[Sis290025]

An unknown gas contains 83% C and 17% H by mass. If effuses at 0.87 times the rate of CO2 gas under the same conditions. What is the molecular formula of the unknown gas?
 
 
a.C4H10    
 
b.C7H17  
 
c.C3H3  
 
d.C2H5  


Well, the rate of effusion is equal to 1/sqrt(m), where m is in amu.

First find CO2 effusion rate, so 1/sqrt(44.011 amu) = 0.150737.
The unknown's effusion rate is 0.87*CO2 effusion rate:
0.87*0.150737 = 0.13114 unknown's rate of effusion

Now find the unknown's total amu.
0.13114 = 1/sqrt(m)

Solve for m = (1/0.13114)^2 = 58.147 amu.

Use percents given:

C = 0.83*58.147 amu = 48.262 amu/12.01 amu = 4.0 C atoms

H = 0.17*58.147 amu = 9.88 amu/1.008 amu = 9.8  Hatoms

Any help is appreciated!

[AWK]

OK -round 9.8 to 10 (C4H10)