Question

0.90 g of an organic compound containing only carbon hydrogen and nitrogen on combustion give 1.1 g CO2 and 0.3g water. What is C% H% and N% in the organic compound

Answer 1

Let given organic compound is $C_xH_yN_z$

so, combustion of $C_xH_yN_z$...

$C_xH_yN_z+O_2\rightarrow xCO_2+\frac{y}{2}H_2O+\textbf{a nitrogen compound}$

here, you can see that one mole of organic compound gives x mole of carbon dioxide and y/2 mole of water.

given weight of CO2 = 1.1g

molar weight of CO2 = 44g

so, mole = 1.1/44 = 0.1/4

similarly, mole of water = 0.3/18 = 0.1/6

mole of organic compound = 0.9/(12x + y+14z)

from reaction,

x × mole of organic compound = mole of CO2

or, 0.9x/(12x+ y + 14z) = 0.1/4

or, 36x = 12x + y + 14z

or, 24x = y + 14z.....(1)

y/2 mole of organic compound = mole of water

or, 0.9(y/2)/(12x + y + 14z) = 0.1/6

or, 27y = 12x + y + 14z

or, 26y = 12x + 14z .......(2)

from equations (1) and (2),

24x - y = 26y - 12x

or, 36x= 27y

or, 4x = 3y ......(3)

putting equation (3) in equation (1),

24 × 3y/4 = y + 14z

or, 17y = 14z

or, y = 14/17 z

or, 3y = 42/17 z

hence, 4x = 3y = 42/17 z

or, 68x = 51y = 42z

so, organic compound will be $C_{68}H_{51}N_{42}$

% of C in compound = (12 × 68)/(12 × 68 + 51 + 42× 14) × 100 = 56%

% of H in compound = 51/(12×68 + 51 + 42 × 14) × 100 = 3.6%

% of N in compound = (42 × 14)/(12× 68 + 51+ 42 × 14) × 100 = 40.4%