Question

In Dumas' method for estimation of nitrogen, ${\text{0}} \cdot {\text{3 g}}$ of an organic compound gave ${\text{50 mL}}$ of nitrogen collected at ${\text{300 K}}$ temperature and ${\text{715 mm}}$ pressure. Calculate the percentage composition of nitrogen in the compound. (Aqueous tension at ${\text{300 K = 15 mm}}$)
A) $17 \cdot 5\% $
B) $28\% $
C) $6 \cdot 25\% $
D) $31\% $

Answer 1

Hint: Analyze the data given in the question and write down the data of each term which will be useful for numerical calculation. Do the needed conversions to fit into the formula to calculate the percentage composition of the nitrogen.

Complete step by step answer:
1) First of all let's analyze the data and the values given in the question one by one as below,
The mass of the organic substance $ = {\text{0}} \cdot {\text{3 g}}$
The volume of the nitrogen collected ${\text{ = 50 mL}}$
Temperature $({T_1}) = 300K$
The vapor pressure of water $ = {\text{ 15 mm }}$
The actual pressure of dry nitrogen $ = {\text{ Vapour pressure of nitrogen - vapour pressure of water (}}{{\text{P}}_1}{\text{) = 715 - 15 = 700 mm}}$
At the conditions of STP i.e. standard temperature and pressure, the value of pressure will be ${{\text{P}}_2}{\text{ = 760 mm}}$
${T_2} = 273K$
We need the value of ${V_2} = ?$
2) Now let's calculate the value ${V_2}$ by using the following formula,
$\dfrac{{{P_1}{V_1}}}{{{T_1}}} = \dfrac{{{P_2}{V_2}}}{{{T_2}}}$
Now, lets put the relevant values of each term in the formula we get,
$\dfrac{{700 \times 50}}{{300}} = \dfrac{{760 \times {V_2}}}{{273}}$
Now let's take the term ${V_2}$ on side of the equation as we need to find out that value,
${V_2} = \dfrac{{700 \times 50 \times 273}}{{300 \times 760}}$
Now let's do the calculation and we get,
${V_2} = 41 \cdot 9{\text{ mL}}$
3) So the volume of nitrogen is ${\text{41}} \cdot {\text{9 mL}}$. Now we need to find out the value of the percentage of nitrogen. As we know the ${\text{22400 mL}}$ of nitrogen at STP weighs as ${\text{28 g}}$ we can find out the weight for the volume of ${\text{41}} \cdot {\text{9 mL}}$ nitrogen as follows,
${\text{Percentage of nitrogen = }}\dfrac{{28 \times 41 \cdot 9 \times 100}}{{22400}}$
${\text{Percentage of nitrogen = 5}} \cdot {\text{23 }} \simeq {\text{ 6}} \cdot {\text{25}}\% $
Therefore, the percentage composition of nitrogen in the compound is $6 \cdot 25\% $ which shows option C as the correct choice.

Note:
The Dumas’ method is generally used for the determination of the molecular weight of the volatile organic substance which is liquid at room temperature. The Dumas’ method is performed by doing combustion of a known mass sample at a very high-temperature range of around $800 - {900^0}C$ in a chamber with the presence of oxygen. This reaction yields nitrogen, carbon dioxide, and water as a product.