Carbon and hydrogen estimation

A chemist hands you an unknown organic liquid and asks: what is it made of? Qualitative tests can confirm carbon and hydrogen are present, but they cannot say how much — and without exact proportions there is no way to write a molecular formula. Carbon and hydrogen estimation answers that "how much" question by combustion: burn a weighed sample completely, trap every atom of carbon as carbon dioxide and every atom of hydrogen as water, weigh those products, and work backward to percentage composition. This article walks through the principle of the combustion method, the arithmetic that turns collected gas masses into percentage carbon and percentage hydrogen, and how that data feeds into empirical and molecular formula calculations — the same reasoning examiners test again and again with only the numbers changed.

The core insight

Burning a known mass of an organic compound in oxygen converts all its carbon to $\text{CO}_2$ and all its hydrogen to $\text{H}_2\text{O}$ , and weighing those absorbed products lets you calculate percentage carbon and hydrogen using the fixed mass fractions $\tfrac{12}{44}$ and $\tfrac{2}{18}$ .

The core idea

The setup. A weighed mass of the organic compound is burned completely in a stream of dry oxygen, in the presence of catalysts such as cupric oxide ( $\text{CuO}$ ), and sometimes lead chromate or platinum, which push the combustion to completion. Every carbon atom ends up in $\text{CO}_2$ and every hydrogen atom ends up in $\text{H}_2\text{O}$ :

\text{C}_x\text{H}_y(\text{compound}) + \text{O}_2 \xrightarrow{\text{CuO, heat}} x\,\text{CO}_2 + \dfrac{y}{2}\,\text{H}_2\text{O}

where: $x$ and $y$ are the number of carbon and hydrogen atoms in one molecule of the compound.

Trapping the products. The hot gases pass through two pre-weighed absorption tubes in a fixed order: first anhydrous calcium chloride (or magnesium perchlorate), which absorbs only water, then potassium hydroxide (or soda-lime), which absorbs the carbon dioxide. Each tube's gain in mass is exactly the mass of water or carbon dioxide produced from the known mass of compound burnt. The order matters — KOH absorbs both $\text{CO}_2$ and $\text{H}_2\text{O}$ , so if it came first it would corrupt the water reading.

From gas mass to element mass. Only a fixed fraction of each product's mass is the element you want: carbon is 12 out of 44 g of $\text{CO}_2$ , hydrogen is 2 out of 18 g of $\text{H}_2\text{O}$ . If $m$ is the mass of compound taken, $m_{\text{CO}_2}$ the mass of $\text{CO}_2$ collected and $m_{\text{H}_2\text{O}}$ the mass of $\text{H}_2\text{O}$ collected:

\%\,\text{C} = \dfrac{12}{44}\times\dfrac{m_{\text{CO}_2}}{m}\times 100, \qquad \%\,\text{H} = \dfrac{2}{18}\times\dfrac{m_{\text{H}_2\text{O}}}{m}\times 100

where: the fractions $\tfrac{12}{44}$ and $\tfrac{2}{18}$ come from the molar masses of $\text{CO}_2$ (44 g/mol, of which 12 g is carbon) and $\text{H}_2\text{O}$ (18 g/mol, of which 2 g is hydrogen).

Figure

Combustion tube burning a sample in oxygen feeding into a calcium chloride tube then a potassium hydroxide tube

Empirical and molecular formula from the data

Once %C and %H are known (and %O by difference, if oxygen is the only other element), each percentage is converted to a mole ratio by dividing by the atomic mass, then the whole ratio is divided by its smallest member to give the simplest whole-number ratio — the empirical formula. Comparing the empirical formula mass to the compound's separately measured molar mass $M$ gives the multiplier $n$ :

n = \dfrac{\text{molecular formula mass}}{\text{empirical formula mass}}, \qquad \text{molecular formula} = n \times (\text{empirical formula})

where: $n$ is always a positive whole number; the molecular formula mass equals $M$ , the experimentally determined molar mass of the compound.

Derivation ATurning collected CO2 and H2O masses into percentage carbon and hydrogen

The goal is to show, from first principles of molar mass and simple proportion, exactly why the fractions $\tfrac{12}{44}$ and $\tfrac{2}{18}$ are the correct multipliers to convert a measured mass of collected $\text{CO}_2$ or $\text{H}_2\text{O}$ into the mass of carbon or hydrogen it represents, and then into a percentage of the original weighed sample — the calculation every combustion-analysis numerical ultimately reduces to.

Step 1. Fix the molar masses involved. One mole of $\text{CO}_2$ has molar mass $12 + 2(16) = 44$ g, of which the carbon atom contributes exactly 12 g. One mole of $\text{H}_2\text{O}$ has molar mass $2(1) + 16 = 18$ g, of which the two hydrogen atoms contribute exactly 2 g. These fixed fractions, $\tfrac{12}{44}$ and $\tfrac{2}{18}$ , hold for any sample, because they come purely from atomic masses, not from the specific compound burnt.

Step 2. Convert collected product mass to element mass. If $m_{\text{CO}_2}$ grams of carbon dioxide were collected, the mass of carbon in it is found by simple proportion: 44 g of $\text{CO}_2$ contains 12 g of C, so $m_{\text{CO}_2}$ g of $\text{CO}_2$ contains $\dfrac{12}{44}\times m_{\text{CO}_2}$ g of carbon. The identical reasoning for water gives mass of hydrogen $= \dfrac{2}{18}\times m_{\text{H}_2\text{O}}$ .

Step 3. Express as a percentage of the original sample. If the weighed sample had mass $m$ grams, the fraction of that sample which is carbon is $\dfrac{12}{44}\times\dfrac{m_{\text{CO}_2}}{m}$ , and multiplying by 100 converts this fraction into a percentage: $\%\,\text{C} = \dfrac{12}{44}\times\dfrac{m_{\text{CO}_2}}{m}\times 100$ . The same three-step logic gives $\%\,\text{H} = \dfrac{2}{18}\times\dfrac{m_{\text{H}_2\text{O}}}{m}\times 100$ .

Step 4. Note what the method assumes. It assumes combustion is complete (no soot, no unburnt compound) and that the absorption tubes capture all the $\text{CO}_2$ and $\text{H}_2\text{O}$ produced with nothing lost — both guaranteed by the catalyst and the tube arrangement described above, which is why apparatus design matters even though only the principle is examined.

Worked example

Question: On complete combustion, 0.246 g of an organic compound gave 0.198 g of $\text{CO}_2$ and 0.1014 g of $\text{H}_2\text{O}$ . Calculate the percentage of carbon and hydrogen, and find the empirical formula if the compound contains only carbon, hydrogen and oxygen.

Step 1. Calculate the mass of carbon from the collected $\text{CO}_2$ : $\text{mass of C} = \dfrac{12}{44}\times 0.198 = 0.0540\ \text{g}$ , so $\%\,\text{C} = \dfrac{0.0540}{0.246}\times 100 = 21.95\%$ .

Step 2. Calculate the mass of hydrogen from the collected $\text{H}_2\text{O}$ : $\text{mass of H} = \dfrac{2}{18}\times 0.1014 = 0.01127\ \text{g}$ , so $\%\,\text{H} = \dfrac{0.01127}{0.246}\times 100 = 4.58\%$ .

Step 3. Find %O by difference, since only C, H and O are present: $\%\,\text{O} = 100 - 21.95 - 4.58 = 73.47\%$ .

Step 4. Convert to mole ratios (per 100 g of compound): $\text{C}: \dfrac{21.95}{12} = 1.83$ , $\text{H}: \dfrac{4.58}{1} = 4.58$ , $\text{O}: \dfrac{73.47}{16} = 4.59$ .

Step 5. Divide by the smallest value, 1.83: $\text{C}: 1.00$ , $\text{H}: 2.50$ , $\text{O}: 2.51 \approx 2.5$ . Since 2.5 is not a whole number, double every ratio to clear the half: $\text{C}: 2$ , $\text{H}: 5$ , $\text{O}: 5$ .

The empirical formula is $\text{C}_2\text{H}_5\text{O}_5$ (empirical formula mass $= 2(12) + 5(1) + 5(16) = 109$ g mol⁻¹; the molecular formula would follow from $n = M/109$ if a molar mass were given).

Competitive stretch

If this same compound's molar mass were separately found to be 109 g mol⁻¹, then $n = \dfrac{109}{109} = 1$ , so the molecular formula equals the empirical formula, $\text{C}_2\text{H}_5\text{O}_5$ . Changing only $M$ (say, to 218) would double every subscript to $\text{C}_4\text{H}_{10}\text{O}_{10}$ — the empirical formula never changes, only the multiplier $n$ does.

Common traps

Using the atomic mass of carbon (12) or hydrogen (2) directly on the collected gas mass, instead of first multiplying by the correct fraction, $\tfrac{12}{44}$ for C from $\text{CO}_2$ and $\tfrac{2}{18}$ for H from $\text{H}_2\text{O}$ .
Forgetting that oxygen (or any element this method does not directly estimate) must be found by difference from 100%, never measured directly by combustion.
Placing the potassium hydroxide (or soda-lime) tube before the calcium chloride tube — KOH absorbs both $\text{CO}_2$ and $\text{H}_2\text{O}$ , so water must be trapped first or the readings are corrupted.
Rounding mole ratios too early. Always divide by the smallest mole value first, and if the result is a recognisable fraction like 1.5 or 2.5, multiply the whole ratio through rather than rounding it blindly to the nearest integer.

Ask the tutor: Want to try another combustion numerical with different starting masses to practice the full percentage-to-molecular-formula chain?

Why do we need two separate absorption tubes instead of one that catches everything at once?

Because the calculation needs the mass of water and the mass of carbon dioxide as two separate numbers, not a combined total. If both gases were trapped in one tube, you would know the total mass gained but have no way to split it back into how much was $\text{H}_2\text{O}$ and how much was $\text{CO}_2$ . Using anhydrous calcium chloride first, which absorbs only water, then potassium hydroxide, which then absorbs only the carbon dioxide that remains, cleanly separates the two measurements so each fraction, $\tfrac{2}{18}$ and $\tfrac{12}{44}$ , can be applied correctly.

Why can this method not tell us the percentage of oxygen directly?

The combustion method only estimates the elements that turn into a product it specifically traps — carbon into $\text{CO}_2$ and hydrogen into $\text{H}_2\text{O}$ . Any oxygen originally in the compound simply combines with the combustion oxygen supply and ends up mixed into those same product gases, so there is no separate measurable product that isolates it. If the compound is known to contain only carbon, hydrogen and oxygen, the oxygen percentage is instead found by subtracting the measured %C and %H from 100%, treating it as everything left over.

My mole ratio came out as 1, 1.5, and 1 — do I round the 1.5 up to 2?

No, rounding a clean fraction like 1.5 to the nearest whole number gives the wrong empirical formula. Instead, multiply every ratio in the set by the same small whole number needed to clear the fraction — here, multiplying by 2 turns 1, 1.5, 1 into 2, 3, 2, which is the correct empirical formula ratio. Only round when the mole values are close to a whole number already, like 1.02 or 2.98, where the small deviation is just experimental rounding error rather than a genuine fraction.