3. $\%\,\text{Cl} = 0.2474\times1.634\times100 = 40.42\%$ . The compound contains approximately 40.42% chlorine by mass.

Common traps

Using Kjeldahl's method on a compound with ring-nitrogen (pyridine) or a nitro/azo/diazo group — it must be Dumas' method instead, since these do not form ammonium sulphate.
Forgetting to correct the collected nitrogen volume to STP before applying the Dumas formula, if it was measured at a different temperature or pressure.
Inverting the mass-ratio direction in Carius' method — always (atomic mass of element) divided by (molar mass of the precipitate), never the reverse.
Treating the Kjeldahl "1.4" as a bare constant rather than remembering it is $\dfrac{14}{1000}\times100$ , which only works when $N$ is in normality and $V$ in mL.

Ask the tutor: Want a fresh Dumas, Kjeldahl, or Carius numerical to practise the formula on your own numbers?

Why does Dumas method use both hot copper gauze and KOH solution rather than just collecting the gas directly?

The initial combustion with CuO produces not pure nitrogen but a mixture containing some nitrogen oxides alongside free N2, plus CO2 and water vapour from carbon and hydrogen in the compound. The hot copper gauze chemically reduces those nitrogen oxides back to N2, so all the compound's nitrogen ends up in gas form. The KOH solution then absorbs the CO2 and water vapour, which are acidic or condensable, leaving only pure, unreactive nitrogen gas whose volume can be measured accurately. Skipping either step would either undercount nitrogen (unreduced oxides) or overcount total gas volume (uncounted CO2).

Why does Kjeldahl's method use back-titration instead of titrating the ammonia directly?

Ammonia gas is difficult to titrate directly because it is volatile and easily lost during handling. Instead, it is absorbed into a known, deliberately excess volume of standard acid, where it reacts completely and is trapped as a stable ammonium salt in solution. The amount of acid left unreacted is then found by titrating that solution with a standard alkali. Subtracting the leftover acid from the total acid taken gives exactly how much acid the ammonia neutralised, which is what the formula needs — an indirect but far more reliable measurement than trying to catch and titrate a gas.

Can Carius' method be used for elements other than halogens?

Yes — the same sealed-tube oxidation with fuming nitric acid also works for sulphur and phosphorus, only the precipitating step changes. Sulphur is oxidised to sulphate and precipitated as barium sulphate using barium chloride solution, then weighed and used in a formula analogous to the halogen one. Phosphorus is oxidised and commonly precipitated as magnesium ammonium phosphate or ammonium phosphomolybdate. In every case the underlying idea is identical: violently oxidise the organic compound until the target element is freed as a simple ion, then trap it as an insoluble, weighable compound.

Nitrogen and halogen estimation

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads