Question

0.7g of an organic compound was Kjeldahlised and ammonia evolved was absorbed in 50ml of semi-molar (0.5M) H​​​​​​₂SO​​​₄. The residual acid required 80ml of 0.5M NaOH. Determine the percentage of nitrogen in the compound.

a.) 80
b.) 40
c.) 20
d.) 10

Answer 1

By applying the Kjeldahl method and subsequent calculations, it was determined that the compound contains 60% nitrogen by mass, which does not match any of the given answer choices. Therefore, the correct answer that should be selected is not provided in the given options (a, b, c, d).

Step-by-step explanation:

To determine the percentage of nitrogen in the compound, we start by understanding the Kjeldahl method. When the compound is Kjeldahlised, it releases ammonia which then reacts with sulphuric acid. The ammonia neutralizes a portion of the acid, and the amount of acid reacted is subsequently determined by titration with a known concentration of sodium hydroxide (NaOH). Here's the process broken down:

• Amount of H₂SO₄ reacted with NH₃: (50 mL of 0.5M) - (80 mL of 0.5M NaOH) = -30 mL of 0.5M, meaning that 30 mL of 0.5M H₂SO₄ reacted with the evolved NH₃.
• Moles of H₂SO₄ that reacted with NH₃: 30mL * 0.5M = 0.015 moles.
• Moles of NH₃: Since 1 mole of H₂SO₄ reacts with 2 moles of NH₃, we have 0.015 * 2 = 0.03 moles of NH₃.
• Moles of nitrogen: 0.03 moles (since 1 mole of NH₃ contains 1 mole of N).
• Mass of nitrogen: 0.03 moles * 14.01 g/mole (atomic weight of N) = 0.4203 g.
• Percentage of nitrogen by mass: (0.4203 g / 0.7 g) * 100 = 60%.

Therefore, the correct answer that should be selected is not provided in the given options (a, b, c, d).