Quantitative Analysis in Organic Chemistry Explained with Methods

We know that organic compounds are primarily composed of carbon, hydrogen, and oxygen. Some contain nitrogen, sulfur, or halogens.

With quantitative analysis, it helps determine the exact percentage composition of these elements. It is important for deducing the empirical formula (simplest ratio of atoms) and molecular formula (actual number of atoms).

For exams like NEET and JEE Main, questions often involve interpreting experimental data from combustion or other methods to find these formulas, making this topic a bridge between theory and practical application.

You can use quantitative analysis using various techniques to measure the amounts of elements in an organic compound. 

1. Determination of Carbon and Hydrogen (Liebigs Method)

This method involves combusting a known mass of an organic compound in a stream of oxygen. The carbon in the compound is oxidized to carbon dioxide $\left(C{O}_2\right)$ , and hydrogen is converted to water $\left(H_{2}O\right)$ . These products are absorbed using specific reagents $C{O}_2$  by potassium hydroxide (KOH) and $H_{2}O$  by anhydrous calcium chloride $\left(CaC{l}_2\right)$ . The increase in mass of these absorbents is measured to calculate the percentages of carbon and hydrogen.

The formulas used are:

$\mathrm{\%}\mathrm{C}=\frac{12}{44}\times \frac{\text{Mass of}CO2\text{formed}}{\text{Mass of compound}}\times 100$

$\mathrm{\%}\mathrm{H}=\frac{2}{18}\times \frac{\text{Mass of}\mathrm{H}2\mathrm{O}\text{formed}}{\text{Mass of compound}}\times 100$

Here, 12 and 2 are the molar masses of carbon and hydrogen, while 44 and 18 are the molar masses of $C{O}_2$  and $H_{2}O$  , respectively.

For a visual understanding, refer to NCERT Class 11 Chemistry, Part II, Chapter 12, Figure 12.5, Page 341, which illustrates the setup for Liebigs combustion method.

2. Determination of Nitrogen

There are two primary methods to estimate nitrogen in organic compounds, both of which are relevant for JEE Main:

1. a) Dumas Method: The organic compound is heated with copper oxide (CuO) in a combustion tube. Nitrogen is converted to $N_2$ gas, while carbon and hydrogen form $C{O}_2$ and $H_{2}O$ . The volume of $N_2$  gas is measured at STP, and the percentage of nitrogen is calculated using:

$\mathrm{\%}\mathrm{N}=\frac{28}{22400}\times \frac{\text{Volume of}N2\left(\text{in}\mathrm{m}\mathrm{L}\text{at STP}\right)}{\text{Mass of compound}\left(\text{in}\mathrm{g}\right)}\times 100$

Here, 28 is the molar mass of $N_2$ , and 22400 mL is the molar volume of a gas at STP.
b) Kjeldahls Method: This method is used for compounds where nitrogen is part of an amino group (e.g., amines, proteins). The compound is heated with concentrated sulfuric acid $\left(H_{2}S{O}_4\right)$ , converting nitrogen to ammonium sulfate $\left({\left(N{H}_4\right)}_{2}S{O}_4\right)$ . This is then treated with a base to release ammonia $\left(N{H}_3\right)$ , which is absorbed in a known volume of standard acid. The amount of $N{H}_3$  is determined by titration, and the percentage of nitrogen is calculated as:

$\mathrm{\%}\mathrm{N}=\frac{1.4\times \text{Normality of acid}\times \text{Volume of acid used}\left(\text{in}\mathrm{m}\mathrm{L}\right)}{\text{Mass of compound}\left(\text{in}\mathrm{g}\right)}$

This method is not suitable for compounds with nitro $(-N{O}_2)$  or diazo $(-N=N-)$  groups.

3. Determination of Oxygen

Oxygen is usually calculated by difference, as direct methods are less common. After determining the percentages of carbon, hydrogen, and nitrogen (if present), the percentage of oxygen is:

$\mathrm{\%}\mathrm{O}=100-(\mathrm{\%}\mathrm{C}+\mathrm{\%}\mathrm{H}+\mathrm{\%}\mathrm{N})$

For example, if a compound has $40\mathrm{\%}$  carbon, $6.67\mathrm{\%}$  hydrogen, and $13.33\mathrm{\%}$  nitrogen, the oxygen content is:

$\mathrm{\%}\mathrm{O}=100-(40+6.67+13.33)=40\mathrm{\%}$

Here are some questions on quantitative analysis that you can expect.  

1. Question: 0.2 g of an organic compound produces 0.44 g of CO2 and 0.18 g of H 2 O on combustion. Calculate the percentage of carbon and hydrogen.

Solution: Using Liebigs method:

$\begin{array}{rr}& \mathrm{\%}\mathrm{C}=\frac{12}{44}\times \frac{0.44}{0.2}\times 100=60\mathrm{\%}\\ & \mathrm{\%}\mathrm{H}=\frac{2}{18}\times \frac{0.18}{0.2}\times 100=10\mathrm{\%}\end{array}$

Tip: Double-check calculations, as JEE often tests accuracy.
2. Question: 0.5 g of a compound gives 560 mL of N 2 at STP (Dumas method). Find the percentage of nitrogen.
Solution:

$\mathrm{\%}\mathrm{N}=\frac{28}{22400}\times \frac{560}{0.5}\times 100=14\mathrm{\%}$