Question

At a temperature of 25 degree C and a pressure of 750 mm Hg, 3.00 g of cyanogen gas occupies 0.714 L. What is the Molar Mass of the Molecule, cyanogen What is the molecular formula of cyanogen if Cyanogen is a gas which contains 46.2% C and 53.8% N by mass.

Answer 1

(A) The ideal gas equation is $PV = \dfrac {W}{M}RT$
P = pressure $= 750 mm Hg = \dfrac {750 mm Hg}{760 mm Hg/atm} = 0.9868$ atm
V = Volume= 0.714 L
W = Mass of cyanogen = 3.00 g
M = Molar mass of cyanogen
R = ideal gas constant = 0.08206 L. atm / mol . K
T = absolute temperature K
Substitute values in the above equation

$PV = \dfrac {W}{M}RT \\ M = \dfrac {W}{PV}RT \\ M = \dfrac {3.00}{0.9869 \times 0.714 } \times 0.08206 \times 298 = 104.1 g/mol$

(B) Molar mass = 104.1 g/mol
Percent of C = 46.2
Mass of C in 1 mole of cyanogen $= 104.1 \times \dfrac {46.2}{100} = 48 g$
Atomic mass of C = 12 g/mol
48 g C $= \dfrac {48 g}{12 g/mol} = 4 mol$
Percent of N = 53.8
Mass of N in 1 mole of cyanogen $= 104.1 \times \dfrac {53.8}{100} = 56$ g
Atomic mass of N = 14 g/mol
56 g N $= \dfrac {56 g}{14 g/mol} = 4$ mol
The molecular formula of cyanogen is $C_4N_4$