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: C_{2}N_{2}

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