Question

In Duma’s method of estimation of nitrogen, 0.35g of an organic compound gave 55ml of nitrogen collected at 300K temperature and 715mm pressure. The percentage composition of nitrogen in the compound would be:
(Aqueous tension at 300K = 15mm)
(A) 16.45%
(B) 17.45%
(C) 14.45%
(D) 15.45%

Answer 1

Hint: Using Ideal gas equation, PV=nRT, find a relationship between the given pressure, temperature and volume and at STP. That relation is used in Duma’s method to calculate the volume at STP and then from the molecular weight of nitrogen which is 28g for 22.4L, cross-multiply and find out the weight of nitrogen at STP. Then find out the percentage.

Complete Answer:
Duma’s method is used to quantitatively analyse the percentage of nitrogen in an organic compound. From Duma’s method, we have the formula,
\[\dfrac{{{P}_{0}}{{V}_{0}}}{{{T}_{0}}}=\dfrac{{{P}_{1}}{{V}_{1}}}{{{T}_{1}}}\text{ }where\text{ }({{P}_{1}}=P-f)\]
where volume of nitrogen collected $={{V}_{1}}\text{ }ml$
room temperature $={{T}_{1}}\text{ }K$
atmospheric pressure $=P\,mm$
aqueous tension at ${{T}_{1}}\,K\text{ }=\text{ }f\text{ }mm$
pressure of dry nitrogen $=P-f={{P}_{1}}\,mm$
${{P}_{0}}$, ${{V}_{0}}$ and ${{T}_{0}}$ is the pressure, volume and temperature respectively of dry nitrogen at STP.
From the question,
Given: \[m=0.35g\], ${{P}_{0}}=760mm$, ${{V}_{0}}=?$, ${{T}_{0}}=273K$, $P=715mm$, $f=15mm$, ${{V}_{1}}=55ml$, ${{T}_{1}}=300K$
$\therefore {{P}_{1}}=P-f=715-15=700mm$
\[\dfrac{{{P}_{0}}{{V}_{0}}}{{{T}_{0}}}=\dfrac{{{P}_{1}}{{V}_{1}}}{{{T}_{1}}}\]
 $\therefore {{V}_{0}}=\dfrac{{{P}_{1}}{{V}_{1}}}{{{T}_{1}}}\times \dfrac{{{T}_{0}}}{{{P}_{0}}}=\dfrac{700\times 55}{300}\times \dfrac{273}{760}=46.099ml$
\[\therefore {{V}_{0}}=46.099ml\]
Now, we know that the molecular weight of nitrogen gas is 28g for 22.4L at STP.
So, 22,400ml of nitrogen at STP weighs 28g.
Therefore, 46.099ml of nitrogen at STP will weigh $\dfrac{28\times 46.099}{22,400}g$
0.35g of organic compound contains $\dfrac{28\times 46.099}{22,400}g$ of nitrogen.
Thus, 100g of compound will contain $\dfrac{28\times 46.099\times 100}{22,400\times 0.35}=16.45%$ of nitrogen.

The percentage of nitrogen in a given organic compound is 16.45%, Option (A).

Note: Remember to convert all units in the same format before doing the calculation. Like in this case, convert all litres to millilitres. Check that all weights are in grams. Remember that molecular weight calculated using atomic mass number is always for 22.4L volume of a gas and for Avogadro’s number of moles.