Explain the principle and estimation method for carbon and hydrogen in an organic compound.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) Principle: Carbon and hydrogen in an organic compound are estimated by burning a known mass of the compound in the presence of excess oxygen and copper $(II)$ oxide $(CuO)$. Carbon is oxidized to carbon dioxide $(CO_2)$ and hydrogen is oxidized to water $(H_2O)$.
Reaction: $C_xH_y + (x + \frac{y}{4})O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O$
Procedure:
$1$. $A$ known mass $(m \ g)$ of the organic compound is placed in a platinum boat and burnt in a combustion tube in the presence of excess oxygen and $CuO$.
$2$. The resulting gases are passed through a weighed $U$-tube containing anhydrous calcium chloride $(CaCl_2)$,which absorbs water.
$3$. The gases are then passed through another weighed $U$-tube containing a concentrated solution of potassium hydroxide $(KOH)$,which absorbs carbon dioxide.
Calculation:
- Let the increase in mass of the $CaCl_2$ $U$-tube be $m_1 \ g$ (mass of $H_2O$ produced).
- Let the increase in mass of the $KOH$ $U$-tube be $m_2 \ g$ (mass of $CO_2$ produced).
Percentage of Hydrogen: Since $18 \ g$ of $H_2O$ contains $2 \ g$ of hydrogen,
$\text{Percentage of } H = \frac{2 \times m_1 \times 100}{18 \times m}$
Percentage of Carbon: Since $44 \ g$ of $CO_2$ contains $12 \ g$ of carbon,
$\text{Percentage of } C = \frac{12 \times m_2 \times 100}{44 \times m}$

Explore More

Similar Questions

$0.53 \ g$ of an organic compound $(x)$ when heated with excess of nitric acid (concentrated) and then with silver nitrate gave $0.75 \ g$ of silver bromide precipitate. $1.0 \ g$ of $(x)$ gave $1.32 \ g$ of $CO_2$ gas on combustion. The percentage of hydrogen in the compound $(x)$ is . . . . . . $\%$. [Nearest Integer]
[Given : Molar mass in $g \ mol^{-1} H : 1, C : 12, Br : 80, Ag : 108, O : 16$; Compound $(x) : C_xH_yBr_z$] (in $\%$)

$2.18 \ g$ of an organic compound containing sulphur produces $0.12 \ g$ of $BaSO_4$. The percentage of sulphur in the compound is?

What is the reason for passing dry air through the combustion tube during the estimation of $C$ and $H$?

$0.1 \ g$ of an organic compound was analyzed by the Kjeldahl method. The $NH_3$ produced was absorbed in $30 \ mL$ of $N/5$ $H_2SO_4$. The remaining acid required $20 \ mL$ of $N/10$ $NaOH$ for neutralization. Calculate the percentage of nitrogen in the organic compound.

Difficult
View Solution

The element whose percentage composition in an organic compound can be determined by the Carius method is

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo