[abcd]

So im doing a lab, and it wants me to calculate the percent of gas that is butane(C4H10). If the Molar mass of the sample gas is less that that of butane, then assume that the lighter gas is a mixture of butane and propane(c3h8). If the Molar Mass of your sample is greater than butane's the nasssume it is butane and pentane.  

So for my sample gas, i found that the molar mass of the lighter gas was 49.7g/mol.
the teacher wants us to use differential equations to find the answer,
can anyone help me?!

[Mitch]

Since there isn't a second order or first order reaction going on in time, I really doubt you'll need a differential equation to figure it out. It seems like straight forward Algebra to me.

(a)(58g/mol) + (1-a)(44g/mol) = 49.7g/mol

Just solve for a. If you don't know what a is, you're in trouble.