Question

$200mg$ of an organic compound containing nitrogen was digested according to Kjeldahl’s method and the evolved $N{H_3}$ gas was absorbed in $30ml$ of $0.2M$ $HCl$ solution. The excess of acid required $20ml$ of $0.2M$ $NaOH$ solution for complete neutralisation. What would be the percentage of nitrogen in the compound?
A) $12\%$
B) $14\%$
C) $20\%$
D) $22\%$

Answer 1

As we know that the percentage composition of any compound is the relative mass of each of the constituent element in $100$ parts of it and mass percentage of an element is given as the ratio of mass of that element in one mole of the compound to the molar mass of the compound.

Complete solution:
As we know that the mass percentage of any element can be calculated with the help of the ratio of the mass of that element present in one mole of the total compound divided by the molecular mass of that whole compound multiplied by $100$ as we need to calculate the percentage.
So, for this we first need to calculate the total moles of ammonia evolved in the reaction. We can calculate these moles using the moles of hydrochloric acid and moles of sodium hydroxide in the solution.
So, let us calculate the moles of hydrochloric acid $(HCl)$, we know that moles are given as the product of molarity and the volume. And we are given that $HCl$ is present as $0.2M$ in $30ml$of volume, so the moles will be equivalent to:
$moles = 30 \times 0.2 = 6\;moles$
Similarly, we can calculate the moles of sodium hydroxide $(NaOH)$as we are given that it is present as $0.2M$ in $20ml$ volume. So the moles will be equivalent to:
$moles = 20 \times 0.2 = 4\;moles$
Now, moles of ammonia will be given by the difference of moles of $(HCl)$ and $(NaOH)$.
So, it will be equivalent to:
$mole{s_{N{H_3}}} = 6 - 4 = 2\;moles$
Now we need to calculate the mass of the element to find out the mass percentage of the element. So the mass of nitrogen will be given by the product of moles and the molecular mass of nitrogen which will be:
$mas{s_{nitrogen}} = 2 \times 14 = 28g$
Lastly, using the formula of mass percentage of nitrogen we can calculate the percentage of nitrogen in the compound which will be given as:
$\% \;nitrogen = \dfrac{{mass\;of\;nitrogen}}{{Molecular\;mass\;of\;compound}} \times 100$
$\% \;nitrogen = \dfrac{{28 \times {{10}^{ - 3}}}}{{200 \times {{10}^{ - 3}}}} \times 100$
$\% \;nitrogen = 14\%$

Therefore, the correct answer is (B).

Note: Always remember that for calculating the mass percentage of any element present in the compound, first we can find out the moles or millimoles of the acid and bases and their difference will give the moles of compound in which our element is present which can then be calculated using mass percentage formula.