0.8 g of an organic compound was analysed by Kjeldahl's method for the estimation of nitrogen. If th

Q: Why do we divide by 14 and not 28 to find the moles of nitrogen?

We are interested in the moles of nitrogen *atoms* ($N$), not nitrogen gas molecules ($N_2$). Each nitrogen atom in the organic compound eventually forms one molecule of ammonia ($NH_3$). Therefore, we use the atomic mass of nitrogen, which is 14.

Q: Why is the mole ratio of ammonia to sulfuric acid 2:1?

Sulfuric acid ($H_2SO_4$) is a diprotic acid, meaning it can donate two protons ($H^+$). Ammonia ($NH_3$) is a base that accepts only one proton to become ammonium ($NH_4^+$). Therefore, it takes two molecules of ammonia to fully neutralize one molecule of sulfuric acid.

Q: Can I use the direct formula for Kjeldahl's method?

Absolutely! The direct formula is $\%N = \frac{1.4 \times M \times V \times \text{basicity}}{W}$. While it's great for quick calculations, understanding the step-by-step stoichiometry builds a stronger foundation and prevents mistakes if the question is twisted.

Q: What if the acid used was Hydrochloric Acid ($HCl$) instead of Sulfuric Acid?

If $HCl$ was used, the reaction would be $NH_3 + HCl \rightarrow NH_4Cl$. Since $HCl$ is monoprotic, the mole ratio would be 1:1. You would need exactly the same number of moles of $HCl$ as you have of ammonia.

Q: Why did we multiply by 1000 at the very end?

The molarity formula ($M = \frac{n}{V}$) inherently calculates volume in Liters. Since the question specifically asks for the answer in milliliters (mL), we must multiply the result in Liters by 1000 to convert it.

Question

0.8 g of an organic compound was analysed by Kjeldahl's method for the estimation of nitrogen. If the percentage of nitrogen in the compound was found to be 42%, then ......... mL of $1	ext{ M } H_2SO_4$ would have been neutralised by the ammonia evolved during the analysis.

Sigma Percentile · Accepted Answer

The Chemistry of Kjeldahl's Method: A Stoichiometric Symphony

Kjeldahl's method is a brilliant piece of analytical chemistry designed to determine the nitrogen content in organic compounds. Imagine you have a mysterious organic powder, and you need to know exactly how much nitrogen is locked inside its molecular structure. You can't just weigh the nitrogen directly. Instead, Kjeldahl's method uses a series of chemical transformations to convert that hidden nitrogen into a measurable quantity: ammonia.

In this problem, we are given a $0.8 	ext{ g}$ sample of an organic compound, and we already know it contains $42\%$ nitrogen by mass. Our mission is to trace this nitrogen as it transforms into ammonia and then determine how much $1	ext{ M }$ sulfuric acid ($H_2SO_4$) is required to neutralize it.

### Decoding the Organic Compound

The first step is to find out exactly how much nitrogen we are dealing with in absolute terms. We know the total mass of the sample is $0.8 	ext{ g}$, and $42\%$ of that mass is pure nitrogen.

By simply multiplying the percentage by the total mass, we find the mass of nitrogen:

$m_N = \frac{42}{100} 	imes 0.8 	ext{ g}$

Instead of calculating this out to a decimal right away, it is often smarter to leave it as a fraction. This prevents rounding errors and makes subsequent calculations much cleaner.

Next, we need to convert this mass into moles. Why? Because chemical reactions happen on a molecule-to-molecule (or mole-to-mole) basis, not gram-to-gram. We divide the mass by the atomic mass of nitrogen ($14 	ext{ g/mol}$):

$n_N = \frac{42 	imes 0.8}{100 	imes 14} = \frac{3 	imes 0.8}{100} = \frac{2.4}{100} 	ext{ mol}$

### The Transformation: Nitrogen to Ammonia

The core principle of Kjeldahl's method is that every single atom of nitrogen in the original organic compound is eventually converted into one molecule of ammonia ($NH_3$). This is a beautiful 1:1 relationship.

Therefore, the moles of ammonia produced will be exactly equal to the moles of nitrogen atoms we just calculated:

$n_{NH_3} = n_N = \frac{2.4}{100} 	ext{ mol}$

### The Climax: The Neutralization Dance

Now comes the titration phase. The ammonia gas produced is bubbled into a solution of sulfuric acid ($H_2SO_4$). Ammonia is a base, and sulfuric acid is, well, an acid. They will neutralize each other. But we must look closely at the balanced chemical equation to understand their relationship:

$2NH_3 + H_2SO_4 ightarrow (NH_4)_2SO_4$

Notice the coefficients! It takes **two** molecules of ammonia to fully neutralize just **one** molecule of sulfuric acid. This is because sulfuric acid is diprotic—it has two acidic protons ($H^+$) to give, while each ammonia molecule can only accept one.

Because of this 2:1 stoichiometric ratio, the number of moles of sulfuric acid required will be exactly half the number of moles of ammonia:

$n_{H_2SO_4} = \frac{n_{NH_3}}{2} = \frac{1.2}{100} 	ext{ mol}$

### The Grand Finale: Calculating the Volume

We are almost there. We know we need $\frac{1.2}{100} 	ext{ mol}$ of sulfuric acid, and we are given that the concentration of our acid solution is $1	ext{ M }$ (one mole per liter).

The formula for molarity is $M = \frac{n}{V}$. Rearranging this to solve for volume gives us $V = \frac{n}{M}$.

Substituting our values:

$V = \frac{1.2/100}{1} 	ext{ L} = 0.012 	ext{ L}$

The question specifically asks for the volume in milliliters (mL). To convert from liters to milliliters, we multiply by 1000:

$V = 0.012 	imes 1000 	ext{ mL} = \mathbf{12 	ext{ mL}}$

And there we have it! By carefully following the atoms from the initial compound through their chemical transformations, we've arrived at the exact volume of acid needed.

Enter Numerical Value:

Visualized Solution

$W = 0.8 g, % N = 42%$

$m_{N} = \frac{% N}{100} \times W$

$m_{N} = \frac{42}{100} \times 0.8 g$

$n_{N} = \frac{m _{N}}{M _{N}}$

$n_{N} = \frac{2.4}{100} mol$

$n_{N H_{3}} = n_{N}$

$2 N H_{3} + H_{2} S O_{4} \to (N H_{4})_{2} S O_{4}$

$n_{H_{2} S O_{4}} = \frac{n _{N H_{3}}}{2}$

$V = \frac{n}{M}$

$V = 12 mL$

\text{Alternative Formula}

The Sigma Insight: Nomenclature and Characterisation

Decoding the Organic Compound

The Transformation

The Climax

The Grand Finale

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