Quantitative Analysis

You already know how to test whether carbon, nitrogen, sulphur or a halogen is present in an organic compound. Quantitative analysis asks the sharper question: how much of each — down to a percentage by mass. That single number, once you have it for every element, is the raw material for the compound's empirical formula, and, given the molar mass, its molecular formula too. Each element gets its own classical method: carbon and hydrogen by Liebig's combustion, nitrogen by either the Dumas or Kjeldahl route, and halogens, sulphur and phosphorus by variants of Carius's sealed-tube oxidation. Every method follows the same three-beat rhythm — convert the element quantitatively into one weighable (or measurable) product, weigh or measure it, then use a fixed mass ratio to work back to the percentage of the original element. Learn that rhythm once and all five methods become the same calculation wearing different reagents.

The core insight

Every quantitative method converts one element into a single measurable product (a gas volume, a titre, or a weighed precipitate) and uses a fixed mass ratio to convert that measurement back into percentage composition, which then feeds the mole-ratio calculation for empirical and molecular formula.

The core idea

Carbon and hydrogen — Liebig's method. Burn a known mass of the compound completely in oxygen over hot $\text{CuO}$ . Every carbon atom ends up as $\text{CO}_2$ , every hydrogen atom as $\text{H}_2\text{O}$ ; the two gases are absorbed separately (in KOH and anhydrous $\text{CaCl}_2$ ) and weighed.

\%\,C = \dfrac{12}{44}\times\dfrac{\text{mass of }CO_2}{\text{mass of compound}}\times100 \qquad \%\,H = \dfrac{2}{18}\times\dfrac{\text{mass of }H_2O}{\text{mass of compound}}\times100

where: $\tfrac{12}{44}$ is the fraction of $\text{CO}_2$ 's molar mass that is carbon, and $\tfrac{2}{18}$ is the fraction of $\text{H}_2\text{O}$ 's molar mass that is hydrogen — every formula in this topic is built from the same idea, "what fraction of the weighed product's mass is the element I actually want."

Nitrogen — two independent routes. In the Dumas method, the compound is heated with $\text{CuO}$ in a $\text{CO}_2$ atmosphere; nitrogen is liberated as free $\text{N}_2$ gas (any nitrogen oxides are reduced back to $\text{N}_2$ over hot copper gauze) and collected over concentrated KOH, which soaks up the accompanying $\text{CO}_2$ and leaves pure $\text{N}_2$ to measure by volume.

\%\,N = \dfrac{28}{22400}\times\dfrac{V_{N_2}\ (\text{mL at STP})}{\text{mass of compound (g)}}\times100

The Kjeldahl method instead digests the compound in concentrated $\text{H}_2\text{SO}_4$ to convert its nitrogen into ammonium sulphate; boiling with excess NaOH liberates ammonia, which is trapped in a known excess of standard acid, and back-titration of the unreacted acid gives the ammonia — and hence the nitrogen.

\%\,N = \dfrac{1.4\times N_{acid}\times V_{acid\ used}}{\text{mass of compound (g)}}

Kjeldahl fails whenever nitrogen cannot be reduced to ammonium sulphate under these conditions — nitro groups, azo groups and ring nitrogen (as in pyridine) all escape it.

Halogens, sulphur and phosphorus — the Carius family. All three heat the compound with fuming nitric acid in a sealed hard-glass Carius tube, which oxidises the element to a form that precipitates cleanly. A halogen becomes silver halide (with $\text{AgNO}_3$ present); sulphur becomes sulphate, precipitated as $\text{BaSO}_4$ with $\text{BaCl}_2$ ; phosphorus becomes phosphoric acid, precipitated as magnesium ammonium phosphate or ammonium phosphomolybdate. Each precipitate is filtered, dried and weighed, then converted to percentage element by the same mass-ratio logic as carbon and hydrogen.

\%\,X = \dfrac{\text{at. mass of }X}{\text{mol. mass of }AgX}\times\dfrac{\text{mass of }AgX}{\text{mass of compound}}\times100 \qquad \%\,S = \dfrac{32}{233}\times\dfrac{\text{mass of }BaSO_4}{\text{mass of compound}}\times100

Figure

Five weighing pans feeding into one shared mass ratio arrow labelled percentage of element

From percentage composition to molecular formula

Once every element's mass percentage is known, divide each by its atomic mass to get relative moles, divide through by the smallest value to get a simplest whole-number ratio — the empirical formula — and then use the compound's known molar mass to find

n = \dfrac{\text{molecular formula mass}}{\text{empirical formula mass}}

where: $n$ is a whole number by which every subscript of the empirical formula is multiplied to give the molecular formula. This final step is exactly the mole-concept arithmetic of Unit I, now applied to combustion and Carius data instead of a simple mass or volume of gas.

Derivation ATurning combustion data into percentage carbon and hydrogen

The goal is to show, step by step and without skipping the unit bookkeeping, exactly how a measured mass of $\text{CO}_2$ and $\text{H}_2\text{O}$ from Liebig combustion becomes the two percentages %C and %H that every empirical-formula question ultimately needs, using nothing beyond the molar masses of $\text{CO}_2$ , $\text{H}_2\text{O}$ , carbon and hydrogen.

Step 1. Recognise what the weighed products actually contain. Every carbon atom in the original compound is now locked inside a $\text{CO}_2$ molecule (molar mass 44 g/mol, of which 12 g/mol is carbon), and every hydrogen atom is locked inside an $\text{H}_2\text{O}$ molecule (molar mass 18 g/mol, of which 2 g/mol is hydrogen — two hydrogen atoms per water molecule).

Step 2. Extract the mass of carbon actually present in the weighed $\text{CO}_2$ . Since carbon is $\tfrac{12}{44}$ of $\text{CO}_2$ 's mass by the ratio of molar masses, mass of $C = \tfrac{12}{44}\times(\text{mass of }CO_2)$ . The same logic gives mass of $H = \tfrac{2}{18}\times(\text{mass of }H_2O)$ .

Step 3. Convert each element's mass into a percentage of the original sample by dividing by the mass of compound burnt and multiplying by 100. This is the ordinary "part over whole, times 100" definition of percentage, applied twice — once for carbon, once for hydrogen — giving exactly the two formulas used throughout this topic.

Step 4. Note what is not measured directly. Any oxygen, nitrogen or halogen in the compound is not captured by this combustion at all; if the sum of the directly-measured percentages falls short of 100%, the shortfall is commonly assigned to oxygen (found "by difference") once other tests rule out nitrogen or halogens being present.

Worked example

On combustion, 0.246 g of an organic compound gave 0.198 g of $\text{CO}_2$ and 0.1014 g of $\text{H}_2\text{O}$ . A separate Carius estimation on a fresh 0.15 g sample of the same class of compound (this portion containing bromine) gave 0.1414 g of $\text{AgBr}$ (molar mass 188 g/mol; atomic mass of Br = 80). Find the percentage of carbon, hydrogen and bromine.

Step 1. Apply the carbon formula to the combustion data. $\%\,C = \dfrac{12}{44}\times\dfrac{0.198}{0.246}\times100 = 0.2727\times0.8049\times100 = 21.95\%$ .

Step 2. Apply the hydrogen formula to the same data. $\%\,H = \dfrac{2}{18}\times\dfrac{0.1014}{0.246}\times100 = 0.1111\times0.4122\times100 = 4.58\%$ .

Step 3. Apply the Carius (halogen) formula to the bromine data. $\%\,Br = \dfrac{80}{188}\times\dfrac{0.1414}{0.15}\times100$ . Computing the two fractions first, $\tfrac{80}{188} = 0.4255$ and $\tfrac{0.1414}{0.15} = 0.9427$ , so $\%\,Br = 0.4255\times0.9427\times100 = 40.11\%$ .

Step 4. State and sanity-check the results. Carbon = 21.95%, hydrogen = 4.58%, bromine = 40.1%. These three numbers, together with a known molar mass, are exactly the inputs the empirical-and-molecular-formula procedure needs: convert each percentage to moles by dividing by its atomic mass (12, 1, 80), find the simplest whole-number ratio, then scale by $n = \tfrac{\text{molecular mass}}{\text{empirical mass}}$ to reach the molecular formula. Any percentage not accounted for by C, H and Br (here $100 - 21.95 - 4.58 - 40.11 = 33.36\%$ ) would point to oxygen or another element requiring its own separate test.

Common traps

Flipping the mass-ratio fraction — it is $\tfrac{12}{44}$ for carbon and $\tfrac{2}{18}$ for hydrogen (element mass over product molar mass), never the reverse.
Forgetting Kjeldahl's blind spot: it cannot estimate nitrogen locked in a nitro group, an azo group, or a ring (pyridine-type nitrogen), because that nitrogen never converts to ammonium sulphate.
Mixing up Dumas (nitrogen measured as a gas volume, using the $\tfrac{28}{22400}$ relation) with Kjeldahl (nitrogen measured by acid-base titration) — the two use structurally different formulas and cannot be interchanged.
Reporting the weighed precipitate's own percentage (of AgX or BaSO4) as the percentage of the element — the precipitate mass must always be converted back to element mass first, via the atomic-mass-to-molar-mass ratio.

Ask the tutor: want to work through a full combustion-plus-Carius problem all the way to the molecular formula, or check your own Dumas/Kjeldahl percentage-nitrogen calculation?

Why does the Dumas method use CO2 as the surrounding atmosphere instead of ordinary air

Air is roughly 78% nitrogen, and any of that atmospheric nitrogen leaking into the collected gas would be measured right alongside the nitrogen coming from the sample, inflating the result. Using pure $\text{CO}_2$ as the sweep gas keeps atmospheric nitrogen out of the system entirely. The $\text{CO}_2$ itself is easy to remove afterward because it is absorbed by the concentrated KOH the collected gas is passed through, leaving only the sample's own $\text{N}_2$ to measure by volume.

If I already know the molecular formula, why bother finding the empirical formula first

Because the two experimental routes — combustion for C and H, Carius for halogens/S/P, Dumas or Kjeldahl for N — only ever hand you mass percentages, and percentages alone can only fix a ratio of atoms, not their absolute number. The empirical formula is that simplest ratio. You need one further piece of information, the molar mass (usually from a separate physical method such as vapour density), to work out the multiplier $n$ and scale the empirical formula up to the true molecular formula.

Why is oxygen never estimated directly by one of these methods

Every method here works by converting the target element into a specific weighable product — carbon into $\text{CO}_2$ , sulphur into $\text{BaSO}_4$ , and so on. Oxygen has no equally clean, unique product to trap and weigh, because oxygen is already present in several of the reagents and in the combustion products themselves, making a direct estimate unreliable. Instead, if all other elements present have been estimated and their percentages fall short of 100%, the missing percentage is simply assigned to oxygen by difference.

Quantitative Analysis

The core idea

From percentage composition to molecular formula

Worked example

Common traps

Before this

Where this leads