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Chemistry · Ch 12 — Organic Chemistry – Some Basic Principles and Techniques

Carbon and Hydrogen

12.10.1

Carbon and Hydrogen

Estimation of Carbon and Hydrogen

Both carbon and hydrogen are determined in a single experiment. A known mass of the organic compound is burnt in excess oxygen in the presence of copper(II) oxide. The copper(II) oxide ensures complete combustion. Carbon and hydrogen in the compound are oxidised to carbon dioxide and water respectively.

The combustion reaction for a hydrocarbon CxHyC_xH_y is:

CxHy+(x+y4)O2→Δx CO2+y2 H2OC_xH_y + \left(x + \frac{y}{4}\right) O_2 \xrightarrow{\Delta} x\,CO_2 + \frac{y}{2}\,H_2O

The mixture of gases produced is passed through a weighed U-tube containing anhydrous calcium chloride. Water vapour is absorbed by the calcium chloride. The carbon dioxide then passes into another U-tube containing a concentrated solution of potassium hydroxide, which absorbs it. These tubes are connected in series, as shown in the figure.

The increase in mass of the calcium chloride tube gives the mass of water produced. The increase in mass of the potassium hydroxide tube gives the mass of carbon dioxide produced. From these masses, the percentages of carbon and hydrogen in the original compound are calculated.

Figure 8.14Estimation of carbon and hydrogen: the combustion train that burns a weighed sample in oxygen and passes the products through a weighed CaCl2 tube (absorbs H2O) then a KOH tube (absorbs CO2).
Fig. 8.14 — Estimation of carbon and hydrogen: the combustion train that burns a weighed sample in oxygen and passes the products through a weighed CaCl2 tube (absorbs H2O) then a KOH tube (absorbs CO2).

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

Figure 8.14 is a schematic of a combustion train — the classic apparatus used to determine the percentages of carbon and hydrogen in an organic compound. The figure does not show a graph or plot; it is a flow diagram of a sequence of connected tubes and furnaces. The key idea is that burning a known mass of the compound in pure oxygen converts all its carbon to CO₂ and all its hydrogen to H₂O. By trapping each product separately and measuring the increase in mass of the absorbents, you can calculate the original masses of carbon and hydrogen.

The diagram shows a horizontal combustion tube made of hard glass, placed inside a furnace. A small porcelain boat containing the weighed organic sample sits inside this tube. A steady stream of pure, dry oxygen enters from one end. The tube is heated strongly, and the compound burns completely. The hot exit gases — a mixture of CO₂, H₂O vapour, and excess O₂ — then pass through a U-tube packed with anhydrous calcium chloride (CaCl₂). This absorbs all the water vapour. The remaining gases then bubble through a second U-tube or set of bulbs containing a concentrated solution of potassium hydroxide (KOH), which absorbs carbon dioxide. The oxygen that is not absorbed simply escapes.

Both absorption units are weighed before and after the experiment. The increase in mass of the CaCl₂ tube gives the mass of water produced; the increase in mass of the KOH tube gives the mass of carbon dioxide produced.

Watch out

A common mistake is to think the KOH tube absorbs only CO₂. In fact, KOH solution also absorbs water vapour from the gas stream. That is why the CaCl₂ tube must come first — to remove water before the gas reaches the KOH. If the order were reversed, the KOH gain would include both CO₂ and H₂O, ruining the calculation.

From these two measured masses, the textbook derives the percentage composition. Let the mass of the organic compound taken be mm grams. Let the increase in mass of the CaCl₂ tube be mH2Om_{\text{H}_2\text{O}} grams, and the increase in mass of the KOH tube be mCO2m_{\text{CO}_2} grams.

The mass of hydrogen in the water is found from the ratio of atomic masses. In one mole of water (18 g), there are 2 g of hydrogen. So:

Mass of hydrogen=218×mH2O=19×mH2O\text{Mass of hydrogen} = \frac{2}{18} \times m_{\text{H}_2\text{O}} = \frac{1}{9} \times m_{\text{H}_2\text{O}}

Similarly, in one mole of CO₂ (44 g), the mass of carbon is 12 g:

Mass of carbon=1244×mCO2=311×mCO2\text{Mass of carbon} = \frac{12}{44} \times m_{\text{CO}_2} = \frac{3}{11} \times m_{\text{CO}_2}

The percentage of each element in the original compound is then:

%H=mass of Hm×100=19×mH2Om×100\% \text{H} = \frac{\text{mass of H}}{m} \times 100 = \frac{1}{9} \times \frac{m_{\text{H}_2\text{O}}}{m} \times 100

%C=mass of Cm×100=311×mCO2m×100\% \text{C} = \frac{\text{mass of C}}{m} \times 100 = \frac{3}{11} \times \frac{m_{\text{CO}_2}}{m} \times 100 …

Watch out

A common mistake is to forget that the masses measured are the increases in mass of the absorption tubes, not the masses of the tubes themselves. The increase in mass of the CaCl2_2 tube is the mass of water absorbed; the increase in mass of the KOH tube is the mass of CO2_2 absorbed.

Calculation of Percentage of Carbon

Let the mass of the organic compound taken be mm grams. Let the mass of carbon dioxide produced be m2m_2 grams.

From the molecular formula of carbon dioxide, CO2CO_2, the molar mass is 4444 g mol−1^{-1}. The mass of carbon in one mole of CO2CO_2 is 1212 g.

Therefore, the mass of carbon that produced m2m_2 grams of CO2CO_2 is:

Mass of carbon=1244×m2\text{Mass of carbon} = \frac{12}{44} \times m_2

The percentage of carbon in the compound is then:

Percentage of carbon=Mass of carbonMass of compound×100\text{Percentage of carbon} = \frac{\text{Mass of carbon}}{\text{Mass of compound}} \times 100

Substituting the expression for the mass of carbon:

% C=12×m2×10044×m\boxed{\%\,C = \frac{12 \times m_2 \times 100}{44 \times m}}

Calculation of Percentage of Hydrogen

Let the mass of water produced be m1m_1 grams.

From the molecular formula of water, H2OH_2O, the molar mass is 1818 g mol−1^{-1}. The mass of hydrogen in one mole of water is 22 g.

Therefore, the mass of hydrogen that produced m1m_1 grams of water is:

Mass of hydrogen=218×m1\text{Mass of hydrogen} = \frac{2}{18} \times m_1

The percentage of hydrogen in the compound is then: …

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