Question

Analysis of a gas indicated that 46.2 % of its mass was carbon and 53.8% of the mass was nitrogen. At 25 oC and 751 torr, 1.05 g of the gas occupies 0.500 L.

What is the molecular formula for the gas?

Answer 1

let total mass be 100 g
C : 46.2 g
N : 53.8 g

divide each by their atomic mass:
C : 46.2/12 = 3.85
N : 53.8/14 = 3.84

divide by smallest number:
C : 3.85/3.84 = 1
N: 3.84/3.84= 1

so empirical formula is CN
empirical formula mass is 14+12 =26 g

let use find the molar mass using:
P*V = n*R*T
P = 751 torr = 751/760 atm = 0.988 atm
T = 25 oC = 298 K

P*V = (mass/MM)*R*T
0.988*0.500 = (1.05/MM)*0.0821*298
MM = 52 g /mol

multiplying factor = MM/empirical formula mass
= 52/26
= 2

so,
molecular formula is 2 (CN) = C2N2
Answer: C₂N₂