Organic Chemistry – Some Basic Principles and Techniques · Lesson 13 of 15

Measuring Carbon, Hydrogen, and Nitrogen

“Derive elemental percentages from combustion products, gas volumes, and acid–base measurements.”

Learning Objectives

• Calculate carbon and hydrogen percentages from carbon dioxide and water masses. • Explain Dumas nitrogen estimation and correct a measured gas volume. • Explain Kjeldahl digestion, ammonia recovery, and back titration. • Choose valid formulas with clearly defined masses, volumes, units, and limitations.

From presence to percentage

A qualitative test tells us that an element is present. Quantitative analysis asks how much of it is present in a known sample mass. The central reasoning is conservation of atoms: convert the element into a measurable product, calculate the mass of the original element represented by that product, then divide by the original sample mass. These percentages support empirical- and molecular-formula determination.

Mass percentage of an elementLaTeX

% element = \frac{mass of that element in the sample}{mass of the sample} \times 100

The numerator is the element’s mass, not the total mass of the compound formed during analysis. Both masses must use the same unit.

Carbon and hydrogen are measured together

A weighed sample is completely oxidised in oxygen with copper(II) oxide. Carbon becomes carbon dioxide and hydrogen becomes water. The gases pass first through a weighed anhydrous calcium-chloride absorber, which retains water, and then through a weighed potassium-hydroxide absorber, which retains carbon dioxide. The mass increases give the product masses. The ordering prevents water being counted along with carbon dioxide.

Combustion analysis pathway— The first mass increase measures water; the second measures carbon dioxide after water removal.

Every 44 g of carbon dioxide contains 12 g of carbon; every 18 g of water contains 2 g of hydrogen atoms. These mass fractions, rather than the full absorber gains, give the element masses. The factors 12/44 and 2/18 come from formula masses and atom conservation, so they can be reconstructed instead of memorised without meaning.

Carbon and hydrogen percentagesLaTeX

% C % H = \frac{12 m _{C O_{2}}}{44 m} \times 100 = \frac{2 m _{H_{2} O}}{18 m} \times 100

m is the original sample mass; the two product masses are the separate measured absorber gains. Use the same mass unit throughout.

Using measured combustion products

Problem
A 0.246 g sample gives 0.198 g carbon dioxide and 0.1014 g water. Find its carbon and hydrogen percentages.

1.Carbon mass is $(12/44) (0.198) = 0.0540 g$ . The carbon percentage is $(0.0540/0.246) 100 = 21.95%$ .
2.Hydrogen mass is $(2/18) (0.1014) = 0.01127 g$ . The hydrogen percentage is $(0.01127/0.246) 100 \approx 4.58%$ .
3.The remaining percentage is not automatically oxygen unless the possible remaining elements are known. The analysis so far measured only carbon and hydrogen.

Working backwards from a percentage

Problem
A 0.20 g compound contains 69% carbon and 4.8% hydrogen. Predict the masses of carbon dioxide and water from complete combustion.

1.Carbon mass is $0.20 (0.69) = 0.138 g$ . Since 12 g carbon gives 44 g CO₂, $m_{C O_{2}} = 0.138 (44/12) = 0.506 g$ .
2.Hydrogen mass is $0.20 (0.048) = 0.0096 g$ . Since 2 g hydrogen gives 18 g water, $m_{H_{2} O} = 0.0096 (18/2) = 0.0864 g$ .
3.The combustion products can weigh more than the original sample because they incorporate oxygen supplied during the experiment.

Dumas analysis measures nitrogen gas

In Dumas analysis the nitrogen-containing compound is oxidised with copper oxide in a carbon-dioxide atmosphere. Carbon and hydrogen form carbon dioxide and water, while nitrogen is recovered as nitrogen gas. Any nitrogen oxides are reduced over hot copper. Potassium hydroxide absorbs carbon dioxide so the remaining measured gas represents nitrogen rather than a mixture dominated by carbon dioxide.

Dumas analysis pathway— Carbon dioxide displaces air and is later absorbed. Hot copper converts nitrogen oxides to nitrogen.

A collected gas volume depends on temperature and pressure. If it is measured over a liquid, part of the total pressure is due to vapour from that liquid. Subtract the given aqueous tension to obtain the nitrogen pressure. Then convert the measured volume to the stated reference conditions using the gas-law relationship. Here the calculation follows the chapter’s 273 K and 760 mm Hg convention.

Correcting a nitrogen volumeLaTeX

P_{N_{2}} = P_{total} - P_{vapour}, V_{0} = V \frac{P _{N_{2}}}{760} \frac{273}{T}

V is the measured volume at T kelvin, pressures are in mm Hg, and V₀ is the volume at 273 K and 760 mm Hg. V and V₀ use the same volume unit.

Nitrogen percentage from a corrected volumeLaTeX

% N = \frac{28 V _{0}}{22400 m} \times 100

With V₀ in mL and m in grams, the chapter uses 22400 mL per mole at the stated reference conditions. One mole of N₂ contains 28 g of nitrogen.

Correcting for aqueous tension

Problem
A 0.300 g sample gives 50.0 mL nitrogen at 300 K and total pressure 715 mm Hg. Aqueous tension is 15 mm Hg. Find the nitrogen percentage.

1.The nitrogen pressure is $715 - 15 = 700 mm Hg$ . Do not use the total pressure as the nitrogen pressure.
2.The corrected volume is $V_{0} = 50.0 (700/760) (273/300) = 41.91 mL$ .
3.The nitrogen mass is $28 (41.91) /22400 = 0.05239 g$ , giving $% N = (0.05239/0.300) 100 = 17.46%$ .

Kjeldahl analysis measures ammonia indirectly

Kjeldahl analysis uses a different nitrogen conversion. Digestion with concentrated sulphuric acid converts suitable organic nitrogen into ammonium sulphate. The digested mixture is then treated with excess alkali, releasing ammonia. The ammonia is distilled into a known amount of standard acid. The amount of acid consumed reveals the ammonia and therefore the nitrogen content.

Ammonia release and captureLaTeX

(N H_{4})_{2} S O_{4} + 2NaOH 2N H_{3} + H_{2} S O_{4} \to N a_{2} S O_{4} + 2N H_{3} + 2 H_{2} O \to (N H_{4})_{2} S O_{4}

Each mole of ammonia contains one mole of nitrogen atoms. One mole of sulphuric acid neutralises two moles of ammonia under complete neutralisation.

Kjeldahl digestion and measurement— Track nitrogen into ammonium salt, then ammonia, then the amount of acid neutralised.

In a back titration, first find the total acid equivalents initially present, then subtract the equivalents of acid left after absorbing ammonia. The leftover acid is measured by titration with standard alkali. The difference equals the moles of ammonia for a monobasic ammonia molecule. Using moles explicitly prevents confusion between acid molarity and the number of acidic protons.

General Kjeldahl back-titration relationshipLaTeX

n (N H_{3}) = \frac{2 M _{a} V _{a} - M _{b} V _{b}}{1000}, % N = \frac{14 n ( N H _{3} )}{m} \times 100

Mₐ and Vₐ are the initial H₂SO₄ molarity and volume in mL. Mᵦ and Vᵦ are NaOH molarity and back-titration volume in mL. m is sample mass in grams. The factor 2 belongs to H₂SO₄; NaOH contributes one equivalent per mole.

Using acid consumed directly

Problem
Ammonia from a 0.500 g sample neutralises 10.0 mL of 1.00 M sulphuric acid. Find the nitrogen percentage.

1.Acid consumed is $1.00 (10.0/1000) = 0.0100 mol$ . Since one mole acid reacts with two moles ammonia, $n (N H_{3}) = 0.0200 mol$ .
2.The nitrogen mass is $0.0200 (14) = 0.280 g$ . The nitrogen percentage is $(0.280/0.500) 100 = 56.0%$ .
3.The given acid volume is already the amount consumed. Do not subtract another acid volume or confuse 0.0100 mol acid with 0.0100 mol nitrogen.

Using a back titration

Problem
Ammonia from a 0.500 g sample is absorbed in 50.0 mL of 0.500 M sulphuric acid. The unused acid requires 60.0 mL of 0.500 M NaOH. Find the nitrogen percentage.

1.Initial acid equivalents are $2 (0.500) (0.0500) = 0.0500 mol$ . Unused acid equivalents equal the NaOH moles: $0.500 (0.0600) = 0.0300 mol$ .
2.The ammonia used the difference, $0.0200 mol$ of equivalents, so it contained $0.0200 mol$ nitrogen.
3.The nitrogen mass is $0.280 g$ and the percentage is $56.0%$ . A positive difference smaller than the initial acid capacity is a useful consistency check.

Kjeldahl’s method in this form is not suitable for nitrogen in nitro groups, azo groups, or certain ring environments such as pyridine, because that nitrogen is not converted quantitatively to ammonium sulphate under the stated conditions. Dumas and Kjeldahl therefore differ both in the measured product and in applicability. A formula cannot rescue a method whose chemical conversion is incomplete.

Common mistake

Use kelvin in gas-volume correction, subtract vapour pressure, and distinguish nitrogen atoms from N₂ molecules. In Kjeldahl analysis, sulphuric acid supplies two acid equivalents per mole. Do not use equal-molarity shortcuts when the acid and alkali molarities differ.

Quick check

Which fraction of carbon-dioxide mass is carbon mass?

Quick check

Why is water removed before the carbon-dioxide absorber in C/H analysis?

Quick check

How is nitrogen pressure obtained for a wet gas when aqueous tension is supplied?

Quick check

How many moles of ammonia react with one mole of sulphuric acid in complete neutralisation?

Quick check

Which nitrogen environment is unsuitable for the stated Kjeldahl procedure?

Practice Problems

Derive the factors 12/44 and 2/18 used in combustion analysis from formula masses.
A 0.300 g sample produces 0.440 g carbon dioxide and 0.180 g water. Calculate its carbon and hydrogen percentages.
A gas sample is measured at 298 K and total pressure 750 mm Hg with aqueous tension 24 mm Hg. Explain how to correct its nitrogen volume to 273 K and 760 mm Hg.
Repeat the Dumas calculation in the worked example and identify the error produced by failing to subtract aqueous tension.
Ammonia from 0.400 g sample consumes 8.00 mL of 0.500 M sulphuric acid. Calculate nitrogen percentage.
Derive the general Kjeldahl back-titration expression for unequal acid and alkali molarities.
Compare Dumas and Kjeldahl by conversion chemistry, measured quantity, calculation, and applicability.

Key Takeaways

• Product masses or gas amounts must first be converted into element mass. • Carbon and hydrogen percentages use the mass fractions in CO₂ and H₂O. • Dumas analysis requires corrected nitrogen pressure and volume. • Kjeldahl analysis uses nitrogen-to-ammonia conversion and acid-equivalent accounting, with stated limitations.

Previous · Lesson 12

Identifying Elements in Organic Compounds

Next · Lesson 14

Measuring Halogens, Sulphur, Phosphorus, and Oxygen

From presence to percentage

Carbon and hydrogen are measured together

Dumas analysis measures nitrogen gas

Kjeldahl analysis measures ammonia indirectly

Quiz

Practice Problems

Key Takeaways