Estimation of nitrogen: There are two methods for the estimation of nitrogen (i) Dumas method and (ii) Kjedahl's method.
i. Dumas method: A known mass of compound is heated with copper oxide (CuO) in an atomsphere of `CO_(2)`, which gives free nitrogen along with `CO_(2) and H_(2)O`
`C_(x)H_(y)N_(z) + (2x + y//2) CuO rarr xCO_(2) + y//2 (H_(2)O) + z//2 (N_(2)) + (2x + y//2) CU`
The gaseous mixture is passed over a heated copper gauze which converts traces of nitrogen oxides formed to `N_(2)`. The gaeous mixture is collected over an aqueous solution of KOH which absorbs `CO_(2)`, and nitrogen is callected in the upper part of the graduated tube.
ii. Kjeldahl's method: A known mass of organic compound (0.5 gm) is mixed with `K_(2)SO_(4)` (10 gm) and `CuSO_(4)`. (1.0 gm) or a drop of mercury (Hg) and conc. `H_(2)SO_(4)` (25ml), and heated in Kjeldahl's flask. `CuSO_(4)` or Hg acts as a catalyst, while `K_(2)SO_(4)` raises the boiling point of `H_(2)SO_(4)`. The nitrogen in the organic compound is quantitatively converted to ammonium sulphate. The resulting mixture is then distilled with excess of NaOH solution and the `NH_(3)` evolved is passed into a known but excess volume of standard HCl or `H_(2)SO_(4)`. The acid left unused is estimated by titration with some standard alkali. The amount of acid used against `NH_(3)` can be thus be known and from this the percentage of nitrogen is calculated.
(a) `C + H + S underset(H_(2)SO_(4))overset("conc.")rarr CO_(2) + H_(2)O + SO_(2)`
(b) `N underset(H_(2)SO_(4))overset("conc.")rarr (NH_(4))_(2)SO_(4)`
(c) `(NH_(4))_(2) SO_(4) + 2NaOH rarr Na_(2)SO_(4) + 2NH_(3) + 2H_(2)O`
(d) `2NH_(3) + H_(2)SO_(4) rarr (NH_(4))_(2) SO_(4)`
(iii) This method is not applicable to compounds containing N in intro and azo groups, and N present in the ring (e.g., pyridine) as N of these compounds does not change to `(NH_(4))_(2)SO_(4)` (ammonium sulphate) under these reaction conditions
0.50 gm of an organic compound was treated according to Kjeldahl's method. The ammonia evolved was absorbed in 50ml of 0,5M `H_(2)SO_(4)`. The residual acid required 60 ml of M/2 NaOH solution. Find the percentage of nitrogen in the compound

Options

A. 50
B. 56
C. 66
D. 40

Answer

To find the percentage of nitrogen in the organic compound using Kjeldahl's method, we will follow these steps:

### Step 1: Calculate the Normality of H₂SO₄
The normality (N) of an acid is calculated using the formula:
\[ \text{Normality} = \text{Molarity} \times n \]
where \( n \) is the number of acidic protons (H⁺) the acid can donate. For sulfuric acid (H₂SO₄), \( n = 2 \) because it is a dibasic acid.

Given:
- Molarity of H₂SO₄ = 0.5 M

Calculating Normality:
\[ \text{Normality of H₂SO₄} = 0.5 \, \text{M} \times 2 = 1 \, \text{N} \]

### Step 2: Calculate the Normality of NaOH
For NaOH, which is a strong base and monobasic, \( n = 1 \).

Given:
- Molarity of NaOH = 0.5 M

Calculating Normality:
\[ \text{Normality of NaOH} = 0.5 \, \text{M} \times 1 = 0.5 \, \text{N} \]

### Step 3: Calculate Milliequivalents of H₂SO₄ and NaOH
Milliequivalents (mEq) can be calculated using the formula:
\[ \text{Milliequivalents} = \text{Normality} \times \text{Volume (mL)} \]

**For H₂SO₄:**
Given volume = 50 mL
\[ \text{Milliequivalents of H₂SO₄} = 1 \, \text{N} \times 50 \, \text{mL} = 50 \, \text{mEq} \]

**For NaOH:**
Given volume = 60 mL
\[ \text{Milliequivalents of NaOH} = 0.5 \, \text{N} \times 60 \, \text{mL} = 30 \, \text{mEq} \]

### Step 4: Calculate the Milliequivalents of Ammonia (NH₃)
The milliequivalents of ammonia can be found by subtracting the milliequivalents of NaOH from the milliequivalents of H₂SO₄:
\[ \text{Milliequivalents of NH₃} = \text{Milliequivalents of H₂SO₄} - \text{Milliequivalents of NaOH} \]
\[ \text{Milliequivalents of NH₃} = 50 \, \text{mEq} - 30 \, \text{mEq} = 20 \, \text{mEq} \]

### Step 5: Calculate the Percentage of Nitrogen
The percentage of nitrogen in the compound can be calculated using the formula:
\[ \text{Percentage of Nitrogen} = \left( \frac{1.4 \times \text{milliequivalents of NH₃}}{\text{weight of organic compound (g)}} \right) \]

Given:
- Weight of organic compound = 0.5 g

Calculating the percentage:
\[ \text{Percentage of Nitrogen} = \left( \frac{1.4 \times 20}{0.5} \right) = \frac{28}{0.5} = 56\% \]

### Final Answer
The percentage of nitrogen in the organic compound is **56%**.

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