Question

In Dumas' method for estimation of nitrogen, $ 0.3g~ $ of an organic compound gave $ 50mL $ of nitrogen collected at $ 300K $ temperature and $ 715mm $ pressure. Calculate the percentage composition of nitrogen in the compound. (Aqueous tension at $ 300K=15mm $ )
(A) $ 15.23% $
(B) $ 17.46% $
(C) $ 19.23% $
(D) $ 22% $

Answer 1

Hint: Using Ideal gas equation, $ PV=nRT $ 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:
In 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}}} $ where $ ({{P}_{1}}=P-f) $
 $ {{P}_{0}},{{V}_{0}} $ and $ {{T}_{0}} $ is the pressure, volume and temperature respectively of dry nitrogen at STP. From the question, given:
 $ \left( m=0.3g \right),\left( {{P}_{0}}=760mm \right),\left( {{V}_{0}}=? \right),\left( {{T}_{0}}=273K \right),\left( P\text{ =}715mm \right)\text{, }\left( f=15mm \right),\left( {{V}_{1}}=50ml \right),\left( {{T}_{1}}=300K \right) $
 $ \therefore {{P}_{1}}=P-f=715-15=700mm $
 $ \dfrac{{{P}_{0}}{{V}_{0}}}{{{T}_{0}}}=\dfrac{{{P}_{1}}{{V}_{1}}}{{{T}_{1}}} $ from here we have to find $ {{V}_{0}} $ ;
 $ {{V}_{0}}=\dfrac{{{P}_{1}}{{V}_{1}}}{{{T}_{1}}}\times \dfrac{{{T}_{0}}}{{{P}_{0}}}=\dfrac{700\times 50}{300}\times \dfrac{273}{760}=41.90ml $
 $ \therefore {{V}_{0}}=41.90ml $ 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, $ 41.90ml $ of nitrogen at STP will weight; $ \dfrac{28\times 41.90}{22,400}g $
 $ 0.30g $ of organic compounds contains $ \dfrac{28\times 41.90}{22,400}g $ of nitrogen.
Thus, $ 100g $ of compound will contain $ \dfrac{28\times 41.90\times 100}{22,400\times 0.3}g=17.458 $ of nitrogen.
The percentage of nitrogen in a given organic compound $ 17.458%\approx 17.46%, $ Option (B).

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.