Solveeit Logo
Login

Question

Question: In Dumas' method for estimation of nitrogen, \({\text{0}} \cdot {\text{3 g}}\) of an organic compoun...

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

Explanation

Solution

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 =0⋅3 g = {\text{0}} \cdot {\text{3 g}}
    The volume of the nitrogen collected  = 50 mL{\text{ = 50 mL}}
    Temperature (T1)=300K({T_1}) = 300K
    The vapor pressure of water = 15 mm  = {\text{ 15 mm }}
    The actual pressure of dry nitrogen = Vapour pressure of nitrogen - vapour pressure of water (P1) = 715 - 15 = 700 mm = {\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 P2 = 760 mm{{\text{P}}_2}{\text{ = 760 mm}}
    T2=273K{T_2} = 273K
    We need the value of V2=?{V_2} = ?
  2. Now let's calculate the value V2{V_2} by using the following formula,
    P1V1T1=P2V2T2\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,
    700×50300=760×V2273\dfrac{{700 \times 50}}{{300}} = \dfrac{{760 \times {V_2}}}{{273}}
    Now let's take the term V2{V_2} on side of the equation as we need to find out that value,
    V2=700×50×273300×760{V_2} = \dfrac{{700 \times 50 \times 273}}{{300 \times 760}}
    Now let's do the calculation and we get,
    V2=41⋅9 mL{V_2} = 41 \cdot 9{\text{ mL}}
  3. So the volume of nitrogen is 41⋅9 mL{\text{41}} \cdot {\text{9 mL}}. Now we need to find out the value of the percentage of nitrogen. As we know the 22400 mL{\text{22400 mL}} of nitrogen at STP weighs as 28 g{\text{28 g}} we can find out the weight for the volume of 41⋅9 mL{\text{41}} \cdot {\text{9 mL}} nitrogen as follows,
    Percentage of nitrogen = 28×41⋅9×10022400{\text{Percentage of nitrogen = }}\dfrac{{28 \times 41 \cdot 9 \times 100}}{{22400}}
    Percentage of nitrogen = 5⋅23 ≃ 6⋅25%{\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⋅25%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 800−9000C800 - {900^0}C in a chamber with the presence of oxygen. This reaction yields nitrogen, carbon dioxide, and water as a product.