What is ultimate analysis of coal and how are carbon, hydrogen, nitrogen, sulphur and oxygen found?

In one line

Ultimate analysis determines the elemental composition of coal (C, H, N, S, ash and O by difference) and is used for calorific value and combustion calculations.

Explained simply

Proximate analysis is like checking a phone's battery and storage. Ultimate analysis is opening it and listing every part. It is more exact and tells a scientist how much air the coal needs and how much heat it can give.

How it works

1. Carbon and hydrogen: burn a known mass of coal in a stream of O₂ in a combustion tube. C → CO₂, H → H₂O. Pass gases through a weighed tube of anhydrous CaCl₂ (absorbs H₂O) and then KOH (absorbs CO₂). Increase in weight gives the masses.
2. Nitrogen (Kjeldahl's method): digest coal with conc. H₂SO₄ and K₂SO₄ (catalyst CuSO₄); N becomes (NH₄)₂SO₄. Add NaOH, distil off NH₃ into a known volume of standard acid, and back-titrate the unused acid.
3. Sulphur: from the washings of the bomb calorimeter (S converts to sulphate); precipitate as BaSO₄ with BaCl₂, filter, dry and weigh.
4. Ash: as in proximate analysis.
5. Oxygen: by difference.

Key formula

• \( \%C = \dfrac{\text{increase in wt. of KOH tube} \times 12}{44 \times \text{wt. of coal}} \times 100 \)
• \( \%H = \dfrac{\text{increase in wt. of CaCl}_2\text{ tube} \times 2}{18 \times \text{wt. of coal}} \times 100 \)
• \( \%N = \dfrac{1.4 \times V \times N}{\text{wt. of coal}} \), V = volume (mL) of acid neutralised by NH₃, N = its normality
• \( \%S = \dfrac{\text{wt. of BaSO}_4 \times 32}{233 \times \text{wt. of coal}} \times 100 \)
• \( \%O = 100 - (\%C + \%H + \%N + \%S + \%\text{ash}) \)

Significance: high C and H mean high calorific value; O lowers calorific value (it is already combined with H); N has no heat value and is undesirable; S forms SO₂, which is corrosive and polluting, and is harmful in metallurgical coke.

Quick example

0.25 g coal: KOH tube gains 0.77 g, CaCl₂ tube gains 0.12 g. %C = 0.77 × 12/44 / 0.25 × 100 = 84.0%. %H = 0.12 × 2/18 / 0.25 × 100 = 5.33%. 1 g coal in Kjeldahl's test neutralises 6 mL of 0.1 N H₂SO₄: %N = 1.4 × 6 × 0.1 / 1 = 0.84%. 1 g coal gives 0.06 g BaSO₄: %S = 0.06 × 32/233 × 100 = 0.82%.

Common mistakes

• Swapping the absorbers: CaCl₂ must come first (KOH solution would also absorb water).
• Using 12/44 for hydrogen; it is 2/18.
• Forgetting that N volume V is acid used by NH₃, i.e. blank minus back-titre.

How MU asks it

"Explain ultimate analysis of coal with significance." (5–6 marks) or numericals asking %C, %H, %N, %S, %O (5 marks). Write each determination with its reaction or principle, the formula, and significance. In numericals list every formula before substituting and finish with %O by difference.