Following Kjeldahl's method, 1 g of organic compound released ammonia, that neutralised 10 mL of 2M . The percentage of nitrogen in the compound is ______%.

Question

Following Kjeldahl's method, 1 g of organic compound released ammonia, that neutralised 10 mL of 2M $H_{2} {SO}_{4}$ . The percentage of nitrogen in the compound is ______%. [2024]

Smart Achievers · Accepted Answer

(56)

$Millimoles of sulphuric acid (n_{H_{2} {SO}_{4}}) = Molarity \times volume in mL$

$= 2 M \times 10 mL = 20 mmol$

$Ammonia is a base, it reacts with sulphuric acid to form salt:$

$2 {NH}_{3} + H_{2} {SO}_{4} \to {({NH}_{4})}_{2} {SO}_{4}$

$By stoichiometry:$

$\frac{n_{N H_{3}}}{2} = \frac{n_{H_{2} S O_{4}}}{1}$

$\frac{n_{N H_{3}}}{2} = \frac{20 mmol}{1}$

$n_{N H_{3}} = 40 mmol = 0.04 mol$

$Moles of N (n_{N}) = Moles of ammonia (n_{N H_{3}}) = 0.04 mol$

$Since amount of nitrogen is conserved, number of moles of N in organic compound (n_{N}) = number of moles of N in ammonia$

$(n_{N H_{3}}) = 0.04 mol$

$Weight of N in organic compound (W_{N}) = n_{N} \times Molar mass of N$

$= 0.04 mol \times 14 {g mol}^{- 1} = 0.56 g$

$Percentage of nitrogen in organic compound:$

$= \frac{W_{N}}{W_{compound}} \times 100 = \frac{0.56}{1} \times 100 = 56 %$

Solution

Q.
Following Kjeldahl's method, 1 g of organic compound released ammonia, that neutralised 10 mL of 2M $H_{2} {SO}_{4}$ . The percentage of nitrogen in the compound is ______%. [2024]

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Solution

Q. Following Kjeldahl's method, 1 g of organic compound released ammonia, that neutralised 10 mL of 2M H2SO4. The percentage of nitrogen in the compound is ______%. [2024]

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Q.
Following Kjeldahl's method, 1 g of organic compound released ammonia, that neutralised 10 mL of 2M $H_{2} {SO}_{4}$ . The percentage of nitrogen in the compound is ______%. [2024]