(N/A) The percentage of oxygen in an organic compound is usually found by the difference between the total percentage composition $(100)$ and the sum of the percentages of all the other elements present.
$\% O = 100 - (\text{sum of } \% \text{ of all other elements})$
$(b)$ $A$ definite mass of an organic compound is decomposed by heating in a stream of nitrogen gas. The mixture of gaseous products containing oxygen is passed over red-hot coke at $1373 \ K$,where all the oxygen is converted to carbon monoxide $(CO)$.
$2C + O_2 \xrightarrow{1373 \ K} 2CO$
This mixture is then passed through warm iodine pentoxide $(I_2O_5)$,where carbon monoxide is oxidized to carbon dioxide,producing iodine $(I_2)$.
$I_2O_5 + 5CO \rightarrow I_2 + 5CO_2$
$(c)$ Calculation:
Let the mass of the organic compound be $m \ g$ and the mass of produced $CO_2$ be $m_1 \ g$.
Since $88 \ g$ of $CO_2$ contains $32 \ g$ of oxygen,the mass of oxygen in $m_1 \ g$ of $CO_2$ is $\frac{32 \times m_1}{88} \ g$.
Therefore,$\% O = \frac{32 \times m_1}{88 \times m} \times 100$.
$(d)$ Modern methods: Currently,estimation is carried out using $CHN$ elemental analyzers,which require only $1-3 \ mg$ of the substance and provide rapid results.