Question

In an effusion experiment, it was determined that nitrogen gas, N 2   , effused at a rate 1.812 times faster than an unknown gas. What is the molar mass of the unknown gas?

Answer 1

1) rate of N2/rate of unknown gas = sqrt(molar mass of unknown gas/molar mass of N2)

or, 1.812 = sqrt(molar mass of unknown gas /28)

or. 3.283 = molar mass of unknown gas /28

or, molar mass of unknown gas = 91.934 g/mole

Answer 2

MASS/RAM: 30.45/14
69.55/16
# of moles: 2.175
4.347
Divide by the smaller: 2.175/2.175=1
4.347/2.175=1.999- approx 2
Ratio: 1:2
Empirical Formula: NO2
Let Molecular Formula = (NO2)x
(NO2)x= 14x +32x= 46
Hence x= 91.98 approx 92/46=2
Therefore Molecular Formula= (NO2)2 = N2O4

These are the calculations, please note that i estimated so therefore the answer is not as precise as it should be. But this answer should be correct to 1 or 2 DP.

Answer 3

MASS/RAM: 30.45/14
69.55/16
# of moles: 2.175
4.347
Divide by the smaller: 2.175/2.175=1
4.347/2.175=1.999- approx 2
Ratio: 1:2
Empirical Formula: NO2
Let Molecular Formula = (NO2)x
(NO2)x= 14x +32x= 46
Hence x= 91.98 approx 92/46=2
Therefore Molecular Formula= (NO2)2 = N2O4