29.5 mg of an organic compound containing nitrogen was digested according to Kjeldahl's method and t

Q: Why do we use back-titration instead of just measuring the ammonia directly?

Ammonia is a highly volatile gas. If we tried to collect and titrate it directly, a significant amount would escape into the atmosphere, leading to massive errors. By bubbling it directly into an excess of strong acid, we trap 100% of it as stable ammonium chloride ($NH_4Cl$).

Q: Why did we multiply by 14 to get the mass of nitrogen? Shouldn't it be 28 since nitrogen gas is $N_2$?

We are calculating the mass of nitrogen *atoms* present in the organic compound, not the mass of nitrogen gas ($N_2$). Since each molecule of ammonia ($NH_3$) contains exactly one atom of nitrogen, we use the atomic mass of a single nitrogen atom, which is 14.

Q: Can I use Normality instead of Molarity for these calculations?

Absolutely! In fact, using Normality is safer. However, in this specific problem, the n-factor for $HCl$, $NaOH$, and $NH_3$ is all 1. Because Normality = Molarity $\times$ n-factor, the Molarity and Normality values are identical here.

Q: What if the acid used in the flask was $H_2SO_4$ instead of $HCl$?

If $H_2SO_4$ was used, you must account for its n-factor of 2. The equivalents of acid would be $2 \times M \times V$. This is why working with equivalents (Normality) is generally preferred in titration problems to avoid missing stoichiometric ratios.

Q: Does Kjeldahl's method work for every single nitrogen-containing organic compound?

No, it doesn't! This is a classic trick question in JEE. Kjeldahl's method fails for compounds with nitro groups ($-NO_2$), azo groups ($-N=N-$), and nitrogen in aromatic rings (like pyridine). Under the digestion conditions, the nitrogen in these compounds does not convert to ammonium sulfate.

Question

29.5 mg of an organic compound containing nitrogen was digested according to Kjeldahl's method and the evolved ammonia was absorbed in 20 mL of 0.1 M HCl solution. The excess of the acid required 15 mL of 0.1 M NaOH solution for complete neutralization. The percentage of nitrogen in the compound is

Options:

(A) 59.0

(B) 47.4

(C) 23.7

(D) 29.5

Sigma Percentile · Accepted Answer

### The Elegance of Back-Titration

Imagine you are a detective trying to figure out exactly how much of a specific chemical was produced in a reaction, but the chemical is a volatile gas—like ammonia ($NH_3$). If you try to measure it directly, it might escape into the air.

This is the exact problem Kjeldahl's method solves. Instead of catching the ammonia directly, we trap it in a known, excessive amount of a strong acid, like Hydrochloric Acid ($HCl$). The ammonia neutralizes some of the acid. To find out how much ammonia was there, we just need to figure out how much acid is *left over*. We do this by titrating the remaining acid with a standard base, like Sodium Hydroxide ($NaOH$). This brilliant indirect measurement is called **back-titration**.

### Setting Up the Math

The core principle of back-titration is the conservation of equivalents. The total amount of acid we started with must equal the amount that reacted with the ammonia plus the amount that reacted with the base during the final titration.

Mathematically, we can write this as:
$$n_{HCl(total)} = n_{HCl(reacted\ with\ NH_3)} + n_{HCl(reacted\ with\ NaOH)}$$

Since the valency factor (n-factor) for $HCl$, $NaOH$, and $NH_3$ is all $1$, we can comfortably work in millimoles ($M 	imes V_{mL}$).

Let's calculate the total millimoles of $HCl$ we took initially:
$$n_{HCl(total)} = 0.1	ext{ M} 	imes 20	ext{ mL} = 2	ext{ mmol}$$

Now, let's calculate the millimoles of $NaOH$ required to neutralize the excess acid:
$$n_{NaOH} = 0.1	ext{ M} 	imes 15	ext{ mL} = 1.5	ext{ mmol}$$

### Finding the Ammonia

If we started with $2	ext{ mmol}$ of acid and $1.5	ext{ mmol}$ were left over to react with the $NaOH$, the difference must be the amount of acid that reacted with our elusive ammonia gas.

$$n_{NH_3} = 2	ext{ mmol} - 1.5	ext{ mmol} = 0.5	ext{ mmol}$$

Here is the crucial chemical link: Every single molecule of ammonia ($NH_3$) contains exactly one atom of nitrogen ($N$). Therefore, the millimoles of nitrogen atoms are exactly equal to the millimoles of ammonia.

$$n_{N} = 0.5	ext{ mmol}$$

To find the mass of this nitrogen, we multiply the millimoles by the atomic mass of nitrogen, which is $14	ext{ g/mol}$ (or $14	ext{ mg/mmol}$).

$$	ext{Mass}_{N} = 0.5	ext{ mmol} 	imes 14	ext{ mg/mmol} = 7	ext{ mg}$$

### The Final Percentage

We now know that our original $29.5	ext{ mg}$ sample of the organic compound contained exactly $7	ext{ mg}$ of pure nitrogen. To find the percentage composition, we simply divide the mass of the nitrogen by the total mass of the compound and multiply by $100$.

$$\% N = \frac{7	ext{ mg}}{29.5	ext{ mg}} 	imes 100 \approx 23.73\%$$

This perfectly matches option (c).

**A Word of Caution:** While Kjeldahl's method is incredibly useful, it has its blind spots. It fails to estimate nitrogen in compounds containing nitro ($-NO_2$) groups, azo ($-N=N-$) groups, or nitrogen present in a ring structure (like pyridine). In these cases, the nitrogen does not convert to ammonium sulfate upon digestion, and we must rely on the Dumas method instead.

Select Answer:

Visualized Solution

$K j e l d ah l^{'} s M e t h o d S e t u p$

$P r in c i pl e o f B a c k T i t r a t i o n$

$C a l c u l a t in g M i l l im o l es$

$M i l l im o l es o f A mm o nia$

$M a ss o f N i t r o g e n$

$E x ec u t in g M a ss C a l c u l a t i o n$

$P er ce n t a g e o f N i t r o g e n$

$T h e W a y F or w a r d$

The Sigma Insight: Nomenclature and Characterisation

The Elegance of Back-Titration

Setting Up the Math

Finding the Ammonia

The Final Percentage

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