Question

A gas composed of methane and ethane has a specific gravity of 0.75. What is the weight percent and volume percent of the methane in the mixture?

Answer 1

To solve this question, you will have to consider all the 3 compounds i.e. methane, ethane and water as ideal gases.NOW
Let us consider that there is V Litre of gas out of which a Litre is methane.
So, volume of ethane=(V-a) L
Now,let that the experiment is carried out at P atm pressure and T Kelvin temperature.So by ideal gas equation, the volume of 1 mole of gas = (RT/P) L ,where R = Ideal Gas Constant.So,
Moles of methane = (Pa/RT)
Mass of methane = (16aP/RT)
Moles of ethane= {P(V-a)/RT}
Mass of ethane
={30P(V-a)/RT}
Now total mass of V Litre gas=(Mass of methane+Mass of ethane)={P(30V-14a)/RT}
Density of gas=Mass/Volume
=(30P/RT)-(14aP/RT)
Density of water by ideal gas equation = (18P/RT)
Specific Gravity
=(Density of gas/Density of water)
=(30/18)-(14a/18V)=0.75 Now, the quantity (a/V) represents the volume fraction of methane.Let us represent it by "b".So our equation is
(30/18) - (14b/18) = 0.75
Solve it to get b = 0.547.So volume fraction = 0.547
Volume percent =
(Volume fraction) * 100=54.7
To find mass percentage:-
0.547 times of total volume is methane and its mass is (16 * 0.547V).Mass of ethane is (30 * 0.453V).Total mass = 22.342V
Mass fraction of methane
=0.3917
Mass percentage = 39.17.
Hope my calculations are right !