# A compound is found to be 62.1% C, 5.21% H, 12.1% N, and 20.7% O. When...?

A compound is found to be 62.1% C, 5.21% H, 12.1% N, and 20.7% O. When 4.94 g is volatilized to a gas at 90°C and 510 torr it fills a 475 mL flask. Find the molecular weight and the molecular formula.

assume 100 g sample
62.1 g C or 62.1/12.011 mole = 5.17 mole C
5.21 g H or 5.21/1.008 mole = 5.17 mole H
12.1 g N or 12.1/14.008 mole = 0.856 mole N
20.7 g O or 20.7/16 mole = 1.29 mole O
find empirical formula, assume N=1
C = 5.17/0.856 = 6 C
H = 5.17/0.856 = 6 H
O = 1.29/0.856 = 1.5 O
N = 1
double everything
C12H12N2O3
The MW of the empirical formula is 232.2
Use ideal gas law to find the moles you have
T=273+90 = 363 deg K
V = 0.475 L
P = 510/760 = 0.671 atm
R = 0.0821
n = 4.94 g / MW
PV=nRT
0.671 x 0.475 = (4.94/MW) x 0.0821 x 363
0.0107 = 4.94/MW
MW = 4.94/0.0107 = 461.9
MW/empirical weight = 461.9/232.2 = 1.99 (must be 2)
Molecular formula = C24H24N4O6 MW = 464.5

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