Animated Solution for Chemistry - Organic Chemistry: For the estimation of nitrogen, $1.4\text{ g}$ of an organic compound was digested by Kjeldahl's method and the evolved ammonia was absorbed in $60\text{ mL}$ of $M/10$ sulphuric acid. The unreacted acid required $20\text{ mL}$ of $M/10$ sodium hydroxide for the complete neutralization. The percentage of nitrogen in the compound is

Quantitative analysis in organic chemistry often feels like solving a mystery. You have an unknown compound, and you need to find out exactly how much of a specific element is hiding inside it. Kjeldahl's method is a classic, elegant technique used to estimate the amount of nitrogen in an organic compound.

Instead of measuring the nitrogen directly, we force it to reveal itself by converting it into ammonia gas ($N{H}_3$). But ammonia is a slippery, volatile gas. If we try to measure it directly, some of it will escape. So, we use a clever trick called back-titration. We trap the ammonia in a known, excess amount of a strong acid, like sulfuric acid ($H_{2}S{O}_4$). Then, we measure how much acid is left over to figure out how much acid reacted with the ammonia.

How Much Acid Do We Have?

Our first step is to determine the total amount of sulfuric acid we started with. We are given $60\text{ mL}$ of $M/10$ (which means $0.1\text{ M}$) $H_{2}S{O}_4$.

To find the millimoles (mmol), we simply multiply the molarity by the volume in milliliters:

So, we have a total pool of $6\text{ mmol}$ of sulfuric acid waiting to trap the ammonia.

Finding the Leftovers

After the ammonia has bubbled through and reacted with the acid, some acid remains unreacted. To find out how much, we titrate this leftover acid with a strong base, sodium hydroxide ($NaOH$). We used $20\text{ mL}$ of $M/10$ $NaOH$.

Let's calculate the millimoles of $NaOH$ used:

Now, we must be very careful with the stoichiometry. The neutralization reaction is:

Notice that it takes two moles of $NaOH$ to neutralize just one mole of $H_{2}S{O}_4$. Therefore, the $2\text{ mmol}$ of $NaOH$ we used must have neutralized exactly half that amount of sulfuric acid:

Unmasking the Ammonia

If we started with $6\text{ mmol}$ of acid and $1\text{ mmol}$ was left over, the amount that actually reacted with the ammonia is simply the difference:

Now, let's look at the reaction between ammonia and sulfuric acid:

Here, one mole of $H_{2}S{O}_4$ can trap two moles of $N{H}_3$. Since $5\text{ mmol}$ of acid reacted, it must have trapped twice as much ammonia:

Every single molecule of ammonia ($N{H}_3$) contains exactly one atom of nitrogen. Therefore, $10\text{ mmol}$ of ammonia means we have $10\text{ mmol}$ of nitrogen atoms.

Let's convert these millimoles into mass. The atomic mass of nitrogen is $14\text{ g/mol}$.

Finally, to find the percentage of nitrogen in the original organic compound, we divide the mass of nitrogen by the total mass of the compound ($1.4\text{ g}$) and multiply by $100$:

And there we have it! The compound contains exactly $10\%$ nitrogen. While Kjeldahl's method is incredibly powerful, always remember its limitations: it struggles with compounds where nitrogen is in a ring (like pyridine) or in nitro/azo groups, as they don't easily convert to ammonium sulfate during digestion.