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✦The Simple Trick for Oxygen

Every element in an organic compound must account for some percentage of its mass, and all percentages must add up to exactly 100. So once you know %C, %H, %N, %S, and %halogen — you're done. %O = 100 − (sum of everything else). Oxygen is the only element routinely estimated "by difference". That's not laziness — it's elegant accounting.

Estimation of Phosphorus

Principle: A known mass of the organic compound is heated with fuming nitric acid. Phosphorus in the compound is oxidised to phosphoric acid ( $\ce{H3PO4}$ ).

The $\ce{H3PO4}$ is then precipitated in two alternative ways:

Method A — Ammonium phosphomolybdate:
Add ammonia and ammonium molybdate. A canary yellow precipitate of ammonium phosphomolybdate forms:
$\ce{(NH4)3PO4.12MoO3} \downarrow$
Molar mass = 1877 g/mol
$\% \text{P} = \frac{31}{1877} \times \frac{m_1}{m} \times 100$

Method B — Magnesium pyrophosphate:
Add magnesia mixture → forms $\ce{MgNH4PO4}$ → ignite → gives $\ce{Mg2P2O7}$
Molar mass of $\ce{Mg2P2O7}$ = 222 g/mol; contains 2 × 31 = 62 g of P
$\% \text{P} = \frac{62}{222} \times \frac{m_1}{m} \times 100$

Where $m$ = mass of compound, $m_1$ = mass of precipitate formed.

Direct estimation of oxygen — pyrolysis in N₂, red-hot coke, I₂O₅ apparatus — 📸 Direct oxygen estimation: compound pyrolysed in N₂ stream → O₂ converted to CO over red-hot coke → CO measured via I₂O₅.

Two Phosphorus Formulas

As ammonium phosphomolybdate $\ce{(NH4)3PO4.12MoO3}$ : $\% \text{P} = \frac{31 \times m_1}{1877 \times m} \times 100$

As magnesium pyrophosphate $\ce{Mg2P2O7}$ : $\% \text{P} = \frac{62 \times m_1}{222 \times m} \times 100$

The 62 in the second formula = 2 phosphorus atoms (31 × 2) because $\ce{Mg2P2O7}$ contains 2 phosphorus atoms per formula unit.

Estimation of Oxygen — By Difference

The percentage of oxygen in an organic compound is usually calculated by difference:
$\% \text{O} = 100 - (\% \text{C} + \% \text{H} + \% \text{N} + \% \text{S} + \% \text{X} + \% \text{P})$

This works because all elemental percentages must sum to 100%.

Direct Estimation of Oxygen

When a direct method is needed, the following process is used:

The organic compound is decomposed by heating in a stream of nitrogen gas. This releases oxygen and other gaseous products from the compound.
The gaseous products (containing $\ce{O2}$ ) are passed over red-hot coke (carbon) at high temperature. All oxygen is converted to carbon monoxide:
$\ce{2C + O2 ->[1373 K] 2CO}$
This gas stream is then passed through warm iodine pentoxide ( $\ce{I2O5}$ ). CO reduces the iodine pentoxide:
$\ce{I2O5 + 5CO -> I2 + 5CO2}$
The iodine liberated is measured (or the $\ce{CO2}$ produced is weighed).

Calculation: Each mole of $\ce{O2}$ from the compound gives 2 moles of CO (from the coke reaction), which gives 2 moles of $\ce{CO2}$ (from the $\ce{I2O5}$ reaction):
$\% \text{O} = \frac{32}{88} \times \frac{m_1}{m} \times 100$
where $m_1$ = mass of $\ce{CO2}$ produced, $m$ = mass of compound
(since 88 g $\ce{CO2}$ = 2 mol $\ce{CO2}$ corresponds to 32 g O)

✦JEE / NEET Exam InsightJEE / NEET

◆Oxygen by difference (always tested):

\% \text{O} = 100 - (\% \text{C} + \% \text{H} + \text{all others})

◆Phosphorus formulas:

◆As ammonium phosphomolybdate:

\% \text{P} = \frac{31 \times m_1}{1877 \times m} \times 100

◆As Mg₂P₂O₇:

\% \text{P} = \frac{62 \times m_1}{222 \times m} \times 100

(62 because 2 P atoms)

◆Direct O₂ estimation steps:

◆Pyrolysis in N₂ → releases O₂

◆O₂ + red-hot coke → CO (

\ce{2C + O2 -> 2CO}

)

◆CO + I₂O₅ → CO₂ + I₂ (

\ce{I2O5 + 5CO -> I2 + 5CO2}

)

◆Measure CO₂: %O = (32/88) × (m₁/m) × 100

◆Modern analysis: CHN elemental analyser determines C, H, N automatically using just 1–3 mg of compound.

Quick Check

Q1.An organic compound contains 40% C, 6.67% H, and 53.33% O (by mass). The % of oxygen was found by: