Oxygen

Estimation of Oxygen in Organic Compounds

The direct determination of oxygen in an organic compound is more involved than the estimation of carbon, hydrogen, nitrogen, or halogens. In many routine analyses, the percentage of oxygen is obtained by difference — that is, you subtract the sum of the percentages of all other elements from 100. This method assumes that no other element is present and that all analytical errors accumulate in the oxygen figure. However, a direct gravimetric method exists, and it is based on a clever sequence of chemical conversions.

Principle of the Direct Method

A known mass of the organic compound is heated strongly in a stream of pure, oxygen-free nitrogen gas. The heat decomposes the compound, and any oxygen present in the compound is liberated, most likely as molecular oxygen ($O_2$) along with other gaseous products. This mixture of gases is then passed over red-hot coke (carbon) maintained at about 1373 K. At this temperature, all the oxygen reacts with carbon to form carbon monoxide.

The reaction occurring over the red-hot coke is:

The stream of gases, now containing carbon monoxide (along with nitrogen and other inert products), is then passed through a tube containing warm iodine pentoxide ($I_2O_5$). Carbon monoxide reduces iodine pentoxide, getting oxidised to carbon dioxide in the process, and liberating free iodine.

The reaction with iodine pentoxide is:

The iodine produced can be collected and weighed, or the carbon dioxide produced can be absorbed and weighed. The amount of oxygen originally present in the compound is then calculated from the mass of either product.

Deriving the Relationship Between Oxygen and Carbon Dioxide

To find a direct stoichiometric link between the oxygen in the compound and the carbon dioxide finally collected, we need to combine the two reactions so that the carbon monoxide produced in the first step exactly matches the carbon monoxide consumed in the second step.

›Proof

Stoichiometric Combination of the Two Reactions

Step 1: The coke reaction produces 2 moles of CO from 1 mole of $O_2$.

$2C + O_2 \rightarrow 2CO$

Step 2: The iodine pentoxide reaction consumes 5 moles of CO to produce 5 moles of $CO_2$.

$I_2O_5 + 5CO \rightarrow I_2 + 5CO_2$

To make the amount of CO produced in Step 1 equal to the amount of CO consumed in Step 2, we need a common multiple of the CO coefficients (2 and 5). The least common multiple is 10.

Multiply Step 1 by 5:

$10C + 5O_2 \rightarrow 10CO$

Multiply Step 2 by 2:

$2I_2O_5 + 10CO \rightarrow 2I_2 + 10CO_2$

Now, the 10 moles of CO produced from 5 moles of $O_2$ are exactly consumed to produce 10 moles of $CO_2$.

Therefore, the overall stoichiometric relationship is:

$5O_2 \equiv 10CO_2$

Simplifying this ratio:

$1 \text{ mole of } O_2 \equiv 2 \text{ moles of } CO_2$

In terms of mass:

• Molar mass of $O_2$ = 32 g
• Molar mass of $CO_2$ = 44 g
• Mass of 2 moles of $CO_2$ = $2 \times 44 = 88$ g

Hence, 32 g of oxygen (as $O_2$) produces 88 g of carbon dioxide.

Calculating the Percentage of Oxygen

With the stoichiometric link established, the calculation follows the standard gravimetric analysis pattern.

Let:

• $m$ = mass of the organic compound taken (in grams)
• $m_1$ = mass of carbon dioxide produced (in grams)

From the derived relationship:

88 g of $CO_2$ is produced from 32 g of $O_2$.

Therefore, 1 g of $CO_2$ is produced from $\frac{32}{88}$ g of $O_2$.

Therefore, $m_1$ g of $CO_2$ is produced from $\frac{32}{88} \times m_1$ g of $O_2$.

This mass of oxygen ($\frac{32}{88} m_1$) was originally present in the $m$ grams of the organic compound.

The percentage of oxygen in the compound is:

Alternative Calculation Using Iodine

The textbook also notes that the percentage of oxygen can be derived from the amount of iodine ($I_2$) produced. From the combined stoichiometry, 5 moles of $O_2$ (160 g) produce 2 moles of $I_2$ (2 $\times$ 254 = 508 g). If the mass of iodine produced is measured, a similar proportionality can be set up to find the mass of oxygen.

Modern Microanalysis …