Question

Duma's method of 0.52g of an organic compound on combustion gave 68.6mL ${N_2}$ at ${27^ \circ }C$ and 76mm pressure. What is the percentage of nitrogen in the compound?
A.12.22%
B.14.93%
C.15.84%
D.16.23%

Answer 1

Hint: We can calculate the percentage of nitrogen in the compound using the amount of nitrogen and the total mass of the organic compound. The amount of nitrogen is calculated using the moles of nitrogen and molar mass. The moles of nitrogen are calculated using the pressure, volume, temperature and gas constant.
Formula used: We can calculate the mass percent using the formula,
${\text{Mass percentage}} = \dfrac{{{\text{Mass}}}}{{{\text{Total mass}}}} \times 100\% $

Complete step by step answer:Given data contains,
Volume of nitrogen is $68.6mL$.
Pressure is $76mmHg = 1atm$
Mass of organic compound is $0.52g$.
Temperature is ${27^ \circ }C$
We have to convert the temperature in degrees Celsius to kelvin.
$K{ = ^ \circ }C + 273$
$K = {27^ \circ }C + 273$
$K = 300K$
The temperature is $300K$.
All the four properties of gas such as pressure, volume, temperature and number of moles are combined into a single equation is said to be Ideal gas law.
$PV = nRT$
Where,
P is the pressure of the gas.
V is the volume of the gas.
The number of moles of the gas is represented by n.
R represents the universal gas constant.
T is the temperature in Kelvin.
We can calculate the moles of nitrogen using the ideal gas law.
On rearranging the ideal gas equation we get,
$PV = nRT$
$n = \dfrac{{PV}}{{RT}}$
Substitute the values of pressure, volume, gas constant and temperature to get the moles of nitrogen gas.
$n = \dfrac{{PV}}{{RT}}$
Substituting the values we get,
$\Rightarrow n = \dfrac{{1atm \times 68.6 \times {{10}^{ - 3}}L}}{{0.0821Latm{K^{ - 1}}mo{l^{ - 1}} \times 300K}}$
$ \Rightarrow $$n = 2.785 \times {10^{ - 3}}mol$
The moles of nitrogen gas is $2.785 \times {10^{ - 3}}mol$.
From the moles of nitrogen gas, we can calculate the amount of nitrogen gas. The amount of nitrogen gas is calculated by multiplying the moles of nitrogen gas and molar mass.
The molar mass of nitrogen is $28g/mol$.
${\text{Amount of nitrogen}} = {\text{Moles}} \times {\text{Molar mass}}$
$ \Rightarrow $${\text{Amount of nitrogen}} = 2.785 \times {10^{ - 3}}{\text{mol}} \times 28\dfrac{g}{{mol}}$
$ \Rightarrow $${\text{Amount of nitrogen}} = 0.078g$
The amount of nitrogen is $0.078g$.
We can calculate the percentage of nitrogen using the amount of nitrogen and total mass of the organic compound.
Substitute the value of the amount of nitrogen and the total mass of the organic compound to get the percentage of nitrogen.
${\text{Mass percentage}} = \dfrac{{{\text{Mass}}}}{{{\text{Total mass}}}} \times 100\% $
Substituting the values we get,
$ \Rightarrow $${\text{Mass percentage}} = \dfrac{{0.078g}}{{0.52g}} \times 100\% $
$ \Rightarrow $${\text{Mass percentage}} = 14.93\% $
The percentage of nitrogen is $14.93\% $.
Therefore, Option (B) is correct.

Note:We can calculate the pressure using the ideal gas law. An example is given below,
The grams of carbon dioxide is $2.0g$.
The temperature is ${25^o}C$
The volume of the cylinder is $1.5L$
The gas constant in $0.0821Latm Mol{l^{ - 1}}{K^{ - 1}}$
We can convert temperature from Celsius to Kelvin.
${0^o}C = 273K$
${25^o}C = 25 + 273K = 298K$
The grams can be converted to moles as,
${\text{grams}} \times \dfrac{{{\text{1mol}}}}{{{\text{molar mass}}}}$
$ \Rightarrow $$2.0g \times \dfrac{{1mol}}{{44.0g}} = 0.0454mol$
The pressure of carbon dioxide can be calculated as,
$PV = nRT$
$PV = \dfrac{{nRT}}{V}$
Substituting the values we get,
$ \Rightarrow $$P = \dfrac{{\left( {0.0454mol} \right)\left( {0.0821Latmmo{l^{ - 1}}{K^{ - 1}}} \right)\left( {298K} \right)}}{{\left( {1.5L} \right)}}$
$ \Rightarrow $$P = 0.7404atm$
The pressure of carbon dioxide gas is $0.7404atm$.