There are many ways to approach this.
First, start with your Ideal Gas Equation.
PV = nRT
You know P,V,T and R, solve for n.
Therefore:
n =0.98 mol
Convert moles to mass of methane
(0.98mol CH
4) * (
16.04 g CH4) = 15.737 g CH
4 1 mol CH
4Now, do the same for air*, assume the same conditions.
n = 0.98 mol
(0.98 mol air) *
( 28.97 g air) = 28.39 g air
1 mol air
Now, you can view it like this.
Given the same quantity (number of moles) and volume as air, methane "weighs" less. The amount of "lift" given is the "weight" of air that would be occupied in that space minus the weight of the substance giving "lift".
In this case, if the balloon was filed with air it would have 28.39 g air, but it is instead filled with 15.737 g of methane, a difference of 12.7 g (and since it is lighter, due to "bouyancey" it is lifting and causes the negative reading on the balance). Now, also the original balloon weighed 4 g so you must account for that, or 12.7-4 = 8.7g difference if the balloon. Now, to get 8.8 it just depends on how accurate you keep your numbers, and what you use for your constants and such and how/if/when you round, etc.
* For info on how to determine the molecular weight of air: (
http://www.engineeringtoolbox.com/molecular-mass-air-d_679.html)
I chose to explain it that way using the ideal gas law, because that is the most important one, in your stage, to work with.
But it can be done equally as easy with densities of the gases.